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Exponential Functions and Logarithms

(Level IV Academic Math)

NSSAL

(Draft)

C. David Pilmer

2009

(Last Updated: Dec 2011)

Use our online math videos. YouTube: nsccalpmath

This resource is the intellectual property of the Adult Education Division of the Nova Scotia Department of Labour and Advanced Education. The following are permitted to use and reproduce this resource for classroom purposes.

• Nova Scotia instructors delivering the Nova Scotia Adult Learning Program • Canadian public school teachers delivering public school curriculum • Canadian nonprofit tuition-free adult basic education programs

The following are not permitted to use or reproduce this resource without the written authorization of the Adult Education Division of the Nova Scotia Department of Labour and Advanced Education.

• Upgrading programs at post-secondary institutions • Core programs at post-secondary institutions • Public or private schools outside of Canada • Basic adult education programs outside of Canada

Individuals, not including teachers or instructors, are permitted to use this resource for their own learning. They are not permitted to make multiple copies of the resource for distribution. Nor are they permitted to use this resource under the direction of a teacher or instructor at a learning institution. Acknowledgments

The Adult Education Division would like to thank the following university professors for reviewing this resource to ensure all mathematical concepts were presented correctly and in a manner that supported our learners.

Dr. Genevieve Boulet (Mount Saint Vincent University)

Dr. Robert Dawson (Saint Mary’s University)

The Adult Education Division would also like to thank the following NSCC instructors for piloting this resource and offering suggestions during its development.

Charles Bailey (IT Campus)

Elliott Churchill (Waterfront Campus)

Barbara Gillis (Burridge Campus)

Barbara Leck (Pictou Campus)

Suzette Lowe (Lunenburg Campus)

Floyd Porter (Strait Area Campus)

Brian Rhodenizer (Kingstec Campus)

Joan Ross (Annapolis Valley Campus)

Jeff Vroom (Truro Campus)

NSSAL i Draft ©2009 C. D. Pilmer

Table of Contents Introduction ………………………………………………………………………………….. ii Negotiated Completion Date ………………………………………………………………… ii Prerequisite Knowledge ……………………………………………………………………... iii The Big Picture ……………………………………………………………………………… iv Course Timelines ……………………………………………………………………………. v Interpreting Graphs of Exponential Functions ………………………………………………. 1 Using the Graphing Calculator to Interpret Exponential Functions ………………………… 4 Exponents, Part 1 ……………………………………………………………………………. 7 Exponents, Part 2 ……………………………………………………………………………. 12 The Most Basic Exponential Functions ……………………………………………………... 21 Exponential Functions and Transformations ………………………………………………... 25 State the Transformations: Exponential Function …………………………………………… 30 Graphing Using Transformations ……………………………………………………………. 31 Equivalent Exponential Functions …………………………………………………………... 38 Recognizing Patterns within Tables of Values ……………………………………………… 40 Investigation: From a Table of Values to an Equation ……………………………………… 42 Determining the Equation Given a Table of Values ………………………………………… 43 More than One Acceptable Answer …………………………………………………………. 47 Determining the Equation Based on the Real World Situation ……………………………... 49 Determining the Equation Given a Graph …………………………………………………… 58 Solving Exponential Equations by Creating the Same Base ………………………………… 60 An Introduction to Logarithms ……………………………………………………………… 66 Solving Exponential Equations Using Logarithms ………………………………………….. 72 Revisiting Application Problems ……………………………………………………………. 84 Putting It Together ………………………………………………………………………….. 86 Post-Unit Reflections ………………………………………………………………………... 95 Additional Practice: Exponents, Part 1 (Optional) …………………………………………... 96 Additional Practice: Exponents, Part 2 (Optional) …………………………………………... 97 Additional Practice: Exponential and Logarithmic Equations (Optional) …………………... 98 Answers ……………………………………………………………………………………… 99

NSSAL ii Draft ©2009 C. D. Pilmer

Introduction So far in this course, we have examined three types of functions.

• In Part 2 of the Bridging Unit we worked with linear functions ( bmxy += ). • In the Sinusoidal Functions Unit we worked with functions of the form

( )( ) dcxaky +°−= cos or ( )( ) dcxaky +°−= sin . • In the Quadratic Functions Units, we worked with functions that could be written as

cbxaxy ++= 2 (standard form) or ( ) dcxky +−= 2 (transformational form). In this unit we will learn about exponential functions. The most basic exponential function is written in the form xby = , where b is a constant that is greater than zero (e.g. xy 3= , xy 5.0= ). In this unit learners will do the following.

• Interpret graphs of real world phenomena. All of the graphs will be exponential functions.

• Evaluate or simplify expressions using the Laws of Exponents.

• Work with integer exponents (e.g. 23− ) and rational exponents (e.g. 23

25 ). • Graph exponential functions using transformations. • Investigate the properties of exponential functions. • Determine the equation of an exponential function given a table of values, graph or

written description of a real world situation. • Solve exponential equations with and without the use of logarithms. • Use the laws of logarithms to solve a variety of logarithmic equations

All of this will be accomplished while looking at a variety of real world applications of exponential functions and revisiting concepts covered in prior units (e.g. domain, range, intercepts,…). Negotiated Completion Date

After working for a few days on this unit, sit down with your instructor and negotiate a completion date for this unit.

Start Date: _________________

Completion Date: _________________

Instructor Signature: __________________________

Student Signature: __________________________

NSSAL iii Draft ©2009 C. D. Pilmer

Prerequisite Knowledge Before you start this unit, it is expected that you are familiar with powers of 2, 3, 4, and 5. You should know that:

422 = and conversely 24 = 823 = and conversely 283 = 1624 = and conversely 2164 = 3225 = and conversely 2325 =

932 = and conversely 39 = 2733 = and conversely 3273 = 8134 = and conversely 3814 =

1642 = and conversely 416 = 6443 = and conversely 4643 =

2552 = and conversely 525 = 12553 = and conversely 51253 = If you are unfamiliar with these, spend a few minutes over next two or three days to learn them. You will need to know this material before you start the section titled Exponents, Part 2.

NSSAL iv Draft ©2009 C. D. Pilmer

The Big Picture The following flow chart shows the optional bridging unit and the eight required units in Level IV Academic Math. These have been presented in a suggested order.

Bridging Unit (Recommended) • Solving Equations and Linear Functions

Systems of Equations Unit • 2 by 2 Systems, Plane in 3-Space, 3 by 3 Systems

Sinusoidal Functions Unit • Periodic Functions, Sinusoidal Functions, Graphing Using

Transformations, Determining the Equation, Applications

Describing Relations Unit • Relations, Functions, Domain, Range, Intercepts, Symmetry

Quadratic Functions Unit • Graphing using Transformations, Determining the Equation,

Factoring, Solving Quadratic Equations, Vertex Formula, Applications

Exponential Functions and Logarithms Unit • Graphing using Transformations, Determining the Equation,

Solving Exponential Equations, Laws of Logarithms, Solving Logarithmic Equations, Applications

Trigonometry Unit • Pythagorean Theorem, Trigonometric Ratios, Law of Sines,

Law of Cosines

Rational Expressions and Radicals Unit • Operations with and Simplification of Radicals and Rational

Expressions

Inferential Statistics Unit • Population, Sample, Standard Deviation, Normal Distribution,

Central Limit Theorem, Confidence Intervals

NSSAL v Draft ©2009 C. D. Pilmer

Course Timelines Academic Level IV Math is a two credit course within the Adult Learning Program. As a two credit course, learners are expected to complete 200 hours of course material. Since most ALP math classes meet for 6 hours each week, the course should be completed within 35 weeks. The curriculum developers have worked diligently to ensure that the course can be completed within this time span. Below you will find a chart containing the unit names and suggested completion times. The hours listed are classroom hours. In an academic course, there is an expectation that some work will be completed outside of regular class time. Unit Name Minimum

Completion Time in Hours

Maximum Completion Time

in Hours Bridging Unit (optional) 0 20 Describing Relations Unit 6 8 Systems of Equations Unit 18 22 Trigonometry Unit 18 20 Sinusoidal Functions Unit 20 24 Quadratic Functions Unit 36 42 Rational Expressions and Radicals Unit 12 16 Exponential Functions and Logarithms Unit 20 24 Inferential Statistics Unit 20 24 Total: 150 hours Total: 200 hours As one can see, this course covers numerous topics and for this reason may seem daunting. You can complete this course in a timely manner if you manage your time wisely, remain focused, and seek assistance from your instructor when needed.

NSSAL 1 Draft ©2009 C. D. Pilmer

Interpreting Graphs of Exponential Functions We are going to begin by examining graphs of real world relationships that can be modeled using exponential functions. For this lesson, there are two terms we need to introduce regarding exponential functions. All exponential functions are either classified as growth curves or decay curves.

An exponential function is called a growth curve if as we proceed from left to right along the curve, the dependent variable increases (i.e. goes up in value).

An exponential function is called a decay curve if as we proceed from left to right along the curve, the dependent variable decreases (i.e. goes down in value).

Example 1 A bacteria population is growing in a Petri dish. The concentration of bacteria per square centimeter is on the vertical axis. The time in hours is on the horizontal axis.

(a) Determine the time when the concentration was 120 bacteria per square centimetre?

(b) Approximate the initial concentration of bacteria. (c) Approximate the bacteria concentration at t = 3 hours. (d) How long does it take for the population of bacteria to

double in size? (e) If the graph extended beyond 4 hours, predict the

bacteria concentration at t = 6 hours. (f) State the domain and range. (g) Is this a growth or decay curve?

Answers: (a) The concentration was 120 bacteria per square centimeter at t = 4 hours. (b) The initial concentration was approximately 30 bacteria per square centimetre. (c) At t = 3 hours, the concentration is approximately 85 bacteria per square centimetre. (d) At t = 0, the concentration is 30 bacteria/cm2. At t = 2, the concentration is 60

bacteria/cm2. At t = 4, the concentration is 120 bacteria/cm2. That means that the population is doubling every 2 hours.

(e) We can’t answer this question directly from the graph because the graph only goes up to 4 hours. By completing question (d), we learned that the population doubles every 2 hours. Using this information and the bacteria concentration at 4 hours, we can determine that the concentration will be 240 bacteria per square centimetre at t = 6 hours.

(f) Domain: { }40 ≤≤ tRtε Range: { }12003 ≤≤ cRcε (g) We are dealing with a growth curve because as we proceed from left to right, the

dependent variable, c, is increasing.

0

10

20

30

40

50

60

70

80

90

100

110

120

130

140

0 1 2 3 4

Time in Hours

Conc

entr

atio

n in

Bac

teri

a pe

r Sq

uare

Cen

timet

re


Questions:

1. After a medication enters your bloodstream, its concentration decreases with time. The concentration of the medication, in micrograms per cubic centimetre ( 3g/cmµ ), is on the vertical axis. The time, in hours, is on the horizontal axis.

Answers (a) What’s the initial concentration of medication in the bloodstream? (b) Approximate the concentration of medication at t = 2.5 hours. (c) When will the concentration of medication be 5 3g/cmµ ? (d) How long does it take for the concentration of medication to be

halved?

(e) Approximate the concentration of medication at t = 0.5 hours. (f) Are we dealing with a growth or decay curve? 2. Kate’s parents put money into an investment

plan at the time of her birth. This plan delivers an annual fixed rate of interest (8%) which is compounded daily. If she does not touch the money in the plan, the following graph shows the relationship between the amount of money in the plan measured in dollars and the time measured in years.

Answers (a) Approximately how much money would be in the plan when Kate

is twenty-five years old?

(b) How old would Kate be when there was $10 000 in the investment plan?

(c) How much money was initially invested at the time of her birth? (d) Are we dealing with a growth or decay curve?

0

5

10

15

20

25

30

35

40

45

0 1 2 3 4

Time

Conc

entr

atio

n of

Med

icat

ion

0

2000

4000

6000

8000

10000

12000

14000

16000

18000

20000

22000

0 5 10 15 20 25 30

time (years)

amou

nt in

pla

n (d

olla

rs)


(e) Approximately how long did it take to double the initial investment?

(f) If she did not touch the plan until she was thirty years old, how much money would she have in the plan?

(g) Thinking about your answers to question (d) and (e), at what age will Kate be when there is $44 000 in the plan.

3. The presence of sediment in water reduces the

distance sunlight can penetrate into the water. The following graph shows the relationship between the light intensity, L, measured in candela/cm2 and the depth, d, of the water measured in metres for a particular body of water on a particular day.

Answers (a) What is the light intensity at a depth of 5 metres? (b) Are we dealing with a growth or decay curve? (c) What is the light intensity right at the surface? (d) Approximately how deep must one go in order for the light

intensity to be cut in half?

(e) At what depth is the light intensity approximately 0.37 candela/cm2?

(f) State the domain. (g) Approximate the range.

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 1 2 3 4 5 6

depth (metres)lig

ht in

tens

ity


Using the Graphing Calculator to Interpret Exponential Functions As mentioned in the Introduction at the start of this unit, the most basic exponential function is written in the form xby = , where b is a constant that is greater than zero (e.g. xy 3= , xy 5.0= ). In this section we will use graphing technology to examine slightly more complicated

exponential functions of the form cx

aby = . In each case we will use the technology to answer a series of questions regarding a real world situation that can be modeled using an exponential function of this form. Example 1 The radioactive isotope radon-222 is a gas that can penetrate a house’s foundation and reach dangerous levels within the structure. Over time

radon-222 gradually decays. The equation

= 925.010

t

A illustrates

the relationship between the amount, A, of radon-222 in a specific situation and the time, t. The amount is measured in milligrams, and the time is measured in hours. Use a graphing calculator and the WINDOW setting shown on the right to answer the following questions. (a) Are we dealing with a growth or decay curve? (b) What was the initial concentration of radon-222 in this particular situation? (c) At what time does the amount of radon gas remaining reach 3 milligrams? (d) How much radon remains after 60 hours? (e) How long does it take to lose half of the original amount of radon? Answers: The hardest part is to correctly enter the

equation into the calculator. It requires one to use the “^” symbol and put parentheses in the appropriate locations.

y = 10(0.5^(x/92)) The TRACE feature can now be used to

answer the questions. (a) It is a decay curve. (b) The initial concentrate can be found by looking at the y-intercept (actually A-intercept).

There was initially 10 milligrams of radon-222. (c) Using the TRACE feature, we learn that x (or t) equals 159 when y (or A) is

approximately equal to 3. Based on this we can say that 3 milligrams remains after 159 hours.

(d) Using the TRACE feature, we learn that y (or A) equals 6.39 when x (or t) equals 60. Based on this we can say that after 60 hours only 6.39 milligrams remains.

(e) We need to find the time corresponding to an amount of 5 milligrams. Using the TRACE feature we learn that the answer is 92 hours.


Questions:

1. A very warm container of water is placed outside on a winter’s day. The temperature of the water decreases over time. The equation

= 23.085

t

T illustrates the relationship between the temperature,

T, of the water in degrees Celsius and the time, t, in hours. Use a graphing calculator and the WINDOW setting shown on the right to answer the following questions.

Answers (a) What was the initial temperature of the water when it was placed

outside?

(b) Are we dealing with a growth or decay curve? (c) What’s the temperature of the water after three hours? (d) How long does it take for the water to reach 46oC?

2. The cane toad, a native of South America, was introduced to

Australia in 1935 as a way to control the destructive sugar cane beetle. It had no natural predators in Australia and it reproduced very rapidly. In 1935, a small number of cane toads were brought to Australia but in less than two years, the population had grown to

64500. The equation

= 21645100

t

P shows the relationship

between the cane toad population, P, and the time, t, measured in months from the toad’s introduction to the country. Use a graphing calculator and the WINDOW setting shown on the right to answer the following questions.

Answers (a) What was the initial population of cane toads introduced to

Australia?

(b) What was the population after 21 months? (c) Approximately how long did it take for the population to reach

25 600?

(d) Are we dealing with a growth or decay curve? (e) Approximately how long did it take for the initial population to

increase by a factor of ten?

