Chapter 31 logarithms

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)n

dollars.

Things to think about:

a How long will it take for Tony’s 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

IGCSE01magentacyan yellow black

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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 = 3

Further examples are: 103 = 1000 , log10 1000 = 3

32 = 9 , log3 9 = 2

41

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 43 = 64.

So, log4 64 = 3 , 43 = 64.


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 7¡1 = 17 g 3¡3 = 1

27 h 271

3 = 3

i 5¡2 = 125 j 2¡

1

2 = 1p2

k 4p2 = 22:5 l 0:001 = 10¡3

2 Write an equivalent exponential statement for:

a log2 8 = 3 b log2 1 = 0 c log2¡12

¢= ¡1 d log2

p2 = 1

2

e log2

³1p2

´= ¡1

2 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 log2¡12

¢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 3

x 18

14

12 1 2 4 8

Notice that:

² the graph of y = log2 x is asymptotic to the y-axis

² the domain of y = log2 x is fx j x > 0g² 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 , f¡1(x) = ax

THE LOGARITHMIC FUNCTION [3.10]B

x

8642

�

�

��

��

y

O

y x��logx

627Logarithms (Chapter 31)

m log7¡

3p7¢

n log2(4p2) o logp2 2 p log2

³1

4p2

´q log10(0:01) r logp2 4 s logp3

¡13

¢t log3

³1

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 = 2¡x 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 = log5¡a2

¢g M = 1

2 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 = 5x2 h w = log3(2x)

i z = 12 £ 3x j y = 1

5 £ 4x k D = 110 £ 2¡x 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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Find the inverse function f¡1(x) for: a f(x) = 5x b f(x) = 2 log3 x

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

) y = log5 x

So, f¡1(x) = log5 x

b y = 2 log3 x has inverse function

x = 2 log3 y

)x

2= log3 y

) y = 3x2

So, f¡1(x) = 3x2

EXERCISE 31B

1 Find the inverse function f¡1(x) for:

a f(x) = 4x b f(x) = 10x c f(x) = 3¡x d f(x) = 2£ 3x

e 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 f¡1(x) = loga x.

x�

�

y

OO

y��x

y x��logx

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 significant

figures. You may need to use the instructions on page 15.


For example, if f(x) = log2 x then f¡1(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) = 2¡x

c log10¡x4

¢= x2 ¡ 2 d log10 x = x¡ 1

e log10 x = 5¡x f log10 x = 3x ¡ 3


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

some p and q.

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

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

x

y=

ap

aq= ap¡q

xn = (ap)n = anp

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

µx

y

¶= p¡ q = loga x¡ loga y

loga(xn) = np = n loga x

loga(xy) = loga x + loga y

loga

µx

y

¶= loga x ¡ loga y

loga(xn) = n loga x


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

µ82

53

¶= log3 8

2 ¡ log3 53

= 2 log3 8¡ 3 log3 5

= 2q ¡ 3p

RULES FOR LOGARITHMS [3.10]C



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 31

2

= log2(7£ 5)¡ log2p3

= log2

³35p3

´b 3¡ log2 5

= log2 23 ¡ log2 5

= log2¡85

¢= log2(1:6)


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EXERCISE 31C

1 Write as a single logarithm:

a log3 2 + log3 8 b log2 9¡ log2 3 c 3 log5 2 + 2 log5 3

d log3 8 + log3 7¡ log3 4 e 1 + log3 4 f 2 + log3 5

g 1 + log7 3 h 1 + 2 log4 3¡ 3 log4 5 i 2 log3 m+ 7 log3 n

j 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 log2¡37

¢c log2 49 d log2 27

e log2¡79

¢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 v

c log5 y = 2 log5 u+ 3 log5 v d log2 y = u+ v

e log2 y = u¡ log2 v f log5 y = ¡ log5 u

g 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 = 1

2 log3 u+ log3 v + 1

4 Without using a calculator, simplify:

alog2 16

log2 4b

logp 16

logp 4c

log5 25

log5¡15

¢ dlogm 25

logm¡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

The logarithm of any positive number can be evaluated using the log key on your calculator. You will

1 Copy and complete: Number Number as a power of 10 log of number

10

100

1000

100 000 105 log(100 000) = 5

0:1

0:001

LOGARITHMS IN BASE 10 [3.10]D

need to do this to evaluate the logarithms in this discovery.


What to do:


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Y:\HAESE\IGCSE01\IG01_31\630IGCSE01_31.CDR Friday, 31 October 2008 9:46:05 AM PETER

2 Copy and complete: Number Number as a power of 10 log of numberp10

3p10

p1000

1p10

3 Can you draw any conclusion from your table? For example, you may wish to comment on when

a logarithm is positive or negative.

