# Chapter 31 logarithms

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Logarithms

31Contents:

A Logarithms in base a [3.10]

B The logarithmic function [3.10]

C Rules for logarithms [3.10]

D Logarithms in base 10 [3.10]

E Exponential and logarithmic

equations [3.10]

In Chapter 28 we answered problems like the one above by graphing the exponential function and using

technology to find when the investment is worth a particular amount.

However, we can also solve these problems without a graph using logarithms.

We have seen previously that y = x2 and y =px are inverse functions.

For example, 52 = 25 andp25 = 5.

If y = ax then we say x is the logarithm of y in base a, and write this as x = loga y.

LOGARITHMS IN BASE a [3.10]A

Opening problem#endboxedheading

Tony invests $8500 for n years at 7:8% p.a. compounding annually. The interest rate is fixed for the

duration of the investment. The value of the investment after n years is given by V = 8500 (1:078)ndollars.

Things to think about:

a How long will it take for Tonys investment to amount to $12 000?

b How long will it take for his investment to double in value?

Logarithms were created to be the .inverse of exponential functions

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Y:\HAESE\IGCSE01\IG01_31\625IGCSE01_31.CDR Tuesday, 18 November 2008 11:10:27 AM PETER

For example, since 8 = 23 we can write 3 = log2 8. The two statements 2 to the power 3 equals 8

and the logarithm of 8 in base 2 equals 3 are equivalent, and we write:

23 = 8 , log2 8 = 3Further examples are: 103 = 1000 , log10 1000 = 3

32 = 9 , log3 9 = 24

1

2 = 2 , log4 2 = 12

The symbol ,is equivalent to

In general, y = ax and x = loga y are equivalent statements

and we write y = ax , x = loga y.

Example 1 Self Tutor

Write an equivalent:

a logarithmic statement for 25 = 32 b exponential statement for log4 64 = 3:

a 25 = 32 is equivalent to log2 32 = 5.

So, 25 = 32 , log2 32 = 5.b log4 64 = 3 is equivalent to 4

3 = 64.

So, log4 64 = 3 , 43 = 64.

Example 2 Self Tutor

Find the value of log3 81:

) 3x = 81

) 3x = 34

) x = 4

) log3 81 = 4

EXERCISE 31A

1 Write an equivalent logarithmic statement for:

a 22 = 4 b 42 = 16 c 32 = 9 d 53 = 125

e 104 = 10000 f 71 = 17 g 33 = 127 h 27

1

3 = 3

i 52 = 125 j 2 1

2 = 1p2

k 4p2 = 22:5 l 0:001 = 103

2 Write an equivalent exponential statement for:

a log2 8 = 3 b log2 1 = 0 c log212

= 1 d log2

p2 = 12

e log2

1p2

= 12 f logp2 2 = 2 g logp3 9 = 4 h log9 3 = 12

3 Without using a calculator, find the value of:

a log10 100 b log2 8 c log3 3 d log4 1

e log5 125 f log5(0:2) g log10 0:001 h log2 128

i log212

j log3

19

k log2(

p2) l log2

p8

The logarithm of in baseis the exponent or powerof which gives .

813

3 81

Let log3 81 = x

626 Logarithms (Chapter 31)

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Y:\HAESE\IGCSE01\IG01_31\626IGCSE01_31.CDR Monday, 27 October 2008 3:01:48 PM PETER

The logarithmic function is f(x) = loga x where a > 0, a 6= 1.

Consider f(x) = log2 x which has graph y = log2 x.

Since y = log2 x , x = 2y, we can obtain the table of values:

y 3 2 1 0 1 2 3x 18

14

12 1 2 4 8

Notice that:

the graph of y = log2 x is asymptotic to the y-axis the domain of y = log2 x is fx j x > 0g the range of y = log2 x is fy j y 2 R g

THE INVERSE FUNCTION OF f(x) = loga x

Given the function y = loga x, the inverse is x = loga y finterchanging x and yg) y = ax

So, f(x) = loga x , f1(x) = ax

THE LOGARITHMIC FUNCTION [3.10]B

x

8642

y

O

y xlogx

627Logarithms (Chapter 31)

m log7

3p7

n log2(4p2) o logp2 2 p log2

1

4p2

q log10(0:01) r log

p2 4 s log

p3

13

t log3

1

9p3

4 Rewrite as logarithmic equations:

a y = 4x b y = 9x c y = ax d y = (p3)x

e y = 2x+1 f y = 32n g y = 2x h y = 2 3a

5 Rewrite as exponential equations:

a y = log2 x b y = log3 x c y = loga x d y = logb n

e y = logm b f T = log5a2

g M = 12 log3 p h G = 5 logbm

i P = logpbn

6 Rewrite the following, making x the subject:

a y = log7 x b y = 3x c y = (0:5)x d z = 5x

e t = log2 x f y = 23x g y = 5

x2 h w = log3(2x)

i z = 12 3x j y = 15 4x k D = 110 2x l G = 3x+1

7 Explain why, for all a > 0, a 6= 1: a loga 1 = 0 b loga a = 1

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Y:\HAESE\IGCSE01\IG01_31\627IGCSE01_31.CDR Tuesday, 18 November 2008 11:12:17 AM PETER