3. The value, v, of a particular vehicle after its purchase depreciates

with time, t, as described by the equation ( )tv 85.021000= . The value is measured in dollars and the time is measured in years. Use a graphing calculator and the WINDOW setting shown on the right to answer the following questions.


Answers (a) Are we dealing with a growth or decay curve? (b) How much was the car when it was purchased? (c) When was the car worth approximately $12 900? (d) How much is the car worth after 5 years? (e) How much is the car worth after 1 year? (f) By what percentage has the value of the car dropped in its first

year?

4. Before the Second World War, many Jewish people from Germany

moved their money to Swiss banks in an attempt to protect their savings from the Nazis. Many of these individuals died due to the Holocaust. After the war, the Swiss banks would not release funds to surviving family members without specific account information. This meant that millions of accounts remained unclaimed. Only in 1998 after much pressure did the banks, who had profited from the original deposits, offer some restitution. Suppose the banks were required to return all funds with accumulated interest from 1938 until 1998. The equation ( )tA 05.15000= describes the potential growth of monies in a particular account during that time period. This is a very conservative growth model because it only assumes a return of 5% per year. The variable A represents the amount of money in the particular account in dollars. The variable t represent the time in years from 1938. This means that t = 0 represents 1938, t = 1 represents 1939, and so on.

Answers (a) How much money was deposited in the account? (b) How much money would be in the account in 1998? (c) By what factor has the amount grown over that 60 year period? (d) How much money would have accumulated in this account by

1978?

(e) In what year would approximately $21 600 have accumulated in the account?

Note: After completing the last two sections of material, we have learned that there are many real world applications of exponential functions. A few such applications have been listed below.

• Radioactive Decay • Unrestricted Growth of Populations • Investments Involving Compound Interest • The Depletion of Medication in the Bloodstream • Cooling of Objects or Liquids • Vehicle Depreciation


Exponents, Part 1 If you look back at the units on sinusoidal and quadratic functions, we started by interpreting graphs of real world situations that could be modeled by either of these functions. We have done the same thing in this unit that is concerned with exponential functions. In the Sinusoidal Functions Unit and Quadratic Functions Unit, we then proceeded to graph very simple functions by generating a table of values. If we did that next in this unit, we would encounter a major problem because our understanding of exponents is limited to natural number (e.g. 1, 2, 3, 4, …) exponents.

This is better explained when we consider the exponential function xy 9= . If we had to complete the following table of values without using technology, we would not know to deal with the negative, fractional, and zero exponents.

xy 9= x y

-2 ?

23

− ?

-1 ?

21

− ?

0 ?

21 ?

1 9 19 ←

23 ?

2 81 29 ←

This means that we are not ready to graph exponential functions at this time. We must first spend some time reviewing and learning specific skills concerned with exponents.

The Laws of Exponents

When given the expression na , we say that na is the power, n is the exponent, and a is the base.

Therefore when given the power 32 , we state that the base is 2 and the exponent is 3. This power can be written as 22223 ××= .


Why does the law of products of powers work?

Consider 32 xx × . The 2x can be represented as xx× . The 3x can be represented as xxx ×× . This means that the product of 2x and 3x can be expressed as ( ) ( )xxxxx ××××

or as 5x . ( ) ( ) 532 xxxxxxxxxxxxx =××××=××××=× Therefore: 532 xxx =×

Why does the law of quotients of powers work?

Consider 35 xx ÷ . The 5x can be represented as xxxxx ×××× . The 3x can be represented as xxx ×× . This means that the quotient can be expressed in the following manner.

23

535 xxx

xxxxxxxx

xxxx =×=

××××××

==÷

Therefore: 235 xxx =÷

Why does the law of powers of powers work? Consider ( )32x . The 2x can be represented as xx× . If this is cubed, it is multiplied by itself

three times. This means that we end up with xxxxxx ××××× or 6x .

The Law of Products of Powers To multiply two powers that have the same base, keep the base and add the exponents. nmnm aaa +=×

Examples: 752 xxx =× 1283 rrrr =×× 734 222 =×

The Law of Powers of Powers To raise a power to a power, keep the same base and multiply the exponents. ( ) nmnm aa ×=

Examples: ( ) 1243 xx = ( ) 1427 hh = ( ) 1535 22 =

The Law of Quotients of Powers To divide two powers that have the same base, keep the base and subtract the exponents. nmnm aaa −=÷

Examples: 325 xxx =÷ 78 rrr =÷ 235 222 =÷


( ) ( ) ( ) ( ) ( ) 6332 xxxxxxxxxxxxxxxx =×××××=×××××=×=

Therefore: ( ) 632 xx = Why does the law of powers of products work? Consider ( )3xy . The xy is being cubed which means that it is multiplied by itself three times.

Using the commutative property we learn that it can be expressed as 33 yx . ( ) 333 yxyyyxxxyxyxyxxyxyxyxy =×××××=×××××=××= Therefore: ( ) 333 yxxy = Important Note: It is important to be able to distinguish between a base and a numerical coefficient when applying the Laws of Exponents. When given the expression 32x , the number 2 is the numerical coefficient and the x is the base. This means that if one is asked to multiply 32x and 74x , this can be accomplished because the powers have the same base, x. We merely multiply the numerical coefficients and multiply the powers (by adding the exponents). 1073 842 xxx =×

You will have similar situations arising with numerical coefficients when dealing with division problems. 549 3721 xxx =÷

The Law of Powers of Products If a product in parentheses is raised to an exponent, the parentheses indicate that each factor must be raised to that exponent. ( ) nnn baab =

Examples: ( ) 666 yxxy = ( ) 615325 yxyx = ( ) 1212334 822 rrr ==

The Law of Power of Quotients If a quotient in parentheses is raised to an exponent, the parentheses indicate that both the numerator and denominator must be raised to that power.

n

nn

ba

ba

=

where 0≠b

Examples:

3

33

yx

yx

=

12

84

3

2

ba

ba

=

1010

22

5

933yyy

==


Examples:

(a) 532 2173 xxx =× (b) 46342 1892 bababa =×

(c) 426 3824 kkk =÷ (d) 27798 9545 dccddc =÷

(e) ( ) 1414227 2555 xxx == (f) ( ) 8128124423 1622 nmnmnm ==

(g) 8

4

8

422

4

2 933vu

vu

vu

==

(h) 15

39

153

3933

5

3

278

32

32

zyx

zyx

zyx

==

(i) ( )

1910

2

2012

2

453

2 8

168

2

yxyxyx

yxyx

=

=

(j)

9

9

910

9

486

9 2

182

36

aab

baab

baba

=

=×

Questions

Simplify each of the following expressions. (a) =× 43 xx (b) =× 5332 baba (c) =×× 234 32 xxx (d) =× 442 46 dcc (e) =÷ 412 yy (f) =÷ 3457 qpqp (g) =÷ 56 412 hh (h) =÷ yxyx 325 444 (i) ( ) =

63a (j) ( ) =625dc

(k) ( ) =

246k (l) ( ) =3683 ba

(m) =

7

dc (n) =

9

4

3

yx

(o) =

4

2

32yx (p) =

2

3

4

23

cab

(q) =÷ 2578 428 dcdc (r) =×× 2875 ppqpq


(s) =

36

2xy (t) ( ) =

237 ba

(The remaining questions will require two steps. Show your work.)

(u) =×

yxxyyx

2

453

844 (v) ( )

=102

364

93

yxyx

(w) ( ) =35

1419

240

qpqp (x) =

×53

452

16412

dccddc

(y) ( ) =×

22

6753

349

abbaba (z) ( )

=× yxyxyx

332

265

46

If you feel that you need additional practice with these types of questions, you can complete the optional questions titled Additional Practice: Exponents, Part 1 found in the appendix of this unit.


Exponents, Part 2 In Exponents, Part 1 we focused on the Laws of Exponents. We did not address the issue of fractional, zero and negative exponents. That will be done in this section of the unit and we will use the Laws of Exponents to accomplish this. Zero Exponents

Consider n

n

xx .

• We know that when we take one quantity and divide it by the same quantity, the quotient

is always 1. For example: 133= , 1

4747

= , 188=

−− , and 1

22

3

3

= . Therefore we can

conclude that 1=n

n

xx . The only exception to this rule is if the two quantities are equal to

zero

≠1

00 .

• Based on the Law of Quotients of Powers, we know that to divide two powers that have the same base, we keep the base and subtract the exponents. Therefore our expression

n

n

xx can now be expressed in the following manner.

0xxxx nn

n

n

== −

• In the first bullet we stated that 1=n

n

xx . In the second bullet we stated that 0x

xx

n

n

= .

Therefore we can conclude 10 =x (as long as x does not equal zero). Examples:

170 = ( ) 14 0 =− 153 0

=

1

79 0

=

−

140 −=− ← (Note: In this question only 4, not -4, is raised to the exponent 0.) 18.0 0 −=− ← (Note: In this question only 0.8, not -0.8, is raised to the exponent 0.)


Negative Exponents

Consider 5

2

xx .

• The expression 5

2

xx can be simplified by dividing through (i.e. canceling out) like

quantities.

3

5

2

1

x

xxxxxxx

xxxxxxx

xx

=

×××/×//×/=

×××××

=

• Using the Law of Quotients of Powers, our expression 5

2

xx can now be expressed in the

following manner.

3525

2−− == xx

xx

• In the first bullet we stated that 35

2 1xx

x= . In the second bullet we stated that 3

5

2−= x

xx .

Therefore we can conclude 3

33 1or 1

=−

xxx . If we research this further we could

generalize that n

n

xx

=− 1 where x is not equal to zero. Therefore the negative exponent

just means that we deal with the reciprocal of the base. Examples:

251

515

22 =

=−

81

212

33 =

=− ( )

811

313

44 =

−=− −

811

313

44 −=

−=− −

4925

75

57 22

=

=

−

827

23

32 33

=

=

−

16221 4

4

==

−

2536

56

65 22

=

−=

−

−

2536

56

65 22

−=

−=

−

−


Fractional Exponents

Part 1: The Fractional Exponent 21

• Suppose we were asked to find two identical numbers that multiplied to give us the product 9. Most people would quickly respond 3 because 933 =× . This mathematical statement could also be written as the following.

999 =× Now let’s work with the variable x. If we needed to find two identical terms that multiply

to give us the product x, we would end up with the following statement. xxx =×

• Again we are going to find two identical terms that multiply to give us the product x but we are not going to rely on radicals (i.e. square roots, cube roots, fourth roots,…). Instead we are going to rely on our understanding of the Law of Products of Powers. We need to find two powers with the base of x and the same exponent whose product is x.

xxx =× ?? The only exponent that makes sense is one half.

xxxx or 121

21

=×

• In the first bullet we stated that xxx =× . In the second bullet we stated

that xxx =× 21

21

. Therefore we can conclude that xx =21

.

Part 2: The Fractional Exponent 31

• Suppose we were asked to find three identical numbers that multiplied to give us the product 8. Most people would quickly respond 2 because 8222 =×× . This mathematical statement could also be written as the following.

8888 333 =×× Now let’s work with the variable x. If we needed to find three identical terms that

multiply to give us the product x, we would end up with the following statement. xxxx =×× 333

• Again we are going to find three identical terms that multiply to give us the product x but we are not going to rely on radicals (i.e. square roots, cube roots, fourth roots,…). Instead we are going to rely on our understanding of the Law of Products of Powers. We need to find three powers with the base of x and the same exponent whose product is x.

xxxx =×× ??? The only exponent that makes sense is one third.

xxxxx or 131

31

31

=××

• In the first bullet we stated that xxxx =×× 333 . In the second bullet we stated that

xxxx =×× 31

31

31

. Therefore we can conclude that 331

xx = .


If xx =21

and 331

xx = , its fair to conclude that nn xx =1

. Examples:

5252521

== 32727 331

== 21616 441

==

65

3625

3625 2

1

==

32

278

278

331

−=−=

−

21

321

321

551

==

number.) negative a ofroot square theecannot tak (Wesolution real no 41

41 2

1

←−=

−

More Challenging Exponents

When dealing with more challenging exponents it is often useful to use the Law of Powers of Powers but in reverse. For example 6x can be expressed as ( )32x or as ( )23x . Examples: Evaluate each of the following.

(a) 21

36−

(b) 21

1649 −

(c) 3

2

27 (d) 43

16−

Answers: In each case, express the original question as a power nested within another power or

powers. Most people find it easiest to put the fractional exponents in the innermost parentheses so that they can be evaluated first.

In conclusion: Zero Exponents: 10 =x where 0≠x Example: 150 =

Negative Exponents: n

n

xx

=− 1 where 0≠x Example:

91

313

22 =

=−

Fractional Exponents: nn xx =1

Example: 288 331

==


(a) We will show two very similar methods to solve this question.

Method 1 (Slightly Easier)

21

36−

We do not know how to deal with an exponent that is both a fraction and a negative number.

1

21

21

3636−

−

=

We will break the exponent that we do not understand into two exponents that we understand. We do this using use the Law of Powers of Powers but in reverse. For example

6x can be expressed as ( )32x or as ( )23x . Using

this logic, the power 21

36−

can be expressed as 1

21

36−

or as ( )2

1136− . We chose the first

version.

( ) 11

21

21

363636−

−−

=

=

We evaluate the power in the parentheses. We learned earlier that raising a number to the exponent one half was the same thing as taking the square root of that number.

( ) 111

21

21

6363636 −−−

−==

= The square root of 36 is 6.

( )616363636 11

1

21

21

===

= −−

−−

We learned earlier that when we raise a number to the exponent -1, we take the reciprocal of the number. The reciprocal of six is one-sixth. This is our final answer.

Method 2

21

36−

( )21

121

3636 −−=

Break the exponent into two exponents using the Powers of Powers in reverse.

21

21

36136

=

−

Evaluate the power in the parentheses. When we raise a number to the exponent -1, we take the reciprocal of the number. The reciprocal of thirty-six is one-thirty-sixth.

=

=

−

361

36136

21

21

We learned earlier that raising a number to the exponent one half was the same thing as taking the square root of that number.

61

361

36136

21

21

=

=

=

− The square root of one-thirty sixth is one-sixth.


(b) 74

47

1649

1649

1649 11

1

21

21

=

=

=

=

−−−

−

(c) Although this question can be done two different ways, we will focus on the easier

method where the fractional exponent is in the innermost parentheses.

32

27 We only know how to deal with fractional exponents where the numerator is 1.

2

31

32

2727

= Break the exponent into two exponents using

the Powers of Powers in reverse.

( )23

2

31

32

272727 =

=

We learned earlier that raising a number to the exponent one third was the same thing as taking the cube root of that number.

( ) 223

2

31

32

3272727 ==

= The cube root of 27 is 3.

( ) 93272727 223

2

31

32

===

= When we square 3, we obtain 9 (the final

answer).

(d) Although this question can be done several different ways, we will focus on the

easier method where the fractional exponent is in the innermost parentheses.

43

16−

13

41

43

1616

−

−

= Break the exponent into three

exponents.

( ) 134

13

41

43

161616−

−

−

=

=

Raising a number to the one fourth is the same as the fourth root of that number.

( ) ( ) 13134

13

41

43

2161616 −−

−

−=

=

= The fourth root of 16 is 2.

( ) ( ) 113134

13

41

43

82161616 −−−

−

−==

=

= When we cube 2, we obtain 8.

( ) ( )8182161616 113

134

13

41

43

===

=

= −−

−−

−

When 8 is raised to the exponent -1, we obtain one-eighth.


Questions:

1. Evaluate each of the following. Do not use a calculator.

(a) =−13 (b) =06

(c) =21

81 (d) =−24

(e) =31

27 (f) =

0

75

(g) =−25 (h) =

−1

32

(i) =

−

0

32 (j) =−43

(k) =

2

1

254 (l) =−42

(m) =

−2

43 (n) =− 09

(o) =31

8 (p) =

2

1

41

(q) =

2

1

169 (r) =

−3

23

(s) =

3

1

827 (t) =

−2

56

2. Express each as a power.

(a) 3 =p (b) =21b

(c) =h (d) =51d


3. Evaluate each of the following. In each case we will express the original question as a power nested within another power or powers. The first few questions have been started for us. Show all your work. Do not use a calculator.

