chapter four of maths in focus

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Exponential equation: Equation where the pronumeral isthe index or exponent such as 3 9

x

=

Exponential function: A function in the form y a x

= wherethe variable x is a power or exponent

Logarithm: A logarithm is an index. The logarithm is thepower or exponent of a number to a certain basei.e. 2 8

x

= is the same as log 8 x 2

=

TERMINOLOGY

Exponentialand Logarithmic

Functions

4



161Chapter 4 Exponential and Logarithmic Functions

DID YOU KNOW?

John Napier (1550–1617), a Scottish theologian and an amateur mathematician, was the first toinvent logarithms. These ‘natural’, or ‘Naperian’, logarithms were based on ‘e’. Napier was also

one of the first mathematicians to use decimals rather than fractions. He invented the notation of

the decimal, using either a comma or a point. The point was used in England, but a few European

countries still use a comma.

Henry Briggs (1561–1630), an Englishman who was a professor at Oxford, decided that

logarithms would be more useful if they were based on 10 (our decimal system). These are called

common logarithms. Briggs painstakingly produced a table of logarithms correct to 14 decimal

places. He also produced sine tables—to 15 decimal places—and tangent tables—to 10 decimal

places.

The work on logarithms was greatly appreciated by Kepler, Galileo and other astronomers

at the time, since they allowed the computation of very large numbers.

INTRODUCTION

THIS CHAPTER INTRODUCES A new irrational number, ‘e’, that has

special properties in calculus. You will learn how to differentiate

and integrate the exponential function.

( ) f x e x

=

The definition and laws of logarithms are also introduced in

this chapter, as well as differentiation and integration involving

logarithms.

Differentiation of Exponential Functions

When differentiating exponential functions ( )f x ax= from first principles, an

interesting result can be seen. The derivative of any exponential function gives

a constant which is multiplied by the original function.

EXAMPLE

Sketch the derivative (gradient) function of y 10x= .

Solution

CONTINUED



162 Maths In Focus Mathematics Extension 1 HSC Course

The graph of y 10x= always has a positive gradient that is becoming steeper.

So the derivative function will always be positive, becoming steeper.

The derivative function of an exponential function will always have a

shape similar to the original function.

We can use differentiation from first principles to find how close this

derivative function is to the original function.

EXAMPLE

Differentiate ( )f x 10x= from first principles.

Solution

( )x

10

=

=

( ) ( )

( )

lim

lim

lim

lim

f h

f x h f x

h

h

h

10 10

10 10 1

10 1

h

h

x h x

h

x h

x

h

h

0

0

0

0

+ −

=−

=−

−

"

"

"

"

+

l

Using the 10 x key on the calculator, and finding values of

h

10 1h−

when h

is small, gives the result:

or

.2 3026 10Z( )x

( ) .dx

d 10 2 3026 10

x

x xZ

f l

Drawing the graphs of . y 2 3026 10x

= and y 10x

= together shows

how close the derivative function is to the original graph.

12

10

8

6

4

2

1 2−1−2

y

y = 2.3026 ×10 x

x

y =10 x

You can explore

limits using a

graphics package

on a computer or a

graphical calculator.

´

´

´




Similar results occur for other exponential functions. In general,

( )dx

d a ka

x x= where k is a constant.

Application

If y a x = then

dx

dy ka

ky

x =

=

This means that the rate of change of y is proportional to y itself. That is, if y is

small, its rate of change is small, but if y is large, then it is changing rapidly.

This is called exponential growth (or decay, if k is negative) and has many

applications in areas such as population growth, radioactive decay, the cooling of

objects, the spread of infectious diseases and the growth of technology.

Different exponential functions have different values of k.

EXAMPLES

1. ( ) .dx

d 2 0 6931 2x xZ .

12

10

8

6

4

2

1 2 3−3 −2 −1

y

y = 2 x

y = 0.6931 × 2 x

x

2. ( ) .dx

d 3 1 0986 3x xZ .

12

10

8

6

4

2

−3

y

y = 3 x

y = 1.0986 × 3 x

x1 32−2−1

You will study exponential

growth and decay in

Chapter 6.

´

´




Notice that the derivative function of y 3x

= is very close to the original

function.

We can find a number close to 3 that gives exactly the same graph for the

derivative function. This number is approximately 2.71828, and is called e.

e x KEY

Use this key to find powers of e.

For example, to find e2:

Press SHIFT e 2 7.389056099e2 x

=

To find e:

Press SHIFT 1 2.718281828e e1 x

=

EXAMPLES

1. Sketch the curve . y ex

=

Solution

Use ex on your calculator to draw up a table of values:

x -3 -2 -1 0 1 2 3

y 0.05 0.1 0.4 1 2.7 7.4 20.1

( )dx

d e ex x=

DID YOU KNOW?

The number e was linked to logarithms before this useful result in calculus was known. It is

a transcendental (irrational) number. This was proven by a French mathematician, Hermite,

in 1873. Leonhard Euler (1707–83) gave e its symbol, and he gave an approximation of e to

23 decimal places. Currently, e is known to about 100 000 decimal places.