Example 3 Self Tutor

Find the inverse function f1(x) for: a f(x) = 5x b f(x) = 2 log3 x

a y = 5x has inverse function x = 5y

) y = log5 x

So, f1(x) = log5 x

b y = 2 log3 x has inverse function

x = 2 log3 y

)x

2= log3 y

) y = 3x2

So, f1(x) = 3x2

EXERCISE 31B

1 Find the inverse function f1(x) for:

a f(x) = 4x b f(x) = 10x c f(x) = 3x d f(x) = 2 3xe f(x) = log7 x f f(x) =

12(5

x) g f(x) = 3 log2 x h f(x) = 5 log3 x

i f(x) = logp2 x

2 a On the same set of axes graph y = 3x and y = log3 x.

b State the domain and range of y = 3x.

c State the domain and range of y = log3 x.

3 Prove using algebra that if f(x) = ax then f1(x) = loga x.

x

y

OO

yx

y xlogx

y x

If f(x) = g(x),

graph y = f(x)

and y = g(x)

on the same set

of axes.

4 Use the logarithmic function log on your graphics calculator

to solve the following equations correct to 3 significantfigures. You may need to use the instructions on page 15.

628 Logarithms (Chapter 31)

For example, if f(x) = log2 x then f1(x) = 2x.

The inverse function y = log2 x is the reflection of y = 2x

in the line y = x.

a log10 x = 3 x b log10(x 2) = 2xc log10

x4

= x2 2 d log10 x = x 1

e log10 x = 5x f log10 x = 3

x 3

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Y:\HAESE\IGCSE01\IG01_31\628IGCSE01_31.CDR Tuesday, 18 November 2008 11:12:59 AM PETER

Consider two positive numbers x and y. We can write them both with base a: x = ap and y = aq, forsome p and q.

) p = loga x and q = loga y ...... (*)

Using exponent laws, we notice that: xy = apaq = ap+q

x

y=

ap

aq= apq

xn = (ap)n = anp

) loga(xy) = p+ q = loga x+ loga y ffrom *gloga

x

y

= p q = loga x loga yloga(x

n) = np = n loga x

loga(xy) = loga x + loga y

loga

x

y

= loga x loga y

loga(xn) = n loga x

Example 5 Self Tutor

If log3 5 = p and log3 8 = q, write in terms of p and q:

a log3 40 b log3 25 c log3

64125

a log3 40

= log3(5 8)= log3 5 + log3 8

= p+ q

b log3 25

= log3 52

= 2 log3 5

= 2p

c log3

64125

= log3

82

53

= log3 8

2 log3 53= 2 log3 8 3 log3 5= 2q 3p

RULES FOR LOGARITHMS [3.10]C

629Logarithms (Chapter 31)

Example 4 Self Tutor

Simplify: a log2 7 12 log2 3 + log2 5 b 3 log2 5

a log2 7 12 log2 3 + log2 5= log2 7 + log2 5 log2 3

1

2

= log2(7 5) log2p3

= log2

35p3

b 3 log2 5

= log2 23 log2 5

= log285

= log2(1:6)

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Y:\HAESE\IGCSE01\IG01_31\629IGCSE01_31.CDR Tuesday, 18 November 2008 11:13:47 AM PETER

EXERCISE 31C

1 Write as a single logarithm:

a log3 2 + log3 8 b log2 9 log2 3 c 3 log5 2 + 2 log5 3d log3 8 + log3 7 log3 4 e 1 + log3 4 f 2 + log3 5g 1 + log7 3 h 1 + 2 log4 3 3 log4 5 i 2 log3 m+ 7 log3 nj 5 log2 k 3 log2 n

2 If log2 7 = p and log2 3 = q, write in terms of p and q:

a log2 21 b log237

c log2 49 d log2 27

e log279

f log2(63) g log2

569

h log2(5:25)

3 Write y in terms of u and v if:

a log2 y = 3 log2 u b log3 y = 3 log3 u log3 vc log5 y = 2 log5 u+ 3 log5 v d log2 y = u+ v

e log2 y = u log2 v f log5 y = log5 ug log7 y = 1 + 2 log7 v h log2 y =

12 log2 v 2 log2 u

i log6 y = 2 13 log6 u j log3 y = 12 log3 u+ log3 v + 14 Without using a calculator, simplify:

alog2 16

log2 4b

logp 16

logp 4c

log5 25

log515

d logm 25logm

15

Logarithms in base 10 are called common logarithms.

y = log10 x is often written as just y = log x, and we assume the logarithm has base 10.

Your calculator has a log key which is for base 10 logarithms.

Discovery Logarithms#endboxedheading