(a) =

=

−−

1

31

31

88

(b) =

=

−−

1

21

21

259

259

(c) =

=

3

21

23

44

(d) =

=

−

−

14

31

34

278

278

(e) =−

21

25

(f) =32

8

(g) =

−21

4916

(h) =

2

3

49

(i) =−

23

16

(j) =

−34

81

(k) =

−32

827


(l) =−

43

81 4. Complete the table of values below for the exponential function xy 9= . Show your work to

the right of the table. xy 9= x y

-2

23

−

-1

21

−

0

21

1

23

2

If you feel that you need additional practice with these types of questions, you can complete the optional questions titled Additional Practice: Exponents, Part 2 found in the appendix of this unit.


The Most Basic Exponential Functions Now that we understand zero, negative and fractional exponents, we can start to generate graphs of exponential functions by generating tables of values. In this section we will limit our examination to very basic exponential functions of the form xby = . Part 1: Functions of the Form xby = where b > 1 Complete the tables of values for the two exponential functions, plot the points, and draw the

resulting curves. You will have to approximate the location of some points. xy 2= xy 3= x y x y -3 -3

-2 -2

-1 -1

0 0

1 1

2 2

3 3

-2

-1

0

1

2

3

4

5

6

7

8

9

-4 -3 -2 -1 0 1 2 3 4

-2

-1

0

1

2

3

4

5

6

7

8

9

-4 -3 -2 -1 0 1 2 3 4

Will either of these two functions ever intersect the x-axis? Yes No Are these growth or decay curves? Growth Decay


Part 2: Functions of the Form xby = where 0 < b < 1 Complete the tables of values for the two exponential functions, plot the points, and draw the

resulting curves.

x

y

=

21

x

y

=

31

x y x y -3 8 -3 27

-2 -2

-1 -1

0 0

1 1

2 2

3 3

-2

-1

0

1

2

3

4

5

6

7

8

9

-4 -3 -2 -1 0 1 2 3 4

-2

-1

0

1

2

3

4

5

6

7

8

9

-4 -3 -2 -1 0 1 2 3 4

Will either of these two functions ever intersect the x-axis? Yes No Are these growth or decay curves? Growth Decay Questions:

1. An asymptote is a straight line that a function approaches but never touches. All four of the exponential functions that we graphed have the same asymptote. Which one of the following best describes the asymptote we encountered?

(a) Vertical Asymptote at x = 0 (b) Horizontal Asymptote at y = 0 (c) Oblique (i.e. inclined) Asymptote at y = x


2. What is the domain of all four of these functions? ___________________________ 3. What is the range of all four of these functions? ___________________________ 4. What point is shared by all four functions? _______ This point is called the focal point. 5. Explain how you can determine if an exponential function produces a growth curve or a

decay curve by just looking at the equation? 6. (a) Look at the table of values for xy 2= and work out the following ratios.

=÷=÷=81

41

121

2 yyyy =÷=÷=

41

21

232

3 yyyy

=÷=÷=21134

3

4 yyyy =÷=÷= 1245

4

5 yyyy

=÷=÷= 24565

6 yyyy =÷=÷= 4867

6

7 yyyy

(b) Is there a common ratio between the successive y-values? YES or NO

7. Look at the table of values for x

y

=

31 and work out the following ratios.

=÷=÷= 279121

2 yyyy =

2

3

yy

=3

4

yy =

4

5

yy

=5

6

yy =

6

7

yy


8. Which one of these statements best describes the pattern observed in all four tables of values. (a) As the x-values change by the same increment, there is a common difference between

successive y-values. (b) As the x-values change by the same increment, there is a common ratio between

successive y-values. (c) As the x-values change by the same increment, there is a common difference at the D2

level between successive y-values. 9. In question 8 we identified which statement corresponded to the pattern we saw in the tables

of values for our four exponential functions. We ignored two of the statements. What type of function (linear, quadratic or sinusoidal) do the other two statements correspond to?


Exponential Functions and Transformations In the previous section we graphed the very basic exponential functions xy 2= , xy 3= ,

x

y

=

21 , and

x

y

=

31 . We learned the following important pieces of information regarding

exponential functions of the form xby = . • Their graphs form curves that approach by never touch the x-axis. This line they

approached is called the horizontal asymptote and it has the equation y = 0. • Their graphs passed through the point (0, 1). It is referred to as the focal point. • The successive y-values in their tables of values exhibit a common ratio when the x-

values are changing by the same increment.

In this section we will examine more complex exponential functions. Specifically we are going to see how various transformations (reflections, stretches, and translations) affect the graph of the exponential function xy 2= . The table of values of this function has been provided below. In addition to this a calculator-generated sketch of the curve and the accompanying WINDOW settings have been provided.

x y

-2 41 or 0.25

-1 21 or 0.5

0 1

1 2

2 4

Investigation:

In each part of this investigation you are going to alter the equation of the exponential function xy 2= and see what effect this has on the graph and table of values. You will also identify the

type of transformation (vertical stretch, horizontal stretch, horizontal translation, or reflection in y-axis) that has occurred. It is important to note that in each part of this investigation, you will be comparing your transformed exponential function to the exponential function y = 2x.

Graph each function using the WINDOW settings shown and generate its table of values using the TABLE feature on a graphing calculator. Note that the TblStart values change throughout this investigation.


Complete the following chart. For the first few parts, the chart is partially completed.

New Exponential Function

Table of Values

Sketch of Graph Mapping Rule and Transformation

(a) xy −= 2 Enter: Y1=2^(-X)

TblStart = 2 ∆Tbl = -1

x y 2 1 0 -1 -2

( ) ( )yxyx ,, −→

We’re dealing with a reflection in the y-axis. This caused our curve to change from a growth curve to a decay curve.

In the Sinusoidal Functions Unit and Quadratic Functions Unit, you encountered reflections in the x-axis, but this is the first time you have encountered reflections in the y-axis. We previously discovered that the base of the exponential governed whether we were dealing with a growth or decay curve. If the base is greater than 1 (e.g. xy 3= ), then we are dealing with a growth curve. If the base of the exponential function is between 0 and 1 (e.g. xy 5.0= ), then we are dealing with a decay curve. The introduction of reflections in the y-axis messes up this rule. Now we have to consider the base and whether a reflection in the y-axis has occurred. For example, the function xy −= 5.0 has a base between 0 and 1 but also exhibits a reflection in the y-axis. This means that the graph will form a growth curve. The function xy −= 4 has a base greater than 1 but also exhibits a reflection in the y-axis. This means that the graph will form a decay curve.


Table of Values


(b) ( )xy 22= Enter: Y1=2(2^X)

TblStart = -2 ∆Tbl = 1

x y -2 -1 0 1 2

( ) ( )yxyx 2 ,, →

We are dealing with a vertical stretch of 2.


New Exponential

Function Table of Values


(c) ( )xy 221

=

Enter: Y1=(1/2)(2^X)


x y -2 -1

( ) ( ) ,, xyx →

(d) x

y 21

2= Enter: Y1=2^(X/2)


x y -4

( ) ( )yyx , , →

We are dealing with a horizontal stretch of 2.

(e) xy 22= Enter: Y1=2^(2X)

TblStart = -1 ∆Tbl = 0.5

x y -1

( )

→ yxyx ,

21,

(f) 12 += xy Enter: Y1=2^(X+1)


x y -3

( ) ( )yyx , , →

We are dealing with a horizontal translation of -1.


New Exponential

Function Table of Values


(g) 12 −= xy Enter: Y1=2^(X-1)


x y -1

( ) ( )yxyx ,1, +→

Summarize Your Findings: Equation Transformation and Affect (a) xy −= 2 We’re dealing with a reflection in the y-axis. This caused our

curve to change from a growth curve to a decay curve.

(b) ( )xy 22= We are dealing with a vertical stretch of 2.

(c) ( )xy 221

=

(d) xy 2

1

2= We are dealing with a horizontal stretch of 2.

(e) xy 22=

(f) 12 += xy We are dealing with a horizontal translation of -1.

(g) 12 −= xy


Conclusions:

If an exponential function is of the form ( ) ( )dxcbay −−=1

, then:

(i) the a indicates that a _____________________________________________________ has

occurred.

(ii) the c indicates that a _____________________________________________________ has

occurred.

(iii) the d indicates that a _____________________________________________________ has

occurred.

(iv) a negative sign in front of the c1 indicates that ____________________________________

_____________ has occurred.

Examples For each of the following, state the transformations and the original function that was transformed. (a) ( ) ( )537 +−= xy (b) ( ) ( )23546.0 −= xy Answers: (a) ( ) ( )537 +−= xy

• Original Function that was Transformed: xy 3= • Vertical Stretch (VS) of 7 • Reflection in the y-axis • No Horizontal Stretch • Horizontal Translation (HT) of -5

(b) ( ) ( )23546.0 −= xy • Original Function that was Transformed: xy 5= • Vertical Stretch (VS) of 0.46 • No Reflection in the y-axis

• Horizontal Stretch (HS) of 31

• Horizontal Translation (HT) of 2


State the Transformations: Exponential Functions 1. In the previous section, you learned how to identify the transformations when an exponential

function is presented in the form ( ) ( )dxcbay −−=1

. In the questions that follow, you will be given the equation of an exponential function and be asked identify the transformations. Three examples have been completed for you.

Function Transformed Function

Vertical Stretch

Reflection in y-axis

Horizontal Stretch

Horizontal Translation

e.g. ( ) ( )325 +−= xy xy 2= 5 yes none -3

e.g. ( ) ( )4

21

73 −= xy xy 7= 3 none 2 4

e.g. ( ) xy 534.0 −= xy 3= 0.4 yes

51 none

(a) ( ) 152 −= xy

(b) ( ) xy −= 36

(c) ( )524 += xy

(d) ( ) xy 4

123 −=

(e) ( ) ( )2

31

65.1 −= xy

(f) ( ) ( )6247.0 +−= xy

(g) ( )3

61

7−−

=x

y

2. Determine whether we are dealing with a growth or decay curve. Remember to consider the

base of the function and whether the function has undergone a reflection in the y-axis. (a) ( )xy 34= (b) ( ) xy −= 25

(c) 3

216

+

=

x

y (d) x

y5

312

−

=

(e) ( )425 −−= xy (f) x

y21

413

−

=


Graphing Using Transformations We have graphed sinusoidal functions and quadratic functions using transformations. We followed a set procedure. This procedure does not change when working with exponential functions; however we must include reflections in the y-axis and will not be dealing with reflections in the x-axis or vertical translations.

Procedure: 1. State the transformations. Separate the transformations that affect the x-values

(reflections in the y-axis, horizontal stretches, and horizontal translations) from those that affect the y-values (vertical stretches).

2. Construct the mapping rule. 3. Using the mapping rule, create the table of values for the desired exponential function by

altering the basic exponential function of the form xby = (e.g. xy 2= , xy 3= ,…). 4. Graph the points and draw the curve. Example 1 Graph the function ( ) ( )123 +−= xy using transformations.

Answer: - Reflection in the y-axis - Vertical Stretch of 3 - Horizontal Translation of -1

The mapping rule is created by examining the transformations that affect the x-values in the table, and examining the transformations that affect the y-values. • The x-values are affected by reflections in the y-axis, horizontal stretches, and horizontal

translations. In this case, the function has undergone a reflection in the y-axis and a horizontal translation of -1. That is why the mapping rule shows that the x-values change to -x - 1.

• The y-values are only affected by vertical stretches. In this case, the function has undergone a vertical stretch of 3. That is why the mapping rule shows that the y-values change to y3 .

• Mapping Rule: ( ) ( )yxyx 3,1, −−→

Old Table ( xy 2= ) New Table ( ( ) ( )123 +−= xy ) x y x y

-2 41 ( ) =−−− 12 1

43

=

413

-1 21 ( ) =−−− 11 0

211

=

213

0 1 ( ) =−− 10 -1 3 ( )13=

1 2 ( ) =−− 11 -2 6 ( )23=

2 4 ( ) =−− 12 -3 12 ( )43=


Notice that there is a still a common ratio between successive y-values in the new table of values.

0

1

2

3

4

5

6

7

8

9

10

11

12

-3 -2 -1 0 1

Example 2

Graph the function ( ) ( )421

35 −= xy using transformations.

Answer: - Horizontal Stretch of 2 - Vertical Stretch of 5 - Horizontal Translation of 4

Mapping Rule: ( ) ( )yxyx 5,42, +→

Old Table ( xy 3= ) New Table ( ( ) ( )421

35 −= xy ) x y x y

-2 91 ( ) =+− 422 0

95

=

915

-1 31 ( ) =+− 412 2

321

=

315

0 1 ( ) =+ 402 4 5 ( )15=

1 3 ( ) =+ 412 6 15 ( )35=

2 9 ( ) =+ 422 8 45 ( )95=



0

5

10

15

20

25

30

35

40

45

0 1 2 3 4 5 6 7 8 9

Example 3

Graph the function ( ) xy 3421 −= using transformations.

Answer:

- Reflection in the y-axis - Vertical Stretch of 21

- Horizontal Stretch of 31

Mapping Rule: ( )

−→ yxyx

21,

31,

Old Table ( xy 4= ) New Table ( ( ) xy 3421 −= )

x y x y

-2 161 ( ) =−− 2

31

32

321

=161

21

-1 41 ( ) =−− 1

31

31

81

=

41

21

0 1 ( ) =− 031 0

21 ( )1

21

=

1 4 ( ) =− 131

31

− 2 ( )421

=

2 16 ( ) =− 231

32

− 8 ( )1621

=



0

1

2

3

4

5

6

7

8

9

-0.7 -0.6 -0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

Questions:

1. This question is partially completed. Complete the question.

Graph ( ) ( )131

25 −−= xy using transformations.

- Reflection in the y-axis - Vertical Stretch of ____ - Horizontal Stretch of ____ - Horizontal Translation of ____

Mapping Rule ( ) ( )_________ , , →yx

x y x y

-2 41 ( ) =+−− 123 7

411

=

415

-1 21 ( ) =+−− 113

=

215

0 1

1

2

2. This question is partially completed. Complete the question.

Graph ( ) ( )234 +−= xy using transformations.

- Reflection in the y-axis - Vertical Stretch of ____ - Horizontal Translation of ____

0123456789

1011121314151617181920

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8


Mapping Rule ( ) ( )_________ , , →yx

x y x y

-2 91 ( ) =−−− 22 0

94

=

914

-1

0

1

2

3. Graph each of the following quadratic functions using transformations.

(a) ( ) 132 += xy Transformations: Mapping Rule:

x y

(b) ( ) ( )321

26 += xy Transformations: Mapping Rule:

x y

0

5

10

15

20

25

30

35

40

-5 -4 -3 -2 -1 0 1

0123456789

1011121314151617181920

-4 -3 -2 -1 0 1 2

0

2

4

6

8

10

12

14

16

18

20

22

24

26

-8 -7 -6 -5 -4 -3 -2 -1 0 1 2


(c) ( )2

41

3+−

=x

y Transformations: Mapping Rule:

x y

(d) ( ) ( )1221 −−= xy

Transformations: Mapping Rule:

x y

(e) ( ) xy 243 −= Transformations: Mapping Rule:

x y

0

1

2

3

4

5

6

7

8

9

10

-10 -8 -6 -4 -2 0 2 4 6 8

0

1

2

3

4

-2 -1 0 1 2 3 4

0

5

10

15

20

25

30

35

40

45

50

-1.5 -1 -0.5 0 0.5 1 1.5


4. Create the mapping rule for each of the following. Mapping Rule

(a) ( ) 657 −= xy

(b) ( ) ( )528 +−= xy

(c) ( )326 += xy

(d) ( ) xy 71

345.0 −=

(e) ( ) ( )439101 −−= xy

5. Graph the function ( ) ( ) ( )531

421 −= xxf using transformations.

6. What was the equation of the horizontal asymptote for all of the exponential functions that

you encountered in this section of the unit? Please view the nsccalpmath video on YouTube titled Overview and Review of Graphing with Transformations.