Euler studied mathematics, theology, medicine, astronomy, physics and oriental languages.

He did extensive research into mathematics and wrote more than 500 books and papers.

Euler gave mathematics much of its important notation. He caused π to become standardnotation and used i for the square root of –1. He first used small letters to show the sides of

triangles and the corresponding capital letters for their opposite angles. Also, he introduced

the symbol for sums and f(x) notation.

A transcendental

number is a number

beyond ordinary

numbers. Another

transcendental

number is π .

e is an irrational numberlike π .




2. Differentiate .e5 x

Solution

`

( )

( ) ( )

dx

d

e e

dx

d e

dx

d e

e

5 5

5

x x

x x

x

=

=

=

3. Find the equation of the tangent to the curve y e3 x= at the point (0, 3).

Solution

dx

dy e3 x

=

`

At ( , ),dx

dy e

m

0 3

3

3

3 0=

=

=

Equation ( )

( )

y y m x x

y x

x

y x

3 0

3

3 3

31 1

− = −

− = −

=

= +

dx

dy gives the gradient of

the tangent.

CONTINUED




4.1 Exercises

1. Find, correct to 2 decimal places,

the value of

(a) e .1 5

(b) e 2−

(c) e2

.0 3

(d)e

1

3

(e) e3 .3 1

−

−

2. Sketch the curve

(a) y e2 x

=

(b) y e x=

−

(c) y ex= −

3. Differentiate

(a) e9 x

(b) ex

−

(c) e xx 2+

(d) x x x e2 3 5 x3 2

− + −

(e) 3( )e 1x

+

(f) 7( )e 5x

+

(g) 2( )e2 3x

−

(h) xex

(i)x

ex

(j) x ex2

(k) ( )x e2 1 x+

(l)x

e

7 3

x

−

(m)e

x5x

4. If ( ) ,f x x x e3 x3= + − find ( )1f l

and ( )f 1m in terms of e.

5. Find the exact gradient of the

tangent to the curve y ex

= at the

point (1, e).


normal to the curve y ex

= at the

point where .x 5=

7. Find the gradient of the tangent

to the curve y e4 x

= at the point

where . ,x 1 6= correct to 2

decimal places.

8. Find the equation of the tangent

to the curve y ex

= − at the point

(1, –e).

4. Differentiate .e

x2 3x

+

Solution

v u

.

( )

( )

( )

( )

dx

dy

v

u v

e

e e x

e

e xe e

e

e xe

e

e x

e

x

2 2 3

2 2 3

2

1 2

1 2

x

x x

x

x x x

x

x x

x

x

x

2

2

2

2

2

=

−

=

− +

=− −

=− −

=

− +

=

− +

l l

This is the quotient rule

from Chapter 8 of the

Preliminary Course book.




Function of a function rule

Remember that the function of a function rule uses the result

.

dx

dy

du

dy

dx

du=

EXAMPLE

Differentiate .ex x5 32

+ −

Solution

Let u x x5 32= + −

Then y eu=

dx

dux2 5= + and

du

dy eu

=

( )

( )

( )

dx

dy

du

dy

dx

du

e xe x

x e

2 5

2 5

2 5

u

x x

x x

5 3

5 3

2

2

=

= +

= +

= +

+ −

+ −

You studied this in Chapter 8 of

the Preliminary Course book.

Can you see a quick way to

do this?

Proof

Let ( )u f x=

Then y eu=

du

dy eu

= and ( )xdx

duf = l

( )x

( )x e

dx

dy

du

dy

dx

du

e f

f ( )

u

f x

=

=

=

l

l

If y e ( )f x= then ( )x e

dx

dy ( )f x

= f l

9. Find the equation of the normal

to the curve y ex

= at the point

where ,x 3= in exact form.

10. Find the stationary point on the

curve y xex

= and determine its

nature. Hence sketch the curve.

11. Find the first and second

derivatives of y e7 x

= . Hence show

that .dx

d y y

2

2

=

12. If , y e2 1

x= +

show that.

dx

d y

y 1

2

2

= −

´

´

´




4.2 Exercises

(m)x

e x

2

3

(n) x e x3 5

(o)x

e

2 5

x2 1

+

+

2. Find the second derivative of

.7( )e 1x2+

3. If ( ) ,f x e x3 2=

− find the exact value

of ( )f 1l and ( ) .f 0m

4. Find the gradient of the tangent

to the curve y e x5

= at the point

where .x 0=

EXAMPLES

1. Differentiate e x5 2−

Solution

( )x e y f

e5

( )f x

x5 2

=

= −

l l

2. Differentiate x e x2 3 .

Solution

u +

dx

dy v v u= l l

. .2 3

( )

x e e x

xe x2 3

x x

x

3 3 2

3

= +

= +

3. Given , y e2 1x3= + show that ( ) .

dx

d y y 9 1

2

2

= −

Solution

( )

( )

( )

y e

dx

dy e

dx

d y e

e

e

y

2 1

6

18

9 2

9 2 1 1

9 1

x

x

x

x

x

3

3

2

2

3

3

3

= +

=

=

=

= + −

= −

This is the product rule from

Chapter 8 of the Preliminary

Course book.

