4.3 - 1 copyright © 2013, 2009, 2005 pearson education, inc. 1 4 inverse, exponential, and...

4.3 - 1Copyright © 2013, 2009, 2005 Pearson Education, Inc. 1

4Inverse, Exponential, and Logarithmic Functions

Copyright © 2013, 2009, 2005 Pearson Education, Inc.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 2

4.3 Logarithmic Functions

• Logarithms• Logarithmic Equations• Logarithmic Functions• Properties of Logarithms


Logarithms

The previous section dealt with exponential functions of the form y = ax for all positive values of a, where a ≠ 1. The horizontal line test shows that exponential functions are one-to-one, and thus have inverse functions.


Logarithms

The equation defining the inverse of a function is found by interchanging x and y in the equation that defines the function. Starting with y = ax and interchanging x and y yields

.yx a


Logarithms

yx a

Here y is the exponent to which a must be raised in order to obtain x. We call this exponent a logarithm, symbolized by the abbreviation “log.” The expression logax represents the logarithm in this discussion. The number a is called the base of the logarithm, and x is called the argument of the expression. It is read “logarithm with base a of x,” or “logarithm of x with base a.”


Logarithm

For all real numbers y and all positive numbers a and x, where a ≠ 1,

if and only if

The expression logax represents the exponent to which the base a must be raised in order to obtain x.

logay x .yx a


Example 1 WRITING EQUIVALENT LOGARITHMIC AND EXPONENTIAL FORMS

The table shows several pairs of equivalent statements, written in both logarithmic and exponential forms.


Logarithmic form: y = loga x

Exponent

Base

Exponential form: ay = x

Exponent

Base

Example 1 WRITING EQUIVALENT LOGARITHMIC AND EXPONENTIAL FORMS

To remember the relationships among a, x, and y in the two equivalent forms y = logax and x = ay, refer to these diagrams.


Example 2 SOLVING LOGARITHMIC EQUATIONS

Solve each equation.

Solution

(a)8

log 327x

8log 3

27x

Write in exponential form.

3 827

x

33 2

3x

38 227 3

23

x Take cube roots



Check8

log 327x

Let x = 2/3.


Original equation

2

?

3

8log 3

27

3 ?2 83 27

8 827 27

True

The solution set is 2.

3




Solution

(b)

4

5log

2x


254 x2 51(4 ) x ( )mn m na a

52 x

4

5log

2x

1 2 2 1 24 (2 ) 2

32 xThe solution set is {32}.

Apply the exponent.




Solution

(c)

Write in exponential form.349 7x

1 32(7 ) 7x 2 1 37 7x

349log 7 x

12

3x

The solution set is

Write with the same base.

Power rule for exponents.

Set exponents equal.

16

x Divide by 2.

1.

6


Logarithmic Function

If a > 0, a ≠ 1, and x > 0, then

defines the logarithmic function with base a.

( ) logax xf


Logarithmic Functions

Exponential and logarithmic functions are inverses of each other. The graph of y = 2x is shown in red. The graph of its inverse is found by reflecting the graph of y = 2x across the line y = x.



The graph of the inverse function, defined by y = log2 x, shown in blue, has the y-axis as a vertical asymptote.



Since the domain of an exponential function is the set of all real numbers, the range of a logarithmic function also will be the set of all real numbers. In thesame way, both the range of an exponential function and the domain of a logarithmic function are the set of all positive real numbers. Thus, logarithms can be found for positive numbers only.


Domain: (0, ) Range: (– , )

LOGARITHMIC FUNCTION ( ) logax xf

x (x)

¼ – 2½ – 11 0

2 1

4 2

8 3 (x) = loga x, for a > 1, is

increasing and continuous on its entire domain, (0, ) .

For (x) = log2 x:




x (x)

¼ – 2½ – 11 0

2 1

4 2

8 3The y-axis is a vertical asymptote

as x 0 from the right.