Example 6 Self Tutor If the base for a

logarithm is not

given then we

assume it is 10.

a 2 b 20

a b

RULES FOR BASE 10 LOGARITHMS

These rules

correspond

closely to the

exponent laws.

log(xy) = logx+ log y

log

µx

y

¶= logx¡ log y

log(xn) = n logx


Write as a single logarithm:

a log 2 + log 7 b log 6¡ log 3 c 2 + log 9 dlog 49

log¡17

¢a log 2 + log 7

= log(2£ 7)

= log 14

b log 6¡ log 3

= log¡63

¢= log 2

c 2 + log 9

= log 102 + log 9

= log(100£ 9)

= log 900

dlog 49

log¡17

¢=

log 72

log 7¡1

=2 log 7

¡1 log 7

= ¡2

The rules for base logarithms are clearly the same rules for general logarithms:10


Use the property a = 10log a to write the following numbers

as powers of 10:

log 2 ¼ 0:301

) 2 ¼ 100:301log 20 ¼ 1:301

) 20 ¼ 101:301


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Y:\HAESE\IGCSE01\IG01_31\631IGCSE01_31.CDR Thursday, 30 October 2008 12:05:29 PM PETER

EXERCISE 31D.1

1

a 8 b 80 c 800 d 0:8 e 0:008

f 0:3 g 0:03 h 0:000 03 i 50 j 0:0005

2 Write as a single logarithm in the form log k:

a log 6 + log 5 b log 10¡ log 2 c 2 log 2 + log 3

d log 5¡ 2 log 2 e 12 log 4¡ log 2 f log 2 + log 3 + log 5

g log 20 + log(0:2) h ¡ log 2¡ log 3 i 3 log¡18

¢j 4 log 2 + 3 log 5 k 6 log 2¡ 3 log 5 l 1 + log 2

m 1¡ log 2 n 2¡ log 5 o 3 + log 2 + log 7

3 Explain why log 30 = log 3 + 1 and log(0:3) = log 3¡ 1

4 Without using a calculator, simplify:

alog 8

log 2b

log 9

log 3c

log 4

log 8d

log 5

log¡15

¢e

log(0:5)

log 2f

log 8

log(0:25)g

log 2b

log 8h

log 4

log 2a

5 Without using a calculator, show that:

a log 8 = 3 log 2 b log 32 = 5 log 2 c log¡17

¢= ¡ log 7

d log¡14

¢= ¡2 log 2 e log

p5 = 1

2 log 5 f log 3p2 = 1

3 log 2

g log³

1p3

´= ¡1

2 log 3 h log 5 = 1¡ log 2 i log 500 = 3¡ log 2

6 74 = 2401 ¼ 2400

Show that log 7 ¼ 34 log 2 +

14 log 3 +

12 .

LOGARITHMIC EQUATIONS

The logarithm laws can be used to help rearrange equations. They are particularly useful when dealing with

exponential equations.


Write the following as logarithmic equations in base 10:

a y = a3b2 b y =mpn

a y = a3b2

) log y = log(a3b2)

) log y = log a3 + log b2

) log y = 3 log a+ 2 log b

b y =mpn

) log y = log

µm

n1

2

¶) log y = logm¡ logn

1

2

) log y = logm¡ 12 logn


Write as powers of 10 using a = 10log a:


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Y:\HAESE\IGCSE01\IG01_31\632IGCSE01_31.CDR Thursday, 30 October 2008 11:59:02 AM PETER


Write these equations without logarithms:

a logD = 2x+ 1 b logN ¼ 1:301¡ 2x

a logD = 2x+ 1

) D = 102x+1

or D = (100)x £ 10

b logN ¼ 1:301¡ 2x

) N ¼ 101:301¡2x

) N ¼ 101:301

102x¼ 20

102x


Write these equations without logarithms:

a logC = log a+ 3 log b b logG = 2 log d¡ 1

a logC = log a+ 3 log b

= log a+ log b3

= log(ab3)

) C = ab3

b logG = 2 log d¡ 1

= log d2 ¡ log 101

= log

µd2

10

¶) G =

d2

10

EXERCISE 31D.2

1 Write the following as logarithmic equations in base 10:

a y = ab2 b y =a2

bc y = d

pp

d M = a2b5 e P =pab f Q =

pm

n

g R = abc2 h T = 5

rd

ci M =

ab3pc

2 Write these equations without logarithms:

a logQ = x+ 2 b log J = 2x¡ 1 c logM = 2¡ x

d logP ¼ 0:301 + x e logR ¼ x+ 1:477 f logK = 12x+ 1

3 Write these equations without logarithms:

a logM = log a+ log b b logN = log d¡ log e

c logF = 2 log x d logT = 12 log p

e logD = ¡ log g f logS = ¡2 log b

g logA = logB ¡ 2 logC h 2 log p+ log q = log s

i ¡ log d+ 3 logm = logn¡ 2 log p j logm¡ 12 logn = 2 logP

k logN = 1 + log t l logP = 2¡ log x



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We have already seen how to solve equations such as 2x = 5 using technology. We now consider an

algebraic method.