1. Differentiate

(a) e x7

(b) e x−

(c) e x6 2−

(d) ex 1

2+

(e) ex x5 73

+ +

(f) e x5

(g) e x2−

(h) e x10

(i) e xx2+

(j) x x e2 x2 1

+ + −

(k) 5( )x e x4

+

(l) xe

x2




Integration of Exponential Functions

Since ( ) ,dx

d e e

x x= then the reverse must be true.


to the curve y e x3x2

= − at the

point (0, 1).


normal to the curve y e x3

= at the

point where .x 1=

7. Find the equation of the tangent to

the curve y ex2

= at the point (1, e).

8. If ( ) ,f x x x e4 3 x3 2 2= + −

− find

( )f 1−m in terms of e.

9. Find any stationary points on

the curve y x e x2 2= and sketch the

curve.

10. If , y e ex x4 4= + − show that

.dx

d y y 16

2

2

=

11. Prove ,dx

d y

dx

dy y 3 2 0

2

2

− + = given

. y e3 x2=

12. Showdx

d y b y

2

2

2= for . y aebx

=

13. Find the value of n if y e x3=

satisfies the equation

y .

dx

d

dx

dy ny 2 0

2

2

+ + =

14. Sketch the curve , y ex x 22

= + −

showing any stationary points

and inflexions.

15. Sketch the curve

e

x y

1x

2

=+

,

showing any stationary points

and inflexions.

e dx e Cx x= +

#

Integration is the inverse of

differentiation.

To find the indefinite integral (primitive function) when the function of

a function rule is involved, look at the derivative first.

EXAMPLE

Differentiate .e x2 1+

Hence find .e dx2 x2 1+ #

Find .e dxx2 1+

Solution

( )dx

d e e

e dx e C

e dx e dx

e C

2

2

2

12

2

1

x x

x x

x x

x

2 1 2 1

2 1 2 1

2 1 2 1

2 1

`

=

= +

=

= +

+ +

+ +

+ +

+

# # #




EXAMPLES

1. Find ( ) .e e dxx x2−

− #

Solution

( )( )

e e dx e e C

e e C

2

1

1

1

2

1

x x x x

x x

2 2

2

− = −

−

+

= + +

− −

−

#

2. Find the exact area enclosed between the curve , y e x3

=

the x-axis andthe lines x 0= and .x 2=

Solution

Area

( )

( ) units

e dx

e

e e

e e

e

3

1

3

1

3

1

3

1

31 1

x

x

3

0

3

0

2

6 0

6 0

6 2

=

=

= −

= −

= −

2

; E #

3. Find the volume of the solid of revolution formed when the curve

y ex= is rotated about the x-axis from x 0= to .x 2=

Solution

( )

y e

y e

e

x

x

x

2 2

2

`

=

=

=

In general

Use index laws to simplify ( ) .e 2 x

e dxa

e C1ax b ax b

= ++ +

Proof

( )dx

d e ae

ae dx e C

e dxa

ae dx

a e C

1

1

ax b ax b

ax b ax b

ax b ax b

ax b

`

=

= +

=

= +

+ +

+ +

+ +

+

# # #




( ) units

π

π

π

π

π

π

V y dx

e dx

e

e e

e

e

2

1

2

1

2

1

2

1

2

1

2 1

a

x

x

2

2

0

2

0

2

4 0

4

4 3

=

=

=

= −

= −

= −

b

2

cc

mm

; E

#

#

4.3 Exercises

(c) ( )e x dx2 3x 5

5

6+ −

+

(d) ( )e t dt t 3 4

0−

+1

(e) ( )e e dxx x4 2

1+

2

4. Find the exact area enclosed by

the curve , y e2 x2= the x-axis and

the lines x 1= and .x 2=

5. Find the exact area bounded by

the curve , y e x4 3

=− the x-axis and

the lines x 0= and .x 1=

6. Find the area enclosed by the

curve , y x e x

= + − the x-axis and

the lines x 0= and ,x 2= correct

to 2 decimal places.

7. Find the area bounded by the

curve , y e x5

= the x-axis and the

lines x 0= and ,x 1= correct to

3 significant figures.

8. Find the exact volume of the

solid of revolution formed when

the curve y ex

= is rotated about

the x-axis from x 0= to .x 3=

9. Find the volume of the solid formed

when the curve y e 1x

= +− is rotated

about the x-axis from x 1= to ,x 2=

correct to 1 decimal place.