Equivalent Exponential Functions As you proceed through this unit, we will not be working with equations of the form

( ) ( )dxcbay −−=1

, as we did in the last few sections. Instead we will work with exponential

functions of the form ( )cx

bay = . This alternate form is generally easier to work with when dealing with more complex questions. We notice that vertical stretch, a, and the horizontal stretch, c, are still present in this alternate form. However, the reflection in the y-axis and horizontal translation, d, are not. This is due to the fact that these transformations can be integrated into the other form of the equation. The easiest way to understand this is to look at specific examples. Example Show that the following exponential functions are equivalent.

(a) ( ) xy −= 53 and x

y

=

513 (b) ( ) 325 += xy and ( )xy 240=

(c) ( ) ( )2223 −−= xy and x

y2

2148

= (d)

( )421

3−−

=x

y and 2

319

x

y

=

Answers:

With each of these questions, we will show that the first function in the ( ) ( )dxcbay −−=1

form is

equivalent to the second function in the ( )cx

bay = form. We will be using our laws of exponents.

(a) ( )( )

x

x

x

y

y

y

=

=

=−

−

513

53

531

(b) ( )( )( )( )( )( )x

x

x

x

y

yyy

240

825225

253

3

=

=

=

= +

(c) ( ) ( )

( )( )( )( ) ( )

x

x

x

x

x

y

y

yy

y

2

21

42

42

22

2148

1623

22323

23

=

=

=

=

=

−

−

+−

−−

(d) ( )

( )

( ) ( )

2

21

1

221

221

421

319

93

33

3

3

x

x

x

x

x

y

y

y

y

y

=

=

=

=

=

−

−

+−

−−


For this course we are not expected to be able to change from the ( ) ( )dxcbay −−=1

form to the

( )cx

bay = form, or vise versa. We only need to be aware that one form can be converted to the other. This explanation was provided only to help explain why this resource abandons one form in favor of another.


Recognizing Patterns within Tables of Values Over the last few months, we have learned how to determine whether a table of values belongs to a linear function, quadratic function, or exponential function based on patterns we see in that table.

• In the Bridging Unit we learned that for linear functions as the x-values change by the same increment, there is a common difference between the successive y-values.

• In the Quadratic Functions Unit we learned that for quadratic functions as the x-values change by the same increment, there is a common difference at the D2 level between the successive y-values.

• In this unit, specifically the section titled The Most Basic Exponential Functions, we learned that for exponential functions as the x-values change by the same increment, there is a common ratio between the successive y-values. To check for a common ratio, work out the following ratios.

Ratio = 1

2

yy Ratio =

2

3

yy Ratio =

3

4

yy Ratio =

4

5

yy

Questions:

For each table of values, identify the type of function it corresponds with and briefly explain how you arrived at that answer (e.g. quadratic: common difference of 4 at the D2 level). Two of these tables do not correspond to linear, quadratic or exponential functions.

1. x y 2. x y 3. x y -3 6 1 4 -10 15 0 12 2 14 -8 12 3 24 3 30 -6 9 6 48 4 52 -4 6 9 96 5 80 -2 3

4. x y 5. x y 6. x y

11.2 15.00 -4 20 4 80 11.4 15.25 -2 6 7 40 11.6 15.50 0 0 10 20 11.8 15.75 2 2 13 10 12.0 16.00 4 12 16 5


7. x y 8. x y 9. x y

-10 1 -0.4 0.9 0 1000 -9 2 0 9 7.2 1100 -8 9 0.4 90 14.4 1210 -7 28 0.8 900 21.6 1331 -6 65 1.2 9000 28.8 1464.1

10. x y 11. x y 12. x y

-4.4 1 8 2.8 -7 10 -3.3 -1 10 3.1 -5 13 -2.2 -5 12 3.4 -4 16 -1.1 -11 14 3.7 -1 19 0 -19 16 4.0 4 22

13. x y 14. x y 15. x y

-22 48.6 100 4 1.5 10000 -20 16.2 200 11 2.0 8000 -18 5.4 300 22 2.5 6400 -16 1.8 400 37 3.0 5120 -14 0.6 500 56 3.5 4096


Investigation: From a Table of Values to an Equation

We have been provided with four equations of the form cx

aby = and their corresponding tables of values. We will examine the equation and corresponding table in an attempt to see a relationship between the two. You may wish to consider:

• the increments by which the x-values change • the y-intercept of each function • the common ratio between successive y-values

( )736x

y = ( )4107x

y = 3

2112

x

y

=

9

5140

x

y

=

x y x y x y x y -7 2 -8 0.07 -9 96 0 40 0 6 -4 0.7 -6 48 9 8 7 18 0 7 -3 24 18 1.6 14 54 4 70 0 12 27 0.32 21 162 8 700 3 6 36 0.064

Questions:

1. Explain in your own words how we can find the equation of an exponential function in the

form cx

aby = by examining its table of values. Your explanation should include a discussion of transformations.

2. Determine the equation of the exponential function that corresponds to the following table of

values. Briefly explain how you accomplished this.

x y -5 7 0 21 5 63 10 189


Determining the Equation Given a Table of Values Examples For each table, determine if it can be represented by an exponential function. If so, determine the equation of the exponential function.

Answers: (a)

x y -10 3.5 -5 7 0 14 5 28 10 56

It is exponential for the following reasons. • The x-values are changing by the same increment. • There is a common ratio between successive y-values.

25.3

7= 2

714

= 21428

=

We need three pieces of information to determine the equation. • y-intercept or a is equal to 14 • common ratio between successive y-values or b is equal to 2 • the x-values are changing by increments of 5, therefore c is

equal to 5

Equation: ( )5214x

y =

(b)

x y 0 4000 3 4400 6 4840 9 5324 12 5856.4


1.140004400

= 1.144004840

= 1.148405324

=

We need three pieces of information to determine the equation. • y-intercept or a is equal to 4000 • common ratio between successive y-values or b is equal to

1.1 • the x-values are changing by increments of 3, therefore c is

equal to 3

Equation: ( )31.14000x

y =

(c)

x y 0 1 2 5 4 13 6 25 8 41

There is no common ratio.

515= 6.2

513

= 92.11325

= 64.12541

=

This is not an exponential function of the form cx

aby = . (It is actually quadratic because it has a common difference at the D2 level.)


Answers: (d)

x y 6 500 12 400 18 320 24 256 30 204.8


8.0500400

= 8.0400320

= 8.0320256

=

We need three pieces of information to determine the equation. • We have to work backwards in the table to determine the y-intercept. The y-intercept or a is equal to 625 (i.e. 8.0500 ÷ ). • common ratio between successive y-values or b is equal to

0.8 • the x-values are changing by increments of 6, therefore c is

equal to 6

Equation: ( )68.0625x

y =

(e)

x y -2.5 0.07 -2 0.21

-1.5 0.63 -1 1.89

-0.5 5.67


37.021.0

= 321.063.0

= 363.089.1

=

We need three pieces of information to determine the equation. • We have to work forwards in the table to determine the y-

intercept. The y-intercept or a is equal to 17.01 (i.e. 367.5 × ). • common ratio between successive y-values or b is equal to 3. • the x-values are changing by increments of 0.5, therefore c is

equal to 0.5 or 21 .

Equation:

With the exponent, we will be taking x and dividing it by 21 .

Remember that dividing by a fraction is the same as multiplying by the fraction’s reciprocal.

xxxx 212

21

21 =×=÷= ( ) xy 2301.17=


Questions:

For each table, determine if it can be represented by an exponential function. If so, determine the equation of the exponential function.

1.

x y -18 6 -9 30 0 150 9 750 18 3750

2.

x y -6 3 0 12 6 48 12 192 18 768

3.

x y -9 120 -6 60 -3 30 0 15 3 7.5

4.

x y -4 3200 -3 800 -2 200 -1 50 0 12.5

5.

x y -10 80 -5 120 0 180 5 270 10 405

6.

x y -3 9 0 7 3 5 6 7 9 9

.7.

x y -15 5 -12 10 -9 20 -6 40 -3 80

8.

x y -20 90000 -16 9000 -12 900 -8 90 -4 9


9.

x y 7 1250 14 500 21 200 28 80 35 32

10.

x y -8 12 -4 10 0 8 4 6 8 4

11.

x y 2 1280 4 1600 6 2000 8 2500 10 3125

12.

x y -0.5 2

0 6 0.5 18 1 54

1.5 162

13.

x y -12 90 -9 60 -6 36 -3 18 0 6

14.

x y -0.5 486 -0.25 162

0 54 0.25 18 0.5 6

15.

x y 0.1 80 0.2 160 0.3 320 0.4 640 0.5 1280

16.

x y -1 3645

-0.8 1215 -0.6 405 -0.4 135 -0.2 45


More than One Acceptable Answer In the last section we learned how to determine the equation of an exponential function given its table of values. It may be surprising to learn that those types of questions have more than one acceptable answer. Consider the following.

Original Table:

x y -1 2.5 0 5 1 10 2 20 3 40 4 80 5 160 6 320 7 640 8 1280 9 2560

We have been supplied with table of values for an exponential function. It has a common ratio of 2, a y-intercept of 5, and the x-values are changing by increments of 1. If we were asked to find the equation based on this table, we would state the following. ( )xy 25= Below you will find two shortened versions of our original table of values. We will examine these and their resulting equations to see how there can be more than one acceptable answer for the equation of the exponential function based on a table of values.

Shortened Table #1

x y 0 5 2 20 4 80 6 320 8 1280

In this case we started at the y-intercept and used every second pair of values from the original table of values. We still have a common ratio. The ratio, however, is 4, rather than 2. If we were asked to find the equation based on this table, we would state the following.

( )245x

y =

Shortened Table #2

x y 0 5 3 40 6 320 9 2560

In this case we started at the y-intercept and used every third pair of values from the original table of values. We still have a common ratio. The ratio, however, is 8, rather than 2. If we were asked to find the equation based on this table, we would state the following.

( )385x

y =

All three of these equations are correct and represent the same curve. You can check this using a graphing calculator. Why is this so? Consider the following.

( )xy 25= They are all the same function.

( )

( )

( )( )x

x

x

x

y

y

y

y

25

45

45

45

21

2

=

=

=

=

( )

( )

( )( )x

x

x

x

yy

y

y

2585

85

85

3

31

3

=

=

=

=


Questions:

For each of these tables of values, produce three different equations that represent the same curve.

1.

x y 0 2 1 6 2 18 3 54 4 162 5 486 6 1458 7 4374 8 13 122 9 39 366 10 118 098 11 354 294 12 1 062 882

2.

x y -2 20 480 -1 10 240 0 5120 1 2560 2 1280 3 640 4 320 5 160 6 80 7 40 8 20 9 10 10 5

3.

x y -9 0.00008 -6 0.0008 -3 0.008 0 0.08 3 0.8 6 8 9 80 12 800 15 8000 18 80 000 21 800 000 24 8 000 000 27 80 000 000


Determining the Equation Based on the Real World Situation In this section we will be given a written description of a real world situation that can be modeled using an exponential function and we will be required to generate the appropriate equation for that function. We will be shown two methods to solve these types of questions. We are permitted to use either of these methods. Example 1 The initial concentration of a bacteria population was 300 per square centimetre. The concentration doubled every forty minutes. Determine the equation of the function that describes the bacteria concentration, C, in terms of time, t.

Answer: Method 1: Directly from Written Description to Equation We are not using the variables x and y in this equation. Since we are using the variables t

and C, the equation will be in the following form.

ct

abC = The initial value, 300, is equivalent to the C-intercept or a in our equation. The

concentration is doubling, therefore, the base, b, of our exponential function will be 2. This doubling occurs every 40 minutes, therefore, the horizontal stretch or c is 40. The resulting equation will be as follows.

( )402300t

C = Method 2: From a Written Description to a Table of Values to an Equation

t C 0 300 40 600 80 1200 120 2400 160 4800

We can use the written description to generate a table of values. Since the initial concentration was 300, we can generate the ordered pair (0, 300). Forty minutes later the concentration doubles which means that our next ordered pair is (40, 600). Another forty minutes later (t = 80), the population doubles again. That gives us our third ordered pair (80, 1200). This process continues and hence we end up with the ordered pairs (120, 2400) and (160, 4800). We can now use the procedure learned in the last two sections to generate the equation. • C-intercept or a is equal to 300 • common ratio between successive C-values or b is equal to

2. • the t-values are changing by increments of 40, therefore c is

equal to 40

Equation: ( )402300t

C =


Example 2 The value, v, of a particular vehicle after its purchase depreciates with time, t. When this car was purchased, it was worth $24 000. The value of the car drops by 15% each year. (a) Determine the equation of the function that describes the value of the car in terms of time. (b) Are we dealing with a growth or decay curve? (c) What is the car worth 6.5 years after its purchase? (d) If this mathematical model only applies until the car is ten years old, state the domain and

range of this function.

Answers: (a) Method 1: Directly from Written Description to Equation Since we are using the variables t and v, the equation will be in the following form.

ct

abv = The initial value of the car was $24 000. That means that the v-intercept or a will be

equal to 24 000. If the value of the car drops by 15%, then it retains 85% of its value. That means that the base, b, of our exponential function will be 0.85. Since this drop occurs every year then c will be equal to 1 (no horizontal stretch).

( )tv 85.024000= Method 2: From a Written Description to a Table of Values to an Equation

t v 0 24 000 1 20 400 2 17 340 3 14 739 4 12528.15

Generate a table of values based on the written description. We can now use the procedure learned in the last two sections to generate the equation. • v-intercept or a is equal to 24 000 • common ratio between successive v-values or b is

equal to 0.85. • the t-values are changing by increments of 1, therefore

c is equal to 1

Equation: ( )tv 85.024000= (b) Since the value of the car is dropping, we know that we are dealing with a decay curve.

We could have also figured this out based on our equation from part (a). Since the base of our equation is 0.85 (between 0 and 1) and there is no reflection in the y-axis, we can predict that we are dealing with a decay curve.

(c) Take our equation, substitute 6.5 in for t, and solve for v. We will need to use a scientific

calculator or graphing calculator to accomplish this.

( )( )

15.8345$3477.024000

85.024000

85.0240005.6

=×=

=

=

vvv

v t


(d) Since this mathematical model only applies until the car is ten years old, we know that our time is restricted between 0 years and 10 years.

Domain: { }100 ≤≤ tRtε We know that the car was initially worth $24 000, however we need to know what it will

be worth in 10 years in order to state the range.

( )( )

99.4724$1969.024000

85.024000

85.02400010

=×=

=

=

vvv

v t

The value of the car is restricted between $4724.99 and $24 000. Range: { }2400099.4724 ≤≤ vRvε Example 3 Montez purchased a rare coin for $280 and he predicts that it will increases in value by 40% every three years. (a) Determine the equation of the function that describes the predicted value, v, of the coin in

terms of time, t. (b) Based on Montez’s mathematical model, what will the coin be worth in eight years? (c) Are we dealing with a growth or decay curve? (d) If this mathematical model only applies for 16 years after the date of purchase, what is the

domain and range of this function?

Answers: (a) Method 1: Directly from Written Description to Equation The initial value of the coin was $280. That means that the v-intercept or a will be equal

to 280. If the value of the coin increases by 40% over and above 100% of its value, then the common ratio or base, b, of our exponential function will be 1.40 (i.e. 40%+100%). Since this increase occurs every three years then c will be equal to 3.

( )34.1280t

v = Method 2: From a Written Description to a Table of Values to an Equation

t v 0 280 3 392 6 548.80 9 768.32

Generate a table of values based on the written description. We can now use the procedure learned in the last two sections to generate the equation. • v-intercept or a is equal to 280 • common ratio between successive v-values or b is

equal to 1.4. • the t-values are changing by increments of 3, therefore

c is equal to 3

Equation: ( )34.1280t

v =


(b) Take our equation, substitute 8 in for t, and solve for v. We will need to use a scientific calculator or graphing calculator to accomplish this.