For (x) = log2 x:




x (x)

¼ – 2½ – 11 0

2 1

4 2

8 3 The graph passes through the points

For (x) = log2 x:

1, 1 , 1,0 , and ,1 .a

a




x (x)

¼ 2½ 11 0

2 – 14 – 28 – 3

(x) = loga x, for 0 < a < 1, is decreasing and continuous on its entire domain, (0, ) .

For (x) = log1/2 x:




x (x)

¼ 2½ 11 0

2 – 14 – 28 – 3

For (x) = log1/2 x:

The y-axis is a vertical asymptote as x 0 from the right.




x (x)

¼ 2½ 11 0

2 – 14 – 28 – 3

For (x) = log1/2 x:

The graph passes through the points 1

, 1 , 1,0 , and ,1 .aa


Characteristics of the Graph of ( ) logax xf

1. The points are on the graph.

2. If a > 1, then is an increasing function. If 0 < a < 1, then is a decreasing function.

3. The y-axis is a vertical asymptote.4. The domain is (0,), and the range is (– , ).

1, 1 , 1,0 , and ,1a

a


Example 3 GRAPHING LOGARITHMIC FUNCTIONS

Graph each function.

Solution

(a) 1 2( ) logx xf

First graph y = (½)x which defines the inverse function of , by plotting points. The graph of (x) = log1/2x is the reflection of the graph ofy = (½)x across the line y = x. The ordered pairs for y = log1/2x are found by interchanging the x- and y-values in the ordered pairs for y = (½)x .



Solution

1 2( ) logx xfGraph each function.

(a)



Solution

(b) 3( ) logx xf

Another way to graph a logarithmic function is to write (x) = y = log3 x in exponential form as x = 3y, and then select y-values and calculate corresponding x-values.




Solution


(b) 3( ) logx xf


Caution If you write a logarithmic function in exponential form to graph, as in Example 3(b), start first with y-values to calculate corresponding x-values. Be careful to write the values in the ordered pairs in the correct order.


Example 4 GRAPHING TRANSLATED LOGARITHMIC FUNCTIONS

Graph each function. Give the domain and range.

Solution

(a) 2( ) log ( 1)x x f

The graph of (x) = log2(x – 1) is the graph of (x) = log2 x translated 1 unit to the right. The vertical asymptote has equation x = 1. Since logarithms can be found only for positive numbers, we solve x – 1 > 0 to find the domain, (1,) . To determine ordered pairs to plot, use the equivalent exponential form of the equation y = log2 (x – 1).




Solution

2( ) log ( 1)x x f

2log ( 1)y x

1 2yx Write in exponential form.

2 1yx Add 1.

(a)




Solution

2( ) log ( 1)x x f

We first choose values for y and then calculate each of the corresponding x-values. The range is (– , ).

(a)




Solution

(b) 3( ) (log ) 1x x f

The function (x) = (log3 x) – 1 has the same graph as g(x) = log3 x translated 1 unit down. We find ordered pairs to plot by writing the equation y = (log3 x) – 1 in exponential form.




Solution

3( ) log ( 1)x x f

3(log ) 1y x

31 logy x Add 1.

13yx Write in exponential form.

(b)




Solution

3( ) log ( 1)x x f

Again, choose y-values and calculate the corresponding x-values. The domain is (0, ) and the range is (– , ).

(b)




Solution

(c) 4( ) log ( 2) 1x x f

The graph of f(x) = log4(x + 2) + 1 is obtained by shifting the graph of y = log4x to the left 2 units and up 1 unit. The domain is found by solving x + 2 > 0, which yields (–2, ). The vertical asymptote has been shifted to the left 2 units as well, and it has equation x = –2. The range is unaffected by the vertical shift and remains (– , ).




Solution

(c) 4( ) log ( 2) 1x x f


Properties of Logarithms

The properties of logarithms enable us to change the form of logarithmic statements so that products can be converted to sums, quotients can be converted to differences, and powers can be converted to products.



For x > 0, y > 0, a > 0, a ≠ 1, and any real number r, the following properties hold.

DescriptionThe logarithm of the product of two numbers is equal to the sum of the logarithms of the numbers.