1. Find these indefinite integrals.

(a) e dxx2 #

(b) e dxx4 #

(c) e dxx− #

(d) e dxx5 #

(e) e dxx2− #

(f) e dxx4 1+ #

(g) e dx3 x5

− # (h) e dt

t 2 # (i) ( )e dx2x7

− # (j) ( )e x dx

x 3+

− #

2. Evaluate in exact form.

(a) e dxx5

0

1

#

(b) e dxx

0

− −

2

#

(c) e dx2 x3 4

1

+4 #

(d) ( )x e dx3 x2 2

2−

3 #

(e) ( )e dx1x2

0+

2 #

(f) ( )e x dxx

1−

2 #

(g) ( )e e dxx x2

0−

−

3 #

3. Evaluate correct to 2 decimal places.

(a) e dxx

1

−

3 #

(b) e dy 2 y 3

0

2 #




10. Use Simpson’s rule with 3

function values to find an

approximation to ,xe dxx

1

2


11. (a) Differentiate.x e

x2

(b) Hence find ( ) .x x e dx2 x+

12. Find x e dxx2 1

3+ # using the

substitution .u x 13= +

13. Use the substitution u x2

= to

evaluate xe dxx

0

22

(give exact

value).

14. The curve y e 1x

= + is rotated

about the x-axis from x 0= to

.x 1= Find the exact volume of

the solid formed.

15. Find the exact area enclosed

between the curve y e x2

= and the

lines y 1= and .x 2=

Application

The exponential function occurs in many fields, such as science and economics.

P P e0

kt = is a general formula that describes exponential growth.

P P e0

kt =

- is a general formula that describes exponential decay.

Logarithms

‘Logarithm’ is another name for the index or power of a number. Logarithms

are related to exponential functions, and allow us to solve equations like

.2 5x

= They also allow us to change the subject of exponential equations such

as y ex

= to x.

Definition

If ,y a x

= then x is called the logarithm of y to the base a.

You will study these formulae in

Chapter 6.

If , y ax= then logx y

a=

Logarithm keys

log is used for log x 10

In is used for log x e




EXAMPLES

1. Find .log 5 310


Solution

log 5.3 0.724275869

0.7 correct to 1 decimal place10

=

=

2. Evaluate log 80e

correct to 3 significant figures.

Solution

.

. correct to 3 significant figures

log 80 4 382026634

4 38e

=

=

3. Evaluate .log 813

Solution

Let

Then (by definition)

i.e.

So .

log

log

x

x

81

3 81

3 3

4

81 4

x

x

3

4

3

`

=

=

=

=

=

4. Find the value of .log 4

12

Solution

Let

Then

So .

log

log

x

x

41

241

2

1

2

2

41 2

x

2

2

2

2

`

=

=

=

=

= −

= −

−

Use the log key.

Use the In key.




1. Evaluate

log(a)2 16

log(b)4 16

log(c)5 125

log(d)3 3

log(e)7 49

log(f)7 7

log(g)5 1

log(h)2 128

2. Evaluate3 log(a)

2 8

log(b)5 25 + 1

3 – log(c)3 81

4 log(d)3 27

2 log(e)10

10 000

1 + log(f)4 64

3 log(g)4 64 + 5

2 + 4 log(h)6 216

(i)log

2

93

(j)log

log

8

64 4

2

8 +

3. Evaluate

(a) log2

12

(b) log 33

log(c)4 2

(d) log25

15

(e) log 77

4

(f) log3

13 3

(g) log2

14

log(h)8 2

(i) log 6 66

(j) log 4

22

4. Evaluate correct to 2 decimal

places.

log(a)10

1200

log(b)10

875

log(c)e 25

ln 140(d)

5 ln 8(e)

log(f)10

350 + 4.5

(g)

log

2

1510

ln 9.8 + log(h)10

17

(i)log

log

30

30

e

10

4 ln 10 – 7(j)

4.4 Exercises

Class Investigation

Sketch the graph of1. log y x2

= .

There is no calculator key for logarithms to the base 2. Use the

definition of a logarithm to change the equation into index form,and the table of values:

x

y –3 –2 –1 0 1 2 3

On the same set of axes, sketch the curve2. 2 y x= and the line y x= .

What do you notice?




5. Write in logarithmic form.

(a) y 3x=

(b) z5x

=

(c) x y 2=

(d) a2b=

(e)b d 3=

(f) y 8x

=

(g) y 6x

=

(h) y ex

=

(i) y ax

=

(j) e=Q x

6. Write in index form.

(a) log x53

=

(b) log x7a

=

(c) log a b3

=

(d) 9log y x=

(e) log b y a

=

(f) log y 62

=

(g) log y x3

=

(h) log y 910

=

(i) 4ln y =

(j) log y x7

=

7. Solve for x, correct to 1 decimal

place where necessary.

(a) log x 610

=

(b) log x 53 =

(c) log 343 3x

=

(d) log 64 6x

=

(e) log x

51

5 =

(f) log 32

1x

=

(g) 3.8ln x =

(h) log x3 2 1010

− =

(i) log x

2

34

=

(j) log 431

x=

8. Evaluate y given that .log 125 3 y

=

9. If . ,log x 1 6510

= evaluate x correct

to 1 decimal place.