( )

( )

80.686$453.2280

4.1280

4.1280

38

3

=×=

=

=

vvv

vt

(c) Since the value of the coin is increasing, we know that we are dealing with a growth

curve. We could have also figured this out based on our equation from part (a). Since the base of our equation is 1.4 (greater than 1) and there is no reflection in the y-axis, we can predict that we are dealing with a growth curve.

(d) Since this mathematical model only applies for 16 years, we know that our time is

restricted between 0 years and 16 years. Domain: { }160 ≤≤ tRtε We know that the coin was initially worth $280, however we need to know what it will

be worth in 16 years in order to state the range.

( )

( )

64.1684$017.6280

4.1280

4.1280

316

3

=×=

=

=

vvv

vt

The value of the coin is restricted between $280 and $1684.64. Range: { }64.1684280 ≤≤ vRvε Example 4 When Catherine drinks a brewed cup of coffee, she ingests 130 mg of caffeine into her system. The half-life of caffeine in a typical adult is 5.5 hours. How much caffeine will be in her system 4 hours after she drinks the cup of coffee?

Answer: Although the question does not specifically ask for the equation of the function, it is required

to complete the question.

Equation: 5.5

21130

t

A

= where A is the amount of caffeine in mg and t is the time in hours.

Now use the equation and solve for A when t = 4.

mg 5.78

6040.013021130

5.54

=×=

=

AA

A


Example 5 Determine the equation of the exponential function that passes through the points (0, 18) and (11, 6).

Answer: When we are only supplied with two points, the question must tell us explicitly what type of

function we are dealing with. Since this question has told us that we are dealing with an exponential function, we know that successive y-values would display a common ratio. These types of questions must also supply us with the y-intercept directly or indirectly.

x y 0 18 11 6

The y-intercept or a will be 18.

Ratio31

186

1

2 ===yy

We know that this ratio is a common ratio because the question states that we are dealing with an exponential function.

Therefore: 31

=b

This common ratio occurs as the x-values change by increments of 11. Therefore c will equal 11.

Equation: 11

3118

x

y

=

Example 6 The presence of sediment in water reduces the distance sunlight can penetrate into the water. In one particular case, the light intensity at the surface measured 0.940 candela/cm2. At a depth of 0.5 m, the light intensity measured 0.846 candela/cm2. The relationship between the depth of the water and light intensity can be described using an exponential function. Determine the equation of the function that describes the light intensity, I, in candela/cm2 in terms of the depth, d, in metres.

Answers:

d I 0 0.940

0.5 0.846

The initial intensity or I-intercept is 0.940. Therefore a equals 0.940.

Ratio 9.0940.0846.0

1

2 ===II

We know that this ratio is a common ratio because the question states that we are dealing with an exponential function. Therefore: 9.0=b

This common ratio occurs as the d-values change by increments of

0.5. Therefore c will equal 0.5 or 21 .

For our exponent, we must divide by a fraction.

dddd 212

21

21 =×=÷=

Equation: ( ) dI 29.0940.0=


Questions

1. Determine the equation of the function based on the following description.

Equation: (a) The y-intercept is 7. As the x-values change by increments of 4, the

y-values triple.

(b) The y-values double every time the x-values increase by 6. The graph intersects (0, 3).

(c) The y-intercept is 15. The y-values halve when the x-values change by 9.

(d) The graph intersects (0, 5). The y-values decrease by 40% as the x-values change by 12.

(e) As the x-values change by 45, the y-values decrease by 65%. The graph intersects the y-axis at (0, 8).

(f) The y-values decrease by 7% as the x-values change by 0.5. The graph’s y-intercept is 2.75.

(g) The y-intercept is 980. The y-values increase by 10% as the x-values increase by 1.

(h) The graph intersects the y-axis at (0, 47). As the x-values increase by 18, the y-values increase by 26%.

(i) The graph of the exponential function has a y-intercept of 5 and passes through the point (4, 15).

(j) The exponential function passes through the points (0, 14) and (8, 7).

(k) The exponential function passes through (9, 10) and (0, 8).

(l) The y-intercept of the exponential function is 6.7. The function also passes through the point (0.25, 5.36).

2. A biologist initially measures 800 yeast cells in a culture. She predicts that that population

will triple every 2 hours. (a) Determine the equation of the function that describes the number, n, of yeast cells in

terms of time, t. (b) How many predicted yeast cells should be in the culture 4.5 hours after the initial

reading? (c) If the mathematical model only applies for the first 12 hours, determine the domain and

range of our function,


3. In some science labs, one can find bell jars hooked up to vacuum pumps. The device is used when scientists wish to conduct experiments involving low air pressure. With one particular device, the pump removes 8% of the air for every 2 seconds the pump is operating.

(a) Find the equation that describes the percentage, P, of air remaining in the jar in terms of the time, t, the pump is in operation.

(b) Are we dealing with a growth or decay curve? (c) What percentage of the air remains in the bell jar at 15

seconds? (d) If the mathematical model only applies for the first 25

seconds, state the domain and range of the function. 4. In 2006 a mint Wayne Gretzky rookie card sold for $80 000. The buyer anticipates that the

value of the card will increase by 25% every three years. (a) Determine the equation of the function that describes the value, v, of the card in terms of

time, t, from its purchase date in 2006. (b) Explain why we are dealing with a growth curve based solely on the equation we

generated in part (a). (c) According to the buyer, what will the card be worth in 7 years? (d) If the mathematical model only applies for the first 10 years, state the domain and range

of the function.


5. Initially there was 15 milligrams of Carbon-14 in a bone fragment. Over time this radioactive isotope decays. It takes 5730 years for half of the Carbon-14 to decay to a more stable element. Based on this we say that Carbon-14 has a half-life of 5730 years.

(a) Determine the equation of the function that describes the amount, A, of Carbon-14 remaining in milligrams in terms of the time, t, in years.

(b) Explain why we are dealing with a decay curve based solely on the equation we generated in part (a).

(c) How much Carbon-14 will remain in the bone fragment after 12 000 years? 6. The preparation, handling and cooking of ground beef must be monitored very carefully to

ensure that people do not ingest E. coli bacteria. Suppose a particular package of ground beef has 5 E coli bacteria per square centimetre. After being left on the counter for 30 minutes, the same package of beef now had 20 bacteria per square centimetre.

(a) Determine the equation that describes the concentration, C, of E coli bacteria in terms of time, t, in minutes.

(b) What will be the concentration of E coli bacteria 100 minutes after the package is left out on the counter?

(c) Are we dealing with a growth or decay curve? (d) If the mathematical model only applies for the first four hours, state the domain and

range of the function.


7. The growth of a particular mutual fund can be described using an exponential function. Initially $6000 was invested in the fund. One year later, the investment had grown to $6390. How much would the investment be worth after 7 years?

8. The initial concentration of a medication in a patient’s bloodstream was 0.08 mg/cm3. The

concentration dropped by 30% every two hours. What was the concentration after 5 hours? 9. A toxin was accidentally released into a freshwater system. As a result the frog population in

this habitat starts to decline. In the beginning, there were approximately 4000 frogs. Three days after the release of the toxin, the estimated population is 1200. If the decline in the frog population can be described using an exponential, determine the frog population, eight days after the release of the toxin.

10. The population in a community has been growing annually by 15%. If the population was

initially 54 000, what will the population be in 4.5 years?


Determining the Equation Given a Graph Example 1 Determine the equation of the exponential function given its graph.

Answer: Construct a table of values using the

designated points from the graph.

x y -4 3 0 6 4 12 8 24

Now use the technique we worked with in the previous two sections of this unit.

Equation: ( )426x

y =

Questions:

Determine the equations of each of these exponential functions given their graphs.

1.

0

25

50

75

100

125

150

175

200

225

-4 -3 -2 -1 0 1 2 3 4 5 6 7

2.

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

2.2

2.4

2.6

2.8

3

-25 -20 -15 -10 -5 0 5 10 15

0

2

4

6

8

10

12

14

16

18

20

22

24

26

-5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9


3.

0

2

46

8

10

12

14

1618

20

22

24

26

2830

32

34

-14 -12 -10 -8 -6 -4 -2 0 2 4

4.

0

0.25

0.5

0.75

1

1.25

1.5

1.75

2

2.25

2.5

2.75

3

3.25

3.5

3.75

4

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

5.

0

10

20

30

40

50

60

70

80

90

100

110

120

130

-0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1

6.

0

2

4

6

8

10

12

14

16

18

20

-0.2 -0.1 0 0.1 0.2 0.3


Solving Exponential Equations by Creating the Same Base Up to this point, we have learned the following.

• Interpretation of Exponential Graphs based on Real World Situations • Evaluation of Powers with Zero, Negative and Fractional Exponents • Graphing Exponential Functions using Transformations • Determining the Equation of Exponential Functions using a Table of Values, a Written

Description, or a Graph • Solving for the Dependent Variable (typically y), given an Exponential Function and a

Value for the Independent Variable (typically x). We only have one other concept to cover in this unit. We need to know how to solve for the independent variable (typically x), given an exponential function and a value for the independent variable (typically y). In this section we will learn how to accomplish this by changing numbers to a common base. In a later section, we will handle these types of questions using logarithms. Example 1

For the exponential function 316x

y = , find x when y is equal to 32.

Answer:

( )

( )

415415

34353

345

:Therefore

22

22

1632

16

34

5

345

3

3

=

=

=

=

=

=

=

=

x

x

x

x

y

x

x

x

x

We substituted 32 in for y and will now solve for x. We want to change 32 and 16 to powers that have the same base. The 32 can be expressed as 52 . The 16 can be expressed as 42 . With powers of powers, we multiply the exponents. That is how

we end up with the exponent 3

4x .

If the powers are equal and the bases are equal, we can assume that the exponents must be equal. We wrote out the word therefore in this example but most math books use the symbol ∴ to represent this word (See examples on the next page.).


Example 2 Solve for x.

(a) x584 = (b) x2

3127

= (c) 413 255 −+ = xx

(d) ( )13

42

412

−+

=

xx

(e) ( ) xx 643 279 =− (f) 6281 += x

Answers: (a)

( )

152152

2222

84

152

532

5

=

=∴=

=

=

x

x

x

x

x

(b)

( )

23

233333

3127

23

213

2

−=

−=∴=

=

=

−

−

x

x

x

x

x

(c)

( )

918238213

5555

255

8213

4213

413

−=−−=−−=+∴

=

=

=

−+

−+

−+

xxx

xx

xx

xx

xx

(d) ( )

( )

007

226262

22

22

412

262

132

42

21

1342

==

−=++−=+∴

=

=

=

+−+

−−

+

−+

xx

xxxx

xx

xx

xx

(e) ( )( ) ( )

326263

8189618986

33

33

33

279

18986

126

23

86

126

21

3432

12643

−=

=−+=−+=−∴

=

=

=

=

+−

+

−

+−

+−

x

xxx

xx

xx

x

x

xx

xx

(f)

( )

217

6211

623

22

22

281

623

621

3

6

−=

=−−

+=−∴

=

=

=

+−

+−

+

x

x

x

x

x

x

Example 3

Given ( ) 4

213

x

xf

= , find:

(a) ( )8f . (b) x when ( ) 12=xf .


Answers: (a)

( )

( )

( )

( )

( )438

4138

2138

2138

213

2

48

4

=

×=

=

=

=

f

f

f

f

xfx

(b) ( )

( )

84

2

22

22

214

3.by sidesboth Divide 21312

213

42

412

4

4

4

−=

−=∴

=

=

=

=

=

−

−

x

x

xf

x

x

x

x

x

Example 4 A bacteria population was initially 300 per square centimetre. Every 15 minutes the population doubled. When will the population reach 9600 bacteria per square centimetre?

Answer: We first need to determine the equation of the exponential function that describes the

number, N, of bacteria per square centimetre in terms of time, t, in minutes.

Equation: ( )152300t

N = Now substitute 9600 in for N and solve for t.

( )

( )

( )

( )

minutes 75155

155

22

232

300.by sidesboth Divide 23009600

2300

155

15

15

15

=×=

=∴

=

=

=

=

tt

t

N

t

t

t

t


Questions:

1. Express each of the following as a power with a base of 2.

(a) 16 = (b) =81

(c) =2 (d) =3 2

(e) =41 (f) =

321

2. Express each of the following as a power with a base of 3.

(a) =27 (b) =91

(c) =3 (d) =811

(e) =3 9 (f) =3

1

(g) =27 (h) =4 27 3. Fill in the blanks to complete the solutions to the following exponential equations. (a)

( ) ( )

147

21262126

55 5

125251

126

4212

421

=−=−

−−=−−−=∴

=

=

=

−

−−−

−−

xx

xxx

x

xx

xx

(b) ( )( ) ( )

18

31532

33

2713

32

564

564

=−=−

−−=−=∴

=

=

=

+

−+

−+

xx

xx

x

xx

xx

4. Solve for x. Please note that in the first two questions we already have a common base. Show all your work. (a) xx 31954 66 −+ = (b) ( ) ( ) 9243 22 +−+

=xx


(c) 424 279 −+ = xx (d) 51 82 −− = xx

(e) 63 366 ++− = xx (f) 212 74 =+x

(g) 82

841 −

+

=

x

x

(h) 13

32

7149

++

=

xx

(i) ( ) 36626=

−x (j) ( )

xx

2684

2515

−+

=


(k) ( ) xx 3637 10100−+ = (l) ( )

xx

−+

=

19363

319

5. Given ( ) 123 −= xxf , find: (a) x when ( ) 27=xf . (b) ( )5.3f . 6. Given ( ) ( ) 385 += xxg , find: (a) ( )2−g . (b) x when ( ) 20=xg . 7. The radioactive isotope Potassium-40 has a very long half-life (1.3 billion years) and is used

extensively by geologists for age dating. If the initial amount of Potassium-40 was 72 milligrams, then determine:

(a) how much Potassium-40 would remain after 500 million years. (b) how long it would take for only 9 milligrams of Potassium-40 to remain.


An Introduction to Logarithms In the last section we learned how to solve exponential equations by creating a common base. We would be supplied with an equation like 2748 −= x and be asked to solve for x. In this case, we would change the 8 and 4 to a power with a base of 2. Suppose, however, that we were given the equation 2747 −= x . We could not use the “common base” strategy to solve for x. We encountered a similar problem in the Quadratic Functions Unit. We initially learned how to solve quadratic equations by factoring only to learn that this technique did not work with most quadratic equations. For this reason we had to go on to learn the quadratic formula. We need to learn about logarithms so that we can solve exponential equations like 2747 −= x where the “common base” strategy does not work. Logarithms

When we first learned how to solve simple linear equations like 753 −=+x , we also learned about inverse operations. Consider the examples below.

371077

107

=−=−+

=+

xxx

The x and 7 are being added together on the left hand side of the equation. In order to solve for x, we had to “undo” that addition. This was accomplished by subtracting 7 from both sides of the equation. Subtraction is the inverse operation to addition. The subtraction reverses the work already done by the addition. We can conclude that: 107 =+x is equivalent to 710 −=x

83

243

3243

=

=

=

x

xx

The 3 and x are being multiplied together. In order to reverse this process of multiplication we used the inverse operation division. We can conclude that:

243 =x is equivalent to 324

=x

In the two examples above, we were able to use some of the basic algebraic operations (i.e. addition, subtraction, multiplication, division, and roots) to solve for x. Unfortunately these same basic algebraic operations can not be used to solve exponential equations. Mathematicians had to create a new inverse operation to handle exponential equations. Complete the following using a calculator.

(a) Evaluate 5.210 , then find the log of that solution. Your Answer: _____

(b) Evaluate 7.010 , then find the log of that solution. Your Answer: _____

(c) Evaluate 85.110− , then find the log of that solution. Your Answer: _____

(d) Did the log button “undo” the work done by the exponent button? Yes or No


Here are the equivalent forms of equations we previously learned.