Property

Product Property

log log loga a axy x y




Quotient Property

log log loga a a

xx y

y

DescriptionThe logarithm of the quotient of two numbers is equal to the difference between the logarithms of the numbers.

Property



Power Property

log logra ax r x

DescriptionThe logarithm of a number raised to a power is equal to the exponent multiplied by the logarithm of the number.


Property



Logarithm of 1log 1 0a

DescriptionThe base a logarithm of 1 is 0.

The base a logarithm of a is 1.


Property

Base a Logarithm of alog 1aa


Example 5 USING THE PROPERTIES OF LOGARITHMS

Rewrite each expression. Assume all variables represent positive real numbers, with a ≠ 1 and b ≠ 1.

Solution

(a) 6log (7 9)

6 6 6log ( ) l7 7og9 9log Product property




Solution

(b) 9

15log

7

9 9 9log log15

15 77

log Quotient property




Solution

(c) 5log 8

51 2

5 5log 8 log (8 ) l12

og 8 Power property




Solution

(d) 2 4loga

mnqp t

2 42 4log log log log log log( )a a a a a a

mnqm n q p t

p t

log log log (2 log 4 log )a a a a am n q p t log log log 2 log 4 loga a a a am n q p t

Use parentheses to avoid errors.

Be careful with signs.




Solution

(e) 3 2loga m

3 2 2 3log l2

og log3a a am m m




Power property

3 51logb m

x yn z

3 51log log log m

b b bx y zn

Product and quotient properties

13 5 3 5

log logn

nb bm m

x y x yz z

1 nn a a

3 5

log nb m

x yz

(f)

Solution




Solution

(f)3 5

log nb m

x yz

Power property 13 log 5 log logb b bx y m z

n

Distributive property3 5

log log logb b b

mx y z

n n n



Write each expression as a single logarithm with coefficient 1. Assume all variables represent positive real numbers, with a ≠ 1 and b ≠ 1.

Solution

(a) 3 3 3log ( 2) log log 2x x

3 3 3 3log log log log2)

( 2)2

( 2x xx x





Solution

(b) 2 log 3 loga am n

32log log l3 og o2 l ga a a an m nm Power property

2

3loga

mn

Quotient property




Solution

(c) 21 3log log 2 log

2 2b b bm n m n

21 3log log 2 log

2 2b b bm n m n

Power property1 22 3 2( )log log 2 logb b bm n m n




Solution

21 3log log 2 log

2 2b b bm n m n


1 2 3 2

2

(2 )logb

m nm n

3 2 1 2

3 2

2logb

nm

Rules for exponents

(c)




Solution

21 3log log 2 log

2 2b b bm n m n

1 23

3

2logb

nm

Rules for exponents

3

8logb

nm

Definition of a1/n

(c)


Caution There is no property of logarithms to rewrite a logarithm of a sum or difference. That is why, in Example 6(a), log3(x + 2) was not written as log3 x + log3 2. The distributive property does not apply in a situation like this because log3 (x + y) is one term. The abbreviation “log” is a function name, not a factor.


Example 7USING THE PROPERTIES OF LOGARITHMS WITH NUMERICAL VALUES

Assume that log10 2 = 0.3010. Find each logarithm.

Solution

(a) 10log 4

102 log 2

102

10log log4 2

2(0.3010)

0.6020


Example 7USING THE PROPERTIES OF LOGARITHMS WITH NUMERICAL VALUES

Assume that log10 2 = 0.3010. Find each logarithm.

Solution

(b) 10log 5

10 10log log10

52

0.6990

0. 01 301

110 0log 10 log 2


Theorem on Inverses

For a > 0, a ≠ 1, the following properties hold.

log (for 0) and loga x xaa x x a x


Theorem on Inverses

The following are examples of applications ofthis theorem.

The second statement in the theorem will be useful in Sections 4.5 and 4.6 when we solve other logarithmic and exponential equations.

7 10log7 ,10 535log ,3 and 1log 1r

kr k

4.3 - 1 copyright © 2013, 2009, 2005 pearson education, inc. 1 4 inverse, exponential, and...

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