10. Evaluate b to 3 significant figures

if . .log b 0 894e =

11. Find the value of log2 1. What is

the value of loga 1?

12. Evaluate log5 5. What is the value

of loga a?

13. (a) Evaluate ln e without a

calculator.

Using a calculator, evaluate(b)

(i) loge e3

(ii) loge e2

(iii) loge e5

(iv) loge e

(v) loge 1

e

(vi) eln 2

(vii) eln 3

(viii) eln 5

(ix) eln 7

(x) eln 1

(xi) eln e

14. Sketch the graph of .log y xe

=

What is its domain and range?

15. Sketch , log y y x10x

10= = and

y x= on the same number plane.

What do you notice about the

relationship of the curves to

the line?

16. Change the subject of log y xe

=

to x.




This corresponds to the

law ( )a a .m n mn

=

,log 36 26

= since .6 362=

Proof

Let x am

= and y an

=

Then logm xa

= and logn y a

=

` (by definition)

÷

log

log log

y

xa a

a

y

xm n

x y

m n

m n

a

a a

=

=

= −

= −

−

b l

log logx n xa

n

a=

Proof

Let x am=

Then logm xa

=

`

( )

(by )log

log

x a

a

x mn

n x

definition

n m n

mn

a

n

a

=

=

=

=

EXAMPLES

1. Evaluate .log log3 126 6

+

Solution

( )log log log

log

3 12 3 12

36

2

6 6 6

6

+ =

=

=

2. Given .log 3 0 685

= and . ,log 4 0 865

= find

(a) log 125

(b) .log 0 755

(c) log 95

(d) log 205

CONTINUED

´




Solution

(a) ( )

. .

.

log log

log log

12 3 4

3 4

0 68 0 86

1 54

5 5

5 5

=

= +

= +

=

(b) .

. .

.

log log

log log

0 754

3

3 4

0 68 0 86

0 18

5 5

5 5

=

= −

= −

= −

(c)

.

.

log log

log

9 3

2 3

2 0 68

1 36

5 52

5

=

=

=

=

(d) ( )

.

.

log log

log log

20 5 4

5 4

1 0 86

1 86

5 5

5 5

=

= +

= +

=

3. Solve log2 12 = log

2 3 + log

2 x.

Solution

log log log

log

x

x

12 3

32 2 2

2

= +

=

So 12 = 3x

4 = x

,log 5 15

= since .5 5 1=

1. Use the logarithm laws to

simplify

log(a)a 4 + log

a y

log(b) a 4 + loga 5log(c)

a 12 – log

a 3

log(d)a b – log

a 5

3 log(e)x y + log

x z

2 log(f)k 3 + 3 log

k y

5 log(g)a x – 2 log

a y

log(h)a x + log

a y – log

a z

log(i)10

a + 4 log10

b + 3 log10

c

3 log(j)3 p + log

3 q – 2 log

3 r

2. Given log7 2 = 0.36 and

log7 5 = 0.83, find

log(a)7 10

log(b) 7 0.4log(c)

7 20

log(d)7 25

log(e)7 8

log(f)7 14

log(g)7 50

log(h)7 35

log(i)7 98

log(j)7 70

4.5 Exercises

´

´

´




3. Use the logarithm laws to

evaluate

log(a)5 50 – log

5 2

log(b)2 16 + log

2 4

log(c)4 2 + log

4 8

log(d) 5 500 – log5 4log(e)

9 117 – log

9 13

log(f)8 32 + log

8 16

3 log(g)2 2 + 2 log

2 4

2 log(h)4 6 – (2 log

4 3 + log

4 2)

log(i)6 4 – 2 log

6 12

2 log(j)3 6 + log

3 18 – 3 log

3 2

4. If loga 3 = x and log

a 5 = y , find an

expression in terms of x and y for

log(a)a 15

log(b) a 0.6log(c)

a 27

log(d)a 25

log(e)a 9

log(f)a 75

log(g)a 3a

log(h)a 5

a

log(i)a 9a

log(j)a a

125

5. If loga x = p and loga y = q, find, interms of p and q.

log(a)a xy

log(b)a y 3

log(c)a x

y

log(d)a x2

log(e)a xy 5

log(f)a y

x2

log(g)a ax

log(h) a y

a

2

log(i)a a3 y

log(j)a ay

x

6. If loga b = 3.4 and log

a c = 4.7,

evaluate

log(a)a b

c

log(b)a bc 2

log(c)a (bc )2

log(d)a abc

log(e) a a2

c log(f)

a b7

log(g)a c

a

log(h)a a3

log(i)a bc 4

log(j)a b4c 2

7. Solve

log(a)4 12 = log

4 x + log

4 3

log(b)3 4 = log

3 y – log

3 7

log(c)a 6 = log

a x – 3 log

a 2

log(d) 2 81 = 4 log2 xlog(e)

x 54 = log

x k + 2 log

x 3

Change of base

Sometimes we need to evaluate logarithms such as log2 7. We use a change of

base formula.

loglog

logx

a

x

a

b

b

=




EXAMPLE

Find the value of ,log 25

correct to 2 decimal places.