Example:

bax =+ is equivalent to abx −= 107 =+x is equivalent to 710 −=x

bax =− is equivalent to abx += 85 =−x is equivalent to 58 +=x

bax = is equivalent to abx = 243 =x is equivalent to

324

=x

bax= is equivalent to bax ×= 4

5=

x is equivalent to 54×=x

axn = is equivalent to n ax = 83 =x is equivalent to 3 8=x Now that we know that logs “undo” the work done by exponents, we can now add a new equivalent form of equations to our list.

Example:

ba x = is equivalent to bx alog= 810 =x is equivalent to 8log10=x

Example 1 Change the following from exponential to logarithmic form. (a) 12553 =

(b) 812 3 =−

(c) 21

93 =

Answers:

(a) 3125log5 = (b) 381log2 −=

(c)

213log9 =

We would say, “Log base 5 of 125 is 3.”

Example 2 Change the following from logarithmic to exponential form. (a) 216log4 =

(b) 2log31

8=

(c) 191log9 −=

Answers:

(a) 1642 = (b) 2831

= (c) 919 1 =−


Example 3 Evaluate each of the following logarithms. (a) 49log7

(b)

161log2

(c) 8log4

Answers: Do not use your log button on your calculator to solve any of these problems. The log button

works only with logarithms with a base of 10. Our questions have the bases 7, 2 and 4. The easiest approach is to make each of these expressions equal to x, change from logarithmic form to exponential form, and then evaluate.

(a)

249log

2497

49log

7

7

=

==

=

x

xx

(b)

4161log

41612

161log

2

2

−=

−=

=

=

x

x

x

(c)

( ) ( )

438log

43

232

22

22

84

8log

4

23

2

21

32

4

=

=

=∴

=

=

=

=

x

x

x

x

x

x

Example 4 Evaluate each of the following logarithms. (a) log 270 (b) log 0.045 (c) log (-100)

Answers: When the base is not supplied on a logarithm, one is to assume that the base is 10. Therefore: xx 10loglog = Since the base is 10, we can use the log button on our calculator to evaluate each of these.

(a) log 270 = 2.431 This could be restated as 27010 431.2 = . The answer 2.431 seems reasonable because

100102 = and 1000103 = . Since 270 is between 100 and 1000, then we would expect our answer to be between 2 and 3.


(b) log 0.045 = -1.347 This could be restated as 045.010 347.1 =− . The answer -1.347 seems reasonable because

10110 1 =− or 0.1, and

100110 2 =− or 0.01. Since 0.045 is between 0.1 and 0.01, then we

would expect our answer to be between -1 and -2. (c) log (-100) no solution The calculator is showing an error because this question can be restated as 10010 −=x .

It is impossible to take 10 and raise it to an exponent such that a negative number is generated.

Example 5 Solve for x.

(a) 3641log −=

x

(b) x8log34=

Answers: With both questions start by changing from logarithmic to exponential form. (a)

4641

3641log

3

=

=

−=

−

x

x

x

(b)

162

8

8

log34

4

4

31

34

8

==

=

=

=

xx

x

x

x

In this section, we have been introduced to logarithms but still do not see how they can be used to solve exponential equations like 2747 −= x where the “common base” strategy does not work. We need to expand our knowledge of logarithms and complete another section of material before we can address this issue.

Questions:

1. Change the following from exponential to logarithmic form.

(a) 8134 = (b) 3464 =

(c) 2391 −= (d)

717 1 =−


(e) 63621

= (f) 927 32

=

(g) 23

481 −= (h)

4132 5

2

=−

2. Change the following from logarithmic to exponential form.

(a) 2144log12 = (b) 16log4 2=

(c) 4811log3 −=

(d)

=−

361log2 6

(e)

=−

31log

21

9 (f) 001.0log3 =−

(g) 23

1251log25 −=

(h) 410000log =

3. Evaluate each of the following logarithms. (a) 9log3 (b) 32log2

(c) 8log8 (d)

811log3

(e)

51log5 (f) ( )5log5

(g) 3.0log (h) ( )3

4 4log

(i)

51log25 (j) 427log


(k) 2log8 (l) ( )32log − 4. Evaluate each of the following logarithms.

(a)

91log27 (b)

2

1log16

5. Solve for x. (a) 38log =x (b) 2log5 −=x

(c) 9log81=x (d)

=−

491log2 x

(e) x2log4 =− (f) ( ) x=3

2 2log 6. Robert used his calculator to evaluate log 3460 and obtained 3.539. Explain why this is a

reasonable answer.


Solving Exponential Equations Using Logarithms In the last section we learned that logs “undo” the work done by exponents. We concluded that logarithms and exponents are inverses of each other. This inverse relationship helps explain the origin of the laws of logarithms. We will start by looking at the laws of exponents (in the left column) and see how they relate to the laws of logarithms (in the right column).

Laws of Exponents Laws of Logarithms When we multiply powers with the same base, the resulting power has the same base and the exponent is obtained by adding the original exponents.

cbcb aaa +=×

Examples: 734 xxx =× 1165 222 =×

When we add logarithms with the same base, the resulting logarithm has the same base and the argument is obtained by multiplying the original arguments.

( )cbcb aaa ×=+ logloglog

Examples: 27log3log9log 222 =+ 4.3log34log1.0log 555 =+

When we divide powers with the same base, the resulting power has the same base and the exponent is obtained by subtracting the original exponents.

cbcb aaa −=÷

Examples: 235 xxx =÷ 729 222 =÷

When we subtract logarithms with the same base, the resulting logarithm has the same base and the argument is obtained by dividing the original arguments.

=−

cbcb aaa logloglog

Examples: 7log3log21log 222 =− 5log9log45log 888 =−

When a power is being raised to an exponent, the resulting power has the same base and the exponent is obtained by multiplying the original exponents.

( ) cbcb aa ×=

Examples: ( ) 1234 xx =

( ) 2172 33 =

When a logarithm is multiplied by a value (i.e. constant or variable), the resulting logarithm has the same base and the argument is being raised to an exponent equal to that constant or variable.

caa bbc loglog =

Examples: 16log4log4log2 72

77 ==

2log8log8log31

531

55 ==

We will take a few minutes to check these laws of logarithms using our calculators. Remember that the log button on our calculators uses a base of 10. Fill in the blanks for the questions on the next page. The first question has been completed for us.


(a) 108log4log27log 101010 =+ (b) 47200log80log590log 101010 =+ 1.431 + 0.602 = 2.033 _______ + _______ = _______ 2.033 = 2.033 _______ = _______ (c) 8log6log48log 101010 =− (d) 14log50log700log 101010 =− _______ - _______ = _______ _______ - _______ = _______ _______ = _______ _______ = _______ (e) 49log7log2 1010 = (f) 8log2log3 1010 = 2× _______ = _______ 3× _______ = _______ _______ = _______ _______ = _______ Below we can see the formal proofs for our three laws of logarithms. We should take a few minutes to read these over but we should not feel that we need to memorize this work. (It will not be on a test.). Law #1: ( )cbcb aaa ×=+ logloglog

Let xba =log and yca =log Take both of these logarithmic expressions and change them into their exponential forms.

ba

xbx

a

=

=log

cayc

ya

=

=log

If we were asked to multiply b and c, we would obtain the following.

yx

yx

acbaacb

+=×

×=×

Now we will take the last exponential expression and change it to its logarithmic form. ( ) yxcba +=×log In our first step we stated that xba =log and yca =log . ( ) cbcb aaa logloglog +=× ← Our First Law of Logarithms

Law #2:

=−

cbcb aaa logloglog

Let xba =log and yca =log Take both of these logarithmic expressions and change them into their exponential forms.

ba

xbx

a

=

=log

cayc

ya

=

=log

If we were asked to take b and divide it by c, we would obtain the following.

yx

y

x

acb

aa

cb

−=

=


Now we will take the last exponential expression and change it to its logarithmic form.

yxcb

a −=

log

In our first step we stated that xba =log and yca =log .

cbcb

aaa logloglog −=

← Our Second Law of Logarithms

Law #3: c

aa bbc loglog = Let xba =log Take the logarithmic expression and change it into its exponential form.

ba

xbx

a

=

=log

If we were asked to take b and raise it the exponent c, we would obtain the following.

( )cxc

cxc

abab

=

=

Now we will take the last exponential expression and change it to its logarithmic form. cxbc

a =log In our first step we stated that xba =log . bcb a

ca loglog = ← Our Third Law of Logarithms

Example 1 Simplify the following logarithmic expressions. (a) 12log5log 22 + (b) 2log4 9 (c) 7log35log 33 − (d) 9log2log 54 + (e) 4log5log8log 666 −+ (f) 3log28log 55 +

Answers: (a)

( )60log

125log12log5log

2

2

22

×+

(b)

16log2log

2log4

9

49

9

(c)

5log735log

7log35log

3

3

33

−

(d) 9log2log 54 + This expression cannot be simplified because the logarithms do not have the same base.

(e)

10log4

58log

4log5log8log

6

6

666

×

−+

(f)

( )72log

98log9log8log3log8log

3log28log

5

5

55

255

55

×++

+


Example 2 Solve for x. (a) 3log4log6log 222 −+=x (b) x=− 54log3log2 66 (c) 14log6log12log =+− xxx (d) xx 3

43 log3log2 −=

Answers: (a)

382

8log3

46log

3log4log6log

2

2

222

==

=

×

=

−+=

x

x

x

x

x

(b)

1616

61log

549log

54log9log54log3log2

6

6

66

66

−=

=

=

=

=−=−

x

x

x

xx

x

(c)

88

18log

16

412log

14log6log12log

1

==

=

=

×

=+−

xx

x

x

xxx

(d)

93

log2

log2

loglog2

log3log2

23

3

4

3

33

43

34

3

==

=

=

−=

−=

xx

xxx

xxxx

Example 3

For

+−=

31log3log227log 333x , solve for x using two different methods.

Answers: Method 1: Use the Laws of Logarithms.

013

1log31

327log

31log3log227log

3

23

333

==

=

×=

+−=

x

x

x

x

x

Method 2: Evaluate each logarithm separately.

( ) ( )( ) ( )

01123

313 33 273

31log3log227log

113

333

=−+−=

===

+−=

−

xx

x


After all this, we still have not answered the question that led us to the topic of logarithms. “How do we solve exponential equations like 2747 −= x where the common base strategy does not work?” With these questions we will be using the log button on our calculators and the third law of logarithms. Example 4 Given the function xy 2= , find x when y equals 9.

Answer:

x

xy292

=

= Substitute 9 into the equation for the y. Since we are dealing with the bases 9 and 2, we cannot use the common base strategy.

x2log9log 1010 = We will log both sides of the equation. We chose logarithms with a base of 10 because our calculators work in this base.

2log9log 1010 x= We will use our third law of logarithms to change the exponent on the argument of our second logarithm to a coefficient of the logarithm.

2log9log

2log2log

2log9log

10

10

10

10

10

10

=

=

x

x

If we are trying to solve for x, we will have to divide both sides of the equation by 2log10 .

170.3301.0954.0

=

=

x

x Using our calculators, we evaluate the two logarithms and complete the question. Since 823 = and 1624 = , it seems reasonable that 92 170.3 = .

Example 5 Solve the following exponential equations.

(a) 167 =x (b) 3512x

= (c) ( ) 1839 12 =+x

Answers: (a)

425.1845.0204.1

7log16log

16log7log16log7log

167

=

=

=

==

=

x

x

x

x

x

x

(b)

632.4

544.13

5log12log

3

5log3

12log

5log12log

512

3

3

=

=

=

=

=

=

x

x

x

x

x

x

(c) ( )

( )

185.0369.02

631.0123log2log12

2log3log122log3log

231839

12

12

12

−=−==+

=+

=+=

=

=

+

+

+

xxx

x

x

x

x

x


Example 6

Given the function ( ) ( )234x

xh = , find: (a) ( )5h . (b) x when ( ) 21=xh .

Answers: (a) ( ) ( )

( ) ( )( )( ) 35.625

59.1545345

34

25

2

=×=

=

=

hhh

xhx

(b) ( ) ( )

( )

018.3

509.12

3log25.5log

2

3log2

25.5log

3log25.5log

325.5

3421

34

2

2

2

2

=

=

=

=

=

=

=

=

x

x

x

x

xh

x

x

x

x

Example 7 The value of a particular vehicle after its purchase depreciates with time. When this car was purchased, it was worth $32 000. The value of the car drops by 12% each year. When will this car be worth $20 000?

Answer: We need to find the equation of the exponential function that describes the value, v, of the car in terms of time, t. If the car drops by 12% each year, then it retains 88% of its value.

Equation: ( )tv 88.032000=

Now substitute 20 000 in for v, and solve for t.

( )( )

88.0log625.0log

88.0log625.0log88.0log625.0log

88.0625.088.03200020000

88.032000

=

==

=

=

=

t

t

v

t

t

t

t

7.3=t years


Example 8 Evaluate 7log3 .

Answer:

771.13log7log

7log3log7log3log

73

7log7log

3

3

=

=

==

=

=

x

x

x

x

x

x

771.17log3 =

We cannot use the log button on our calculator because the base is 3 (not 10). To evaluate this expression, set it equal to x, change from logarithmic to exponential form, and solve by using the procedure shown in the last few examples.

Questions:

1. Simplify each of the following logarithmic expressions.

(a) 2log8log 77 + (b) 5log2 4 (c) 5log10log 33 − (d) 5log8log 49 + (e) 2log3log10log 888 ++ (f) 3log3 7

(g) 3log6log5log 222 −+ (h) 25log21

(i)

+−

21log3log36log 444 (j)

+−+

31log4log6log8log 5555


(k) 25log10log2 1212 − (l) 4log2log3 + (m) 2log43log 66 + (n) 6log2144log 77 −

(o) 16log213log 22 + (p) 4log8log

3112log 999 −+

2. Solve for x.

(a) 24log9log =+ xx (b) 8log2log 44 +=x (c) 3log2log 4

34 =− xx (d) x=− 4log500log 55

(e) 34log2log8log =++ xxx (f) 2

88 loglog51 xx −=


(g) 40log2log3 55 −=x (h) 10log24log2 xx −=−

3. For 16log218log2log2 222 −+=x , solve for x using two different methods

4. Solve the following exponential equations. (a) 125 =x (b) x3121 =

(c) 9.04 =x (d) 262 3 =x


(e) 2576.1x

= (f) 5.4732 =x (g) 1249 −= x (h) ( ) 2157 24 =+x

(i) ( ) 3525 3 =x

(j) ( ) 49.0216 += x

(k) ( ) 85.020 4 =x

(l) ( ) 523106 += x


5. Solve the exponential equation 3432 −= x using: (a) the common base method (b) logarithms 6. Given ( ) ( ) xxf 354= , find: (a) x when ( ) 51=xf . (b) ( )8.0f 7. The population in a community has been growing annually by 20%. If the population was

initially 48 000, when will the population reach 60 000? 8. The initial concentration of a medication in a patient’s bloodstream was 0.15 mg/cm3. The

concentration dropped by 25% every three hours. When will the concentration reach 0.045 mg/cm3?


9. Evaluate each of the following logarithms. (a) 16log5 (b) 5.0log7 (c) 8.3log2 (d) 5.22log3 10. Evaluate 6log4log3log 777 −+ . If you feel that you need additional practice with these types of questions, you can complete the optional questions titled Additional Practice: Exponential and Logarithmic Equations found in the appendix of this unit.


Revisiting Application Problems Now that we know how to solve exponential equations using logarithms, we will look at a few more application problems. 1. Back in the 1960s Gordon Moore, a Caltech professor and later the cofounder of Intel,

predicted that with continued technological advances the number of transistors on a computer chip would double every 2 years.

(a) If in 1971 (t = 0), the Intel 4004 chip contained 2250 transistors, determine the equation of the exponential function that predicts the number, n, of transistors on a chip in terms of the time, t, in years since 1971.

(b) Based on your mathematical model from part (a), how many transistors should be present on a chip built in 2000.

(c) In 2000, the Pentium 4 chip was comprised of 42 million transistors? How far off was our mathematical model?

(d) Based on our model, when would one expect a chip to contain 200 000 transistors? 2. At sea level (elevation = 0 m), air pressure is 101.325 kilopascals (kPa). As we proceed to

higher elevations there is less air and therefore the air pressure drops. As we climb the air pressure decreases by 12% per 1000 m.