Solution

.0 430676558Z

(by change of base)

.

loglog

log2

5

2

0 43

5 =

=

Proof

Let log y xa

=

Then x a y

=

Take logarithms to the base b of both sides of the equation:

`

log log

log

log

log

log

x a

y a

a

x y

x

b b

y

b

b

b

a

=

=

=

=

You can use the change of base formula to find the logarithm of any number,

such as .log 25

You change it to either log x10

or ,log xe

and use a calculator.

Exponential equations

You can also use the change of base formula to solve exponential equations

such as .5 7x

=

You studied exponential equations such as 2 8x

= in the Preliminary

Course. Exponential equations such as 2 9x

= can be solved by taking

logarithms of both sides, or by using the definition of a logarithm and the

change of base formula.

You can use either log or In




EXAMPLES

1. Solve 5 7x

= correct to 1 decimal place.

Solution

5 7x

=

Using the definition of a logarithm, this means:

(using change of base formula)

.

log

log

log

x

x

x

7

5

7

1 2

5 =

=

=

If you do not like to solve the equation this way, you can use the

logarithm laws instead.

Taking logs of both sides:

.

log log

log log

log

log

x

x

5 7

5 7

5

7

1 2 1correct to decimal place

x

`

=

=

=

=

2. Solve 4 9 y 3

=− correct to 2 decimal places.

Solution

y 3−

4 9=

Using the logarithm definition and change of base:

.

log

log

log

log

log

y

y

y

y

9 3

4

93

4

93

4 58

4 = −

= −

+ =

=

Using the logarithm laws:

Taking logs of both sides:

( )

.

log loglog log

log

log

log

log

y

y

y

4 93 4 9

34

9

4

93

4 58 2correct to decimal place

y 3 =

− =

− =

= +

=

−

You can use either

log or ln.

Use log x n log x a

n

a=




1. Use the change of base formula

to evaluate to 2 decimal places.

log(a)4 9

log(b) 6 25log(c)

9 200

log(d)2 12

log(e)3 23

log(f)8 250

log(g)5 9.5

2 log(h)4 23.4

7 – log(i)7 108

3 log(j)11

340

2. By writing each equation as a

logarithm and changing the base,solve the equation correct to


(a) 4 9x

=

(b) 3 5x

=

(c) 7 14x

=

(d) 2 15x

=

(e) 5 34x

=

(f) 6 60x

=

(g) 2 76x

=

(h) 4 50x

=

(i) 3 23x

=

(j) 9 210x

=

3. Solve, correct to 2 decimal places.

(a) 2 6x

=

(b) 5 15 y =

(c) 3 20x

=

(d) 7 32m

=

(e) 4 50k=

(f) 3 4t =

(g) 8 11x

=

(h) 2 57 p=

(i) .4 81 3x

=

(j) .6 102 6n

=

4. Solve, to 1 decimal place.

(a) 3 8x 1

=+

(b) n35 71=

(c) 2 12x 3

=−

(d) 4 7n2 1

=−

(e) 7 11x5 2

=+

(f) .8 5 7n3=

−

(g) .2 18 3x 2

=+

(h) k7 3−

.3 32 9=

(i) 29 50=

x

(j) .6 61 3 y 2 1

=+

5. Solve each equation correct to


(a) e 200x

=

(b) e 5t 3=

(c) e2 75t =

(d) e45 x

=

(e) e3000 100 n

=

(f) e100 20 t 3

=

(g) e2000 50 . t 0 15

=

(h) e15 000 2000 . k0 03

=

(i) Q Qe3 . t 0 02

=

(j) . M Me0 5 . k0 016

=

4.6 Exercises




Derivative of the Logarithmic Function

Drawing the derivative (gradient) function of a logarithm function gives a

hyperbola.

EXAMPLE

Sketch the derivative function of .log y x2

=

Solution

The gradient is always positive but is decreasing.

If log y xe

= then1

dx

dy

x=

Proof

dx

dy

dy

dx

1=

Given log y xe

=

Then y x e=

`

y

dy

dxe

dx

dy

e

x

1

1

y

=

=

=

dx

dy

dy

dx

1= is a special result that

can be proved by differentiating

from first principles.




Function of a function rule

If .( ), then ( )( ) ( )

( )log y f x

dx

dy f x

f x f x

f x1e

= = =ll

Proof

Let ( )u f x=

Then log y ue

=

`

Also

.

.

.

( )

( )

( ) ( )

du

dy

u

dx

duf x

dx

dy

du

dy

dx

du

u f x

f x f x

1

1

1

=

=

=

=

=

l

l

l

EXAMPLES

1. Differentiate ( 3 1) .log x x2

e− +

Solution

[ ( )]logdx

d x x

x x

x3 1

3 1

2 3e

2

2− + =

− +

−

2. Differentiate3 4

1.log

x

x

e−

+

Solution

Let

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

log

log log

y x

x

x x

dx

dy

x x

x x

x x

x x

x x

x x

3 4

1

1 3 4

1

1

3 4

3

1 3 4

1 3 4 3 1

1 3 4

3 4 3 3

1 3 4

7

e

e e

=−

+

= + − −

=+

−−

=

+ −

− − +

=

+ −

− − −

=

+ −

−




3. Find the gradient of the normal to the curve ( )log y x 5e

3= − at the

point where 2.x =

Solution

dxdy

xx5

33

2

=

−

When 2,x =

( )

dx

dy

m

2 5

3 2

4

3

2

1

=

−

=

The normal is perpendicular to the tangent

m m

m

m

1

4 1

4

1

i.e.1 2

2

2`

= −

= −

= −

4. Differentiate .log y x2

=

Solution

log

log

log

log log

log

log

y x

x

x

dxdy

x

x

2

2

1

21 1

2

1

e

e

ee

e

e

2=

=

=

=

=

5. Find the derivative of 2x.

Solution

( )

ln

ln

ln

e

e

e

dx

dy e

2

2

2

2 2

2 2

2

ln

ln

ln

ln

x x

x

x

x

x

2

2

2

`

=

=

=

=

=

=

This result comes from the

Preliminary Course.

´

´

´




1. Differentiate

(a) logx xe

+

(b) log x1 3e

−

(c) ( )ln x3 1+

(d) ( )log x 4e

2−

(e) ( )ln x x5 3 93+ −

(f) ( )log x x5 1e

2+ +

(g) lnx x x3 5 5 42

+ − +

(h) ( )log x8 9 2e

− +

(i) (2 4) (3 1)log x xe

+ −

(j)2 7

4 1log

x

x

e−

+

(k) ( )log x

1 e

5+

(l) ( )ln x x 9

−

(m) ( )log xe

4

(n) ( )logx xe

2 6+

(o) logx xe

(p)log

x

xe

(q) ( ) logx x2 1e

+

(r) ( )logx x 1e

3+

(s) ( )log log xe e

(t) lnx

x

2−

(u)log x

e

e

x2

(v) lne xx

(w) ( )log x5e

2

2. If ( ) 2 ,logf x xe

= − find ( )f 1l .

3. Find the derivative of .log x10

4. Find the equation of the tangentto the curve log y xe

= at the

point ( , )log2 2e

.


to the curve ( 1)log y xe

= − at the

point where 2x = .

6. Find the gradient of the normal

to the curve ( )log y x xe

4= + at the

point ( , )log1 2e

.

7. Find the exact equation of the

normal to the curve log y xe

= at

the point where 5x = .


to the curve ( )log y x5 4e

= + at

the point where 3x = .

9. Find the point of inflexion on the

curve .log y x x xe

2= −

10. Find the stationary point on the

curveln

x

x y = and determine its

nature.

11. Sketch, showing any stationary

points and inflexions.

(a) log y x xe

= −

(b) ( )log y x 1e

3= −

(c) ln y x x=

12. Find the derivative of ( )log x2 53

+ .

13. Differentiate

(a) 3x

(b) 10x

(c) 2 x3 4−


to the curve y 4x 1

= + at the point

(0, 4).

15. Find the equation of the normal

to the curve log y x3

= at the

point where 3x = .

4.7 Exercises




logx

dx

x dx x C

1e

= = + #

Integration and the Logarithmic Function

EXAMPLES

1. Find the area enclosed between the hyperbola , y

x

1= the x-axis and

the lines 1x = and 2x = , giving the exact value.

Solution

log

log log

log

Ax

dx

x

1

2 1

2

e

e e

e

1

1

2

=

=

= −

=

2

7 A

#

So area is log 2e

units2.

2. Find .x

xdx

73

2

+

#

Solution

( )log

x

xdx

x

xdx

x C

7 3

1

7

3

31

7e

3

2

3

2

3

+

=

+

= + +

#

( )

( )( )log

f x

f xdx f x C

e= +

l

CONTINUED

Integration is the inverse of

differentiation.




1. Find the indefinite integral

(primitive function) of

(a)2 5

2

x +

(b)x

x

2 1

4

2+

(c)2

5

x

x

5

4

−

(d)2

1

x

(e)

2

x

(f)3

5

x

(g)3

2 3

x x

x

2−

−

(h)x

x

22+

(i)x

x

7

3

2+

(j)x x

x

2 5

1

2+ −

+

2. Find

(a)x

dx4 1

4

−

#

(b)x

dx

3+ #

(c)x

xdx

2 73

2

−

#

(d)x

xdx

2 56

5

+

#

(e)x x

xdx

6 2

3

2 + +

+

#

3. Evaluate correct to 1 decimal

place.