(a) Determine the equation that describes the air pressure, P, in kilopascals in terms of the elevation, E, in metres.

(b) What will the air pressure be 750 m above sea level? (c) At what elevation will the air pressure be 60 kPa? (d) How do we know that we are dealing with a decay curve based solely on the equation

you generated in part (a)?


3. In the previous question we learned that air pressure decreases with higher elevations. This decrease in air pressure also affects the takeoff distances of aircraft. At higher elevations, a plane requires a longer distance to takeoff. This relationship between elevation and takeoff distances can be described using an exponential function. The following data shows the takeoff distances for a particular aircraft for different elevations.

Elevation in metres 0 305 610 915 1220 Takeoff Distance in metres 204 224 245 269 295

(a) Determine the equation that describes the takeoff distance, D, in metres in terms of the elevation, E, in metres.

(b) The airport at the highest elevation in the world is located in Peru. It sits at an elevation of 4396 metres. What will be the takeoff distance for our particular aircraft at this elevation?

(c) The airport at the lowest elevation in the world is located in Death Valley, California. It sits at an elevation of -64 metres. What will be the takeoff distance for our particular aircraft at this elevation?

(d) Based your answers to parts (b) and (c), state the domain and range of our exponential function.

(e) If the takeoff distance is 255 m for this particular aircraft, at what elevation is it?


Putting It Together Complete the following review questions. 1. Simplify each of the following expressions.

(a) =× 243 47 dcdc (b) =÷ 2346 840 yxyx

(c) ( ) =523ba (d) =

3

4

52yx

(e) ( ) =×

22

3852

3321

abbaba (f) ( )

=× yxxy

yx33

256

84

2. Evaluate each of the following. Do not use a calculator.

(a) =09 (b) =−211

(c) =21

64 (d) =41

16

(e) =−19 (f) =

0

35

(g) =−34 (h) =

2

1

491

(i) =

−

0

72 (j) =

−2

47

(k) =

2

1

3625 (l) =

3

1

81

(m) =− 012 (n) =

−3

32


(o) =

−1

611 (p) ( ) =− 2

19

(q) =23

25

(r) =−

32

64

(s) =

−21

49100

(t) =

−23

259

3. Graph ( ) ( )141

23 +−= xy using transformations.


4. For each table, determine if it can be represented by an exponential function. If so, determine the equation of the exponential function.

(a)

x y -5 9 0 18 5 36 10 72 15 144

(b)

x y -0.2 486 -0.1 162

0 54 0.1 18 0.2 6

(c)

x y -9 3 -6 4 -3 9 0 18 3 31

(d)

x y -5 4 -4 6 -3 9 -2 13.5 -1 20.25

5. The initial concentration of a bacteria population was 150 per square centimetre. The

concentration doubled every thirty minutes. Determine the equation of the function that describes the bacteria concentration, C, in terms of time, t.

6. Jake purchased a rare toy for $90 on eBay. He predicts that it will increases in value by 30%

every two years. Determine the equation of the function that describes the predicted value, v, of the toy in terms of time, t.

7. Initially there was 18 milligrams of the radioactive isotope Radium-226. Over time this

radioactive isotope decays. It takes 1600 years for half of the Radium-226 to decay to a more stable element. Based on this we say that Radium-226 has a half-life of 1600 years. Determine the equation of the function that describes the amount, A, of Radium-226 remaining in milligrams in terms of the time, t, in years.


8. Determine the equation of the exponential function given its graph.

0123456789

1011121314151617181920212223242526272829

0 0.25 0.5 0.75 1

9. Solve the following exponential equations without using logarithms.

(a) 327 25125 −+ = xx (b) 824

931 +

+

=

x

x

(c) ( )x

x−

−

=

462

4917 (d) ( ) 54540 += x


10. Change from exponential to logarithmic form.

(a) 4917 2 =− (b) 3

1

1255 =

11. Change from logarithmic to exponential form.

(a) 324log8 = (b)

=−

641log3 4

12. Simplify each of these logarithmic expressions. (a) 27log9log2 55 − (b) 20log2log35log 777 −+ 13. Solve for x.

(a) x=125log5 (b) 4811log −=

x

(c) x4log2 =− (d) x=15log (e) 2log8log 44 +=x (f) 75 =x (g) 2100log4log −=− xx (h) 45log6=x


(i) xx 22 log2log6 += (j) 372log3log2 −=− xx (k) ( )x7224 = (l) 54log2log 33 −=x (m) 1.2log =x (n) x=3log8

(o) ( ) 325 4 =x

(p) x=+ 7log2log3 55


14. Given ( ) ( )27.04x

xh = , find: (a) ( )6.4h (b) x when ( ) 32=xh . 15. There were initially 25 insects of a particular type per square metre infesting Vivian’s lawn.

Every three weeks the concentration of these insects increased by a factor of 4. (a) Determine the equation that describes the insect concentration, C, in terms of time, t. (b) Based on our mathematical model, what will be the concentration of insects after 7

weeks? (c) When will the concentration reach 80 insects per square metre? (d) How do we know that we are dealing with a growth curve by just looking at the equation

of our exponential function?


16. Kadeer shocks his swimming pool with chlorine to kill the bacteria. After the water is shocked, the initial chlorine concentration is 2 ppm (parts per million), the highest level for swimmers to safely use the water. The chlorine then dissipates at a rate of 30% every two days.

(a) Determine the equation that describe the chlorine concentration, C, in terms of the time, t, from initial shocking.

(b) When will the chlorine concentration reach 1 ppm, the lowest level for swimmers to safely use the water?

(c) What will be the chlorine concentration after 1.5 days? (d) If we restrict the use of this mathematical model to chlorine concentrations which are

safe to swimmers, then state the domain and range. 17. The amount of water vapor, V, in the air with respect to the temperature, T, can be modeled

using an exponential function. At 0oC, the saturation level of water in the air is 4.9 ml/m3. The saturation level increases by approximately 6.2% for every one degree rise in air temperature.

(a) Determine the equation that describes the amount of water vapor in terms of temperature. (b) What is the amount of water vapor at 15oC? (c) At what temperature will the amount of water vapor be 5.3 ml/m3?


18. The growth of a particular mutual fund can be described using an exponential function. Initially $3000 was invested in the fund. One year later, the investment had grown to $3165.

(a) Determine the equation of the exponential function that describes the amount, A, of money in the fund with respect to time, t.

(b) How much would the investment be worth after 2.5 years? (c) When will the investment be worth $3618? (d) If this mathematical model only applies for 5 years, determine the range and domain for

our exponential function? 19. Dave took a desk job and within one year he was 20 pounds over his optimum weight. He

realized that he was consuming too many calories and not exercising enough. He decided to change some of his eating practices and attend the local gym with the objective of returning to his optimum weight. As stated he was initially 20 pounds over his optimum weight. After three weeks of exercise and new eating practices, he was only 15 pounds over his optimum weight. Assume that this weight decreases exponentially with time.

(a) Determine the equation that describes the number, N, of pounds over the optimum weight in terms of time, t.

(b) According to the mathematical model, when will Dave be 4 pounds over his optimum weight?

(c) According to this mathematical model, how many pounds will Dave be over his optimum weight in 5 weeks?

(d) How do we know that we are dealing with a decay curve by just looking at the equation of our exponential function?


Post-Unit Reflections What is the most valuable or important thing you learned in this unit?

What part did you find most interesting or enjoyable?

What was the most challenging part, and how did you respond to this challenge?

How did you feel about this math topic when you started this unit?

How do you feel about this math topic now?

Of the skills you used in this unit, which is your strongest skill?

What skill(s) do you feel you need to improve, and how will you improve them?

How does what you learned in this unit fit with your personal goals?


Additional Practice: Exponents, Part 1 Simplify each of the following expressions.

(a) =× 83 aa (b) =× 6243 yxyx (c) =×× 423 54 kkk (d) =× 42 32 ddc (e) =÷ 912 yy (f) =÷ 3247 baba (g) =÷ 26 927 pp (h) =÷ 269 742 xyyx (i) ( ) =

53k (j) ( ) =645ba

(k) ( ) =

342 p (l) ( ) =258 yx

(m) =

4

3

2

yx (n) =

2

4

32

75

cba

(The remaining questions will require two steps. Show your work.)

(o) =×

3

34

1065

pqqqp (p) ( )

=5

232

48

xyyx

(q) ( ) =×

22

64

336yx

xyyx (r) ( )=

× 32

253

4510

qppqp

(s) ( )( ) =32

253

24

xyyx (t) =

××

22

563

4368abba

baba


Additional Practice: Exponents, Part 2 1. Evaluate each of the following. Do not use a calculator.

(a) =21

64 (b) =023

(c) =−115 (d) =

2

1

964

(e) =

−2

75 (f) =

0

113

(g) =31

125 (h) =

−3

34

(i) ( ) =− 08 (j) =−27

(k) =51

32 (l) =

3

1

12527

(m) =− 06 (n) =

2

1

4936

2. Evaluate each of the following. Show all your work. Do not use a calculator.

(a) =52

32

(b) =−

21

49

(c) =−

34

8

(d) =

−23

94

(e) =

−32

6427


Additional Practice: Exponential and Logarithmic Equations Solve for x.

(a) 1306 52 =−x (b) 40log5log 22 −=x (c) 32log54log =− xx (d) 3loglog4 3

55 =− xx (e) 2log372log 33 −=x (f) 15log6=x (g) 4log4log6 22 =− xx (h) ( ) 12512 += x (i) x=8.0log2

(j) 5log12log 44 +=x (k) 24log6log2 =+ xx (l) ( ) 64.08 2 =x


Answers Interpreting Graphs (pages 1 to 3) 1. (a) 40 3g/cmµ (b) 7.5 3g/cmµ (c) 3 hours (d) 1 hour (e) 30 3g/cmµ (f) decay 2. (a) $15 000 (b) 20 years old (c) $2000 (d) growth (e) 8 years (f) $22 000 (g) 38 years old 3. (a) 0.3 2candela/cm (b) decay (c) 0.9 2candela/cm (d) 3 m (e) 4 m (f) { }60 ≤≤ dRdε

(g) { }9.023.0 ≤≤ LRLε Using the Graphing Calculator to Interpret Exponential Functions (pages 4 to 6) 1. (a) 85oC (b) decay (c) 14oC (d) 1 hour 2. (a) 100 cane toads (b) 64 500 cane toads (c) 18 months (d) growth (e) 7.5 months 3. (a) decay (b) $21 000 (c) 3 years (d) $9318 (e) $17850 (f) 15% 4. (a) $5000 (b) $93 395.93 (c) 18.68 times larger (d) $35 199.94 (e) 1968 Exponents, Part 1 (pages 7 to 11) (a) 7x (b) 85ba (c) 96x (d) 4624 dc (e) 8y (f) 23qp (g) h3 (h) yx211 (i) 18a (j) 1230dc


(k) 836k (l) 182427 ba

(m) 7

7

dc (n) 36

27

yx

(o) 8

1216yx (p) 6

82

49

cba

(q) 537 dc (r) 9435 qp

(s) 3

18

8xy (t) 2649 ba

(u) 822 yx (v) 8103 yx (w) 1145 qp (x) 43d (y) 784 ba (z) 859 yx Exponents, Part 2 (pages 12 to 20)

1. (a) 31 (b) 1

(c) 9 (d) 161

(e) 3 (f) 1

(g) 251 (h)

23

(i) 1 (j) 811

(k) 52 (l)

161

(m) 9

16 (n) 1−

(o) 2 (p) 21

(q) 43 (r)

278

(s) 23 (t)

3625

2. (a) 31

p (b) 2−b

(c) 21

h (d) 5−d


3. (a) 21 (b)

35

(c) 8 (d) 1681

(e) 51 (f) 4

(g) 47 (h)

827

(i) 641 (j) 16

(k) 94 (l)

271

4. x y

-2 811

23

− 271

-1 91

21

− 31

0 1

21 3

1 9

23 27

2 81


The Most Basic Exponential Functions (pages 21 to 24)

Part 1: Functions of the Form xby = where b > 1 xy 2= xy 3= x y x y

-3 81 -3

271

-2 41 -2

91

-1 21 -1

31

0 1 0 1

1 2 1 3

2 4 2 9

3 8 3 27 Will either of these two functions ever intersect the x-axis? No

Are these growth or decay curves? Growth Part 2: Functions of the Form xby = where 0 < b < 1

x

y

=

21

x

y

=

31

x y x y -3 8 -3 27

-2 4 -2 9

-1 2 -1 3

0 1 0 1

1 21 1

31

2 41 2

91

3 81 3

271

Will either of these two functions ever intersect the x-axis? No

Are these growth or decay curves? Decay


1. (b) Horizontal Asymptote at y = 0 2. Domain: { }Rxε 3. Range: { }0 >yRyε 4. Shared Point: (0, 1) 5. If the base of the exponential function is greater than 1 (b > 1), then the resulting graph is a

growth curve. If the base of the exponential function is between 0 and 1 (0 < b < 1), then the resulting graph is a decay curve.

6. (a) All the ratios equal 2. (b) Yes

7. All the ratios equal 31 .

8. (b) 9. Statement (a): Linear Statement (c): Quadratic Exponential Functions and Transformations (pages 25 to 29)


Table of Values


(a) xy −= 2 Enter: Y1=2^(-X)

TblStart = 2 ∆Tbl = -1

x y 2 0.25 1 0.5 0 1 -1 2 -2 4

( ) ( )yxyx ,, −→

We’re dealing with a reflection in the y-axis. This caused our curve to change from a growth curve to a decay curve.


(b) ( )xy 22= Enter: Y1=2(2^X)


x y -2 0.5 -1 1 0 2 1 4 2 8

( ) ( )yxyx 2 ,, →

We are dealing with a vertical stretch of 2.

(c) ( )xy 221

=

Enter: Y1=(1/2)(2^X)


x y -2 0.125 -1 0.25 0 0.5 1 1 2 2

( )

→ yxyx

21 ,,

We are dealing with a

vertical stretch of 21 .

(d) x

y 21

2= Enter: Y1=2^(X/2)


x y -4 0.25 -2 0.5 0 1 2 2 4 4

( ) ( )yxyx ,2, →

We are dealing with a horizontal stretch of 2.

(e) xy 22= Enter: Y1=2^(2X)

TblStart = -1 ∆Tbl = 0.5

x y -1 0.25

-0.5 0.5 0 1

0.5 2 1 4

( )

→ yxyx ,

21,

We are dealing with a horizontal stretch of

21 .


(f) 12 += xy Enter: Y1=2^(X+1)


x y -3 0.25 -2 0.5 -1 1 0 2 1 4

( ) ( )yxyx ,1, −→

We are dealing with a horizontal translation of -1.

(g) 12 −= xy Enter: Y1=2^(X-1)


x y -1 0.25 0 0.5 1 1 2 2 3 4

( ) ( )yxyx ,1, +→

We are dealing with a horizontal translation of 1.

Summarize Your Findings:

Equation Transformation and Affect (a) xy −= 2 We’re dealing with a reflection in the y-axis. This caused our

curve to change from a growth curve to a decay curve.

(b) ( )xy 22= We are dealing with a vertical stretch of 2.

(c) ( )xy 221

= We are dealing with a vertical stretch of21 .

(d) xy 2

1

2= We are dealing with a horizontal stretch of 2.

(e) xy 22= We are dealing with a horizontal stretch of 21 .

(f) 12 += xy We are dealing with a horizontal translation of -1.

(g) 12 −= xy

We are dealing with a horizontal translation of 1.

Conclusions:

If an exponential function is of the form ( ) ( )dxcbay −−=1

, then:

(i) the a indicates that a vertical stretch of a has occurred.


(ii) the c indicates that a horizontal stretch of c has occurred.

(iii) the d indicates that a horizontal translation of d has occurred.

(iv) a negative sign in front of the c1 indicates that a reflection in the y-axis has occurred.