(a)x

dx2 5

2

1

3

+

(b)x

dx

12 +

5

(c)x

xdx

23

2

1

7

+

(d)x x

xdx

2 1

4 1

20

3

+ +

+

(e)x x

xdx

2

1

23

4

−

−

4. Find the exact area between the

curve , y x

1= the x-axis and the

lines 2x = and 3.x =

5. Find the exact area bounded by

the curve , y x 1

1=

−

the x-axis and

the lines 4x = and 7.x =

6. Find the exact area between the

curve , y x1

= the x-axis and the

lines y x= and 2x = in the first

quadrant.

7. Find the area bounded by the

curve , y x

x

12

=

+

the x-axis and

the lines 2x = and 4x = , correct

to 2 decimal places.

4.8 Exercises

3. Find .x x

xdx

4

1

2+ +

+ #

Solution

( )

( )log

x x

x

dx x x

x

dx

x x

xdx

x x C

2 4

1

2

1

2 4

2 1

2

1

2 4

2 2

2

12 4

e

2 2

2

2

+ +

+

=+ +

+

=

+ +

+

= + + +

# # #





solid formed when the curve

y x

1= is rotated about the x-axis

from 1x = to 3x = .

9. Find the volume of the solid

formed when the curve

y x2 1

2=

−

is rotated about the

x-axis from 1x = to 5x = , giving

an exact answer.

10. Find the area between the curve

ln y x= , the y -axis and the lines

2 y = and 4 y = , correct to 3

significant figures.


solid formed when the curve

log y xe

= is rotated about the

y -axis from 1 y = to 3 y = .

12. (a) Show that

9

3 3

3

1

3

2

x

x

x x2−

+=

++

−

.

(b) Hence findx

xdx

9

3 3

2−

+.

13. (a) Show thatx

x

x1

6

1

51

−

−

−

−

= .

(b) Hence findx

xdx

1

6

−

−

.

14. Find the indefinite integral

(primitive function) of 3 x2 1− .

15. Find, correct to 2 decimal places,

the area enclosed by the curve

log y x2

= , the x-axis and the

lines 1x = and 3x = by using

Simpson’s rule with 3 functionvalues.




Test Yourself 41. Evaluate to 3 significant figures.

(a) e 12−

(b) log 9510

(c) log 26e

(d) 7log4

(e) 3log4

(f) ln50

(g) 3e +

(h) ln

e

4

5 3

(i) eln 6

(j) eln 2

2. Differentiate

(a) e x5

(b) e2 x1−

(c) 4log xe

(d) ( )ln x4 5+

(e) xex

(f)ln

x

x

(g) 10( )e 1x

+

3. Find the indefinite integral (primitive

function) of

(a) e x4

(b)x

x

92−

(c) e x−

(d)4

1

x +

4. Find the equation of the tangent to the

curve y e2 x3

= + at the point where 0x = .

5. Find the exact gradient of the normal to

the curve y x e x

= −− at the point where

2x = .

6. Find the exact area bounded by the curve

y e x2

= , the x-axis and the lines 2x = and

5x = .

7. Find the volume of the solid formed if

the area bounded by y e x3

= , the x-axis

and the lines 1x = and 2x = is rotated

about the x-axis.

8. If .log 2 0 367

= and .log 3 0 567

= , find the

value of

(a) log 67

(b) log 87

(c) .log 1 57

(d) log 147

(e) .log 3 57

9. Find the area enclosed between the curve

ln y x= , the y -axis and the lines 1 y = and

3 y = .

10. (a) Use Simpson’s rule with 3 function

values to find the area bounded by the

curve ln y x= , the x-axis and the lines

2x = and 4x = .

(b) Change the subject of ln y x= to x.

(c) Hence find the exact area in part (a).

11. Solve

(a) 3 8x

=

(b) 2 3x3 4

=−

(c) log 81 4x

=

(d) log x 26

=

(e) e12 10 . t 0 01

=

12. Evaluate

(a) e dx3 x2

0

1

(b)x

dx

3 21 −

4

(c)x

x x xdx

2 5 33 2

1

2 − + + #

13. Find the equation of the tangent to the

curve y ex

= at the point ( , )e4 4 .

14. Evaluate log 89

to 1 decimal place.




14. Use Simpson’s rule with 3 function

values to find the area enclosed by thecurve , y e

x2= the y -axis and the line 3 y = ,

correct to 3 significant figures.

15. Find the derivative of .log

e

x x

x

e

16. Use the substitution u x3 12

= + to find

xe dxx3 1

2+ .

17. If y e ex x

= + − , show

dx

d y y

2

2

= .

18. Use the substitution u x 23

= − to

evaluate x e dxx2 2

0

3−

2 # .

19. Provedx

d y

dx

dy y 4 5 10 0

2

2

− − − = , given

y e3 2x5

= − .

20. Find the equation of the curve that has

( )f x e12 x2=m and a stationary point at

( , )0 3 .

21. Sketch ( )log y x xe

2= − .

22. A curve has xedx

dy x2

= and passes through

the point ,021c m . Use the substitution

u x2

= to find the equation of the curve.

chapter four of maths in focus

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