State the Transformations: Exponential Functions (page 30) Function Transformed

Function Vertical Stretch

Reflection in y-axis

Horizontal Stretch

Horizontal Translation

1. (a) ( ) 152 −= xy xy 5= 2 none none 1

(b) ( ) xy −= 36 xy 3= 6 yes none none

(c) ( )524 += xy xy 4= none none 21 -5

(d) ( ) xy 41

23 −= xy 2= 3 yes 4 none

(e) ( ) ( )231

65.1 −= xy xy 6= 1.5 none 3 2

(f) ( ) ( )6247.0 +−= xy xy 4= 0.7 yes 21 -6

(g) ( )361

7−−

=x

y xy 7= none yes 6 3

2. (a) growth (b) decay (c) decay (d) growth (e) decay (f) growth


Graphing Using Transformations (pages 31 to 37)

1. ( ) ( )131

25 −−= xy

- Reflection in the y-axis - Vertical Stretch of 5 - Horizontal Stretch of 3 - Horizontal Translation of 1

Mapping Rule ( ) ( )yxyx 5 ,13, +−→

x y x y

-2 41 ( ) =+−− 123 7

411

=

415

-1 21 ( ) =+−− 113 4

212

=

215

0 1 1 5

1 2 -2 10

2 4 -5 20

2. ( ) ( )234 +−= xy

- Reflection in the y-axis - Vertical Stretch of 4 - Horizontal Translation of -2

Mapping Rule ( ) ( )yxyx 4,2, −−→

x y x y

-2 91 ( ) =−−− 22 0

94

=

914

-1 31 -1

311

0 1 -2 4

1 3 -3 12

2 9 -4 36


3. (a) ( ) ( )yxyx 2,1, −→ (b) ( ) ( )yxyx 6,32, −→

x y x y

-3 92 -7

211

-2

32 -5 3

-1 2 -3 6

0 6 -1 12

1 18 1 24

(c) ( ) ( )yxyx ,24, −−→ (d) ( )

+−→ yxyx

21,1,

x y x y

6 91 3

81

2 31 2

41

-2 1 1 21

-6 3 0 1

-10 9 -1 2

(e) ( )

−→ yxyx 3,

21,

x y

1 163

21

43

0 3

21

− 12

-1 48


4. (a) ( ) ( )yxyx 7,6, +→ (b) ( ) ( )yxyx 8,5, −−→

(c) ( )

−→ yxyx ,3

21, (d) ( ) ( )yxyx 45.0,7, −→

(e) ( )

+−→ yxyx

101,4

31,

5. ( )

+→ yxyx

21,53,

x y

-1 321

2 81

5 21

8 2

11 8

6. y = 0 Recognizing Patterns within Tables of Values (pages 40 to 41) 1. Exponential: common ratio of 2 2. Quadratic: common difference of 6 at the D2 level 3. Linear: common difference of -3 4. Linear: common difference of 0.25 5. Quadratic: common difference of 8 at the D2 level 6. Exponential: common ratio of 0.5 7. Not Linear, Quadratic or Exponential 8. Exponential: common ratio of 10 9. Exponential: common ratio of 1.1


10. Quadratic: common difference of -2 at the D2 level 11. Linear: common difference of 0.3 12. Not Linear, Quadratic or Exponential (Check the x-values)

13. Exponential: common ratio of 31

14. Quadratic: common difference of 4 at the D2 level 15. Exponential: common ratio of 0.8 Investigation: From a Table of Values to an Equation (page 42)

1. All references to a, b, and c are in relation to the equation cx

aby = . • The y-intercept, which occurs when x equals zero, is equal to a in the equation. For

example, the function ( )726x

y = has a y-intercept 6 as designated by the point (0,6). We previously learned that a can also be called the vertical stretch. In the case of the first function, the vertical stretch would be 6. That would explain how the traditional y-intercept or focal point would change from (0, 1) to (0, 6).

• The incremental changes of x-values or common differences between the successive x-values give us the value of c in the equation. For example, the incremental changes of x-

values for the function ( )726x

y = were 7. We previously learned that when an equation

is written in the form x

caby1

= or cx

aby = , the c represents the horizontal stretch. A horizontal stretch would account for changes to the x-values. Consider the first function. The x-values were changing by increments of 7. When we first looked at equations of the form, xby = , the x-values changed by 1. A vertical stretch of 7 would explain how the incremental changes in x-values went from 1 to 7.

• The common ratio between successive y-values is equal to b, the base of the exponential

function. For example, the function ( )726x

y = had a common ratio of 2 between successive y-values. We are not dealing with a transformation because this is merely base of the function.


• To recap:

2. Equation: ( )5321x

y = Reason: y-intercept or a is equal to 21 common ratio between successive y-values or b is equal to 3 the x-values are changing by increments of 5, therefore c is equal to 5 Determining the Equation Given a Table of Values (pages 43 to 46)

1. ( )95150x

y = 2. ( )6412x

y =

3. ( )35.015x

y = 4. ( )xy 25.05.12=

5. ( )55.1180x

y = 6. not exponential

7. ( )32160x

y = 8. ( )41.09.0x

y =

9. ( )74.03125x

y = 10. not exponential

11. ( )225.11024x

y = 12. ( ) xy 236=

13. not exponential 14. x

y4

3154

=

15. ( ) xy 10240= 16. x

y5

3115

=

cx

aby =

common ratio or base (not a transformation)

y-intercept or vertical stretch

incremental changes of x or horizontal stretch


More than One Acceptable Answer (pages 47 to 48)

1. ( )xy 32= , ( )292x

y = , ( )3272x

y = , ( )4812x

y = ,…

2. ( )xy 5.05120= , ( )225.05120x

y = , ( )3125.05120x

y = , ( )40625.05120x

y = ,…

3. ( )31008.0x

y = , ( )610008.0x

y = , ( )9100008.0x

y = , ( )121000008.0x

y = Determining the Equation Based on the Real World Situation (pages 49 to 57)

1. (a) ( )437x

y = (b) ( )623x

y =

(c) 9

2115

x

y

= (d) ( )126.05

xy =

(e) ( )4535.08x

y = (f) ( ) xy 293.075.2=

(g) ( )xy 10.1980= (h) ( )1826.147x

y =

(i) ( )435x

y = (j) 8

2114

x

y

=

(k) ( )925.18x

y = (l) ( ) xy 48.07.6=

2. (a) ( )23800t

n = (b) 9476 yeast cells (c) Domain: { }120 ≤≤ tRtε Range: { }200 583800 ≤≤ nRnε

3. (a) ( )292.0100x

P = (b) Decay Curve (c) 54% (d) Domain: { }250 ≤≤ tRtε Range: { }10035 ≤≤ nRnε

4. (a) ( )325.180000t

v = (b) The base is greater than 1 and there is no reflection in the y-axis. (c) $134 652.17 (d) Domain: { }100 ≤≤ tRtε Range: { }21.16831580000 ≤≤ vRvε

5. (a) 5730

2115

t

A

=


(b) The base is between 0 and 1 and there is no reflection in the y-axis. (c) 3.51 mg

6. (a) ( )3045t

C = (b) 508 bacteria per square centimetre (c) Growth Curve (d) Domain: { }2400 ≤≤ tRtε Range: { }3276005 ≤≤ CRCε 7. ( )tA 065.16000= $9323.92

8. ( )27.008.0t

C = 0.033 mg/cm3

9. ( )33.04000t

P = 161 frogs 10. ( )tP 15.154000= 101 282 people

Determining the Equation Given a Graph (pages 58 to 59)

1. ( )35.0100x

y = 2. ( )1038.1x

y =

3. ( )625.02x

y = 4. ( )525.1x

y =

5. x

y2

3290

= 6. ( ) xy 105.112=

Solving Exponential Equations by Creating the Same Base (pages 60 to 65) 1. (a) 42 (b) 32−

(c) 21

2 (d) 31

2 (e) 22− (f) 52− 2. (a) 33 (b) 23−

(c) 21

3 (d) 43−

(e) 32

3 (f) 21

3−


(g) 23

3 (h) 43

3 3. (a)

( ) ( )

2147

2126212622

5555

125251

12622

42312

421

=−=−

−−=−−−=+−∴

=

=

=

−+−

−−−

−−

xx

xxxx

xx

xx

xx

(b) ( )

( )

1818

3153231532

33

33

2713

31532

53

64

21

564

=−=−

−−=−+−=+∴

=

=

=

+−+

−−

+

−+

xx

xxxx

xx

xx

xx

4. Hint: Answer: (a) None Supplied 2 (b) None Supplied -6 (c) ( ) ( ) 42342 33 −+

=xx 5

(d) ( ) 531 22 −− =xx 7

(e) ( ) 623 66 ++− =xx -3

(f) 174 22 −+ =x -2 (g) ( ) ( ) 8322 22 −+− =

xx 4 (h) ( ) ( ) 131322 77 +−+

=xx -1

(i) 2

26

21

66 =

−x

1

(j) ( ) x

x262

84

21

55 −−

+

=

8

(k) ( )

xx

36

31

72 1010−

+

= -4

(l) ( ) x

x−−

+

=

191

36

32

33 -7

5. (a) 2 (b) 729

6. (a) 40 (b) 37

−


7. 3.1

2172

t

A

=

(a) 55.15 mg (b) 3.9 billion years An Introduction to Logarithms (pages 66 to 71) 1. (a) 481log3 = (b) 364log4 =

(c) 291log3 −=

(d) 1

71log7 −=

(e) 216log36 = (f)

329log27 =

(g) 23

81log4 −=

(h)

52

41log32 −=

2. (a) 144122 = (b) 1624 =

(c) 8113 4 =− (d)

3616 2 =−

(e) 319 2

1

=−

(f) 001.010 3 =−

(g) 125

125 23

=−

(h) 10000104 =

3. (a) 2 (b) 5 (c) 1 (d) -4

(e) -1 (f) 21

(g) -0.523 (h) 31

(i) 21

− (j) 2.630

(k) 31 (l) no solution

4. (a) 32

− (b) 81

−

5. (a) 2 (b) 251

(c) 21 (d) 7


(e) 161 (f)

31

6. log 3460 = 3.539 This could be restated as 346010 539.3 = . The answer 3.539 seems reasonable because

1000103 = and 10000104 = . Since 3460 is between 1000 and 10000, then we would expect our answer to be between 3 and 4.

Solving Exponential Equations Using Logarithms (pages 72 to 83) 1. (a) 16log7 (b) 25log4 (c) 2log3 (d) Cannot be simplified. (e) 60log8 (f) 27log7 (g) 10log2 (h) 5log (i) 6log4 (j) 4log5 (k) 4log12 (l) 32log (m) 48log6 (n) 4log7 (o) 12log2 (p) 6log9 2. (a) 6 (b) 2 (c) 64 (d) 3 (e) 4 (f) 2 (g) -1 (h) 5

3. Hints:

×=

21

2

2

16

82logx ( ) ( )421312 −+=x

Answer: 3=x 4. (a) 1.544 (b) 4.365 (c) -0.076 (d) 14.101 (e) 0.702 (f) 1.757 (g) 1.292 (h) -0.329 (i) 8.422 (j) -23.736 (k) 5.288 (l) -2.732

5. Hints: (a) ( ) 325 22 −=

x (b) 4log

32log3 =−x

Answer: x =2

11 or 5.5


6. (a) 0.527 (b) 190.365 7. Hint: ( )tP 2.148000= Answer: 1.2 years

8. Hint: ( )375.015.0t

C = Answer: 12.6 hours 9. (a) 1.723 (b) -0.356 (c) 1.926 (d) 2.834 10. Hint: Simplify using the laws of logarithms and then use the same technique that was used

in the previous question. Answer: 0.356 Revisiting Application Problems (pages 84 to 85)

1. (a) ( )222250t

n = (b) 52.1 million (rounded off) (c) 10.1 million (d) almost 13 years later (1984)

2. (a) ( )100088.0325.101E

P = (b) 92.1 kPa (c) 4099 m (d) Base is between 0 and 1, and no reflection in the p-axis

3. (a) ( )30510.1204E

D = (b) 806 m (c) 200 m (d) Domain: { }439664 ≤≤− EREε Range: { }806200 ≤≤ DRDε (e) 714 m Putting It Together (pages 86 to 94) 1. (a) 3728 dc (b) 235 yx

(c) 1015ba (d) 12

158yx

(e) 487 ba (f) 682 yx


2. (a) 1 (b) 1211

(c) 8 (d) 2

(e) 91 (f) 1

(g) 641 (h)

71

(i) 1 (j) 4916

(k) 65 (l)

21

(m) -1 (n) 827

(o) 116 (p) no solution

(q) 125 (r) 161

(s) 107 (t)

27125

3. ( ) ( )yxyx 3 ,14, −−→ x y

7 43

3

211

-1 3

-5 6

-9 12

4. (a) ( )5218x

y = (b) x

y10

3154

=

(c) not exponential (d) ( )xy 5.1375.30=

5. ( )302150t

C =


6. ( )23.190t

v =

7. 1600

2118

t

A

=

8. x

y4

3227

=

9. (a) 27 (b) -4

(c) 5 (d) 27

−

10. (a) 2491log7 −=

(b)

315log125 =

11. (a) 4832

= (b) 6414 3 =−

12. (a) 3log5 (b) 2log7 13. Hint: Answer: (a) Change to its exponential form. 3

(b) Change to its exponential form. 3

(c) Change to its exponential form. 161

(d) Use the log button on your calculator. 1.176

(e) 16log4=x 2

(f) Log (base 10) both sides of the equation. 1.209

(g) 2251log −=

x 5

(h) Change from logarithmic to exponential form and then log (base 10) both sides of the equation. 2.125

(i) 32log6 x= 4

(j) 381log −=

x 2

(k) Divide both sides of the equation by 2 and then log (base 10) both sides of the equation. 1.277


13. Hint: Answer:

(l)

=

271log3x -3

(m) Change from logarithmic to exponential and then use your calculator. 125.9

(n) Change from logarithmic to exponential form and then log (base 10) both sides of the equation. 0.528

(o) Divide both sides of the equation by 5 and then log (base 10) both sides of the equation. -2.948

(p) Simplify using the laws of logarithms, change from logarithmic to exponential form, and then log (base 10) both sides of the equation.

2.501

14. (a) 1.761 (b) -11.660

15. (a) ( )3425t

C = (b) 635 insects per square metre (c) 2.5 weeks (d) The base is greater than 1, and there is no reflection in the y-axis.

16. (a) ( )27.02t

C = (b) 3.9 days (c) 1.53 ppm (d) Domain: { }9.30 ≤≤ tRtε Range: { }21 ≤≤ CRCε 17. (a) ( )TV 062.19.4= (b) 12.1 ml/m3 (c) 1.3oC 18. (a) ( )tA 055.13000= (b) $3429.67 (c) 3.5 years (d) Domain: { }50 ≤≤ tRtε Range: { }88.39200 300 ≤≤ ARAε

19. (a) ( )375.020t

N = (b) 16.8 weeks (c) 12.4 pounds (d) The base is between 0 and 1, and there is no reflection in the y-axis.


Additional Practice: Exponents, Part 1 (page 96)

(a) 11a (b) 105 yx (c) 920k (d) 526 dc (e) 3y (f) ba5 (g) 43p (h) 486 yx (i) 15k (j) 2430ba (k) 128p (l) 21064 yx

(m) 12

8

yx (n) 8

64

4925

cba

(o) qp33 (p) yx316 (q) 52xy (r) 735 qp (s) 432 yx (t) 454 ba Additional Practice: Exponents, Part 2 (page 97) 1. (a) 8 (b) 1

(c) 151 (d)

38

(e) 2549 (f) 1

(g) 5 (h) 6427

(i) 1 (j) 491

(k) 2 (l) 53

(m) -1 (n) 76

2. (a) 4 (b) 71

(c) 161 (d)

827

(e) 9

16


Additional Practice: Exponential and Logarithmic Equations (page 98) (a) 3.858 (b) -3 (c) 3 (d) 125 (e) 2 (f) 1.511 (g) 4 (h) 0.263 (i) -0.322 (j) 2.953 (k) 12 (l) 0.628

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