By definition, the exact solution is x = log2 5, but we need to know how to evaluate this number.

We therefore consider taking the logarithm of both sides of the original equation:

log(2x) = log 5

) x log 2 = log 5 flogarithm lawg) x =

log 5

log 2

We conclude that log2 5 =log 5

log 2.

In general: the solution to ax = b where a > 0, b > 0 is x = loga b =log b

log a:


Use logarithms to solve for x, giving answers correct to 3 significant figures:

a 2x = 30 b (1:02)x = 2:79 c 3x = 0:05

a 2x = 30

) x =log 30

log 2

) x ¼ 4:91

b (1:02)x = 2:79

) x =log(2:79)

log(1:02)

) x ¼ 51:8

c 3x = 0:05

) x =log(0:05)

log 3

) x ¼ ¡2:73


Show that log2 11 =log 11

log 2. Hence find log2 11.

Let log2 11 = x

) 2x = 11

) log(2x) = log 11

) x log 2 = log 11

) x =log 11

log 2

EXPONENTIAL AND LOGARITHMICEQUATIONS [3.10]

E

) log2 11 =log 11

log 2¼ 3:46



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To solve logarithmic equations, we can sometimes write each side as a power of 10.


Solve for x: log3 x = ¡1

log3 x = ¡1

)log x

log 3= ¡1 f loga b =

log b

log ag

) log x = ¡1 log 3

) log x = log(3¡1)

) log x = log¡13

¢) x = 1

3

EXERCISE 31E

1 Solve for x using logarithms, giving answers to 4 significant figures:

a 10x = 80 b 10x = 8000 c 10x = 0:025

d 10x = 456:3 e 10x = 0:8764 f 10x = 0:000 179 2

2 Solve for x using logarithms, giving answers to 4 significant figures:

a 2x = 3 b 2x = 10 c 2x = 400

d 2x = 0:0075 e 5x = 1000 f 6x = 0:836

g (1:1)x = 1:86 h (1:25)x = 3 i (0:87)x = 0:001

j (0:7)x = 0:21 k (1:085)x = 2 l (0:997)x = 0:5

3 The weight of bacteria in a culture t hours after it has been established

is given by W = 2:5£ 20:04t grams.

After what time will the weight reach:

a 4 grams b 15 grams?

4 The population of bees in a hive t hours after it has

been discovered is given by P = 5000£ 20:09t.

After what time will the population reach:

a 15 000 b 50 000?

5 Answer the Opening Problem on page 625.

6 Show that log5 13 =log 13

log 5. Hence find log5 13.

7 Find, correct to 3 significant figures:

a log2 12 b log3 100 c log7 51 d log2(0:063)

8 Solve for x:

a log2 x = 2 b log5 x = ¡2 c log2(x+ 2) = 2 d log5(2x) = ¡1



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Review set 31A#endboxedheading

1 a On the same set of axes, sketch the graphs of y = 2x and y = log2 x.

b What transformation would map y = 2x onto y = log2 x?

c State the domain and range of y = log2 x.

2 Copy and complete:

a loga ax = b if y = bx then x = , and vice versa.

3 Find the value of:


¢c log2

p32 d log4 8

4 Write the following in terms of logarithms:

a y = 5x b y = 7¡x

5 Write the following as exponential equations:

a y = log3 x b T = 13 log4 n

6 Make x the subject of:

a y = log5 x b w = log(3x) c q =72x

3

7 Find the inverse function, f¡1(x) of:

a f(x) = 4£ 5x b f(x) = 2 log3 x

8 Solve for x, giving your answers correct to 5 significant figures:

a 4x = 100 b 4x = 0:001 c (0:96)x = 0:013 74

9 The population of a colony of wasps t days after discovery is given by P = 400£ 20:03t:

a How big will the population be after 10 days?

b How long will it take for the population to reach 1200 wasps?


a log 12¡ log 2 b 2 log 3 + log 4 c 2 log2 3 + 3 log2 5

11 Write as a logarithmic equation in base 10:

a y =a3

b2b M = 3

qab

12 Write as an equation without logarithms:

a logT = ¡x+ 3 b logN = 2 log c¡ log d

13 If log2 3 = a and log2 5 = b, find in terms of a and b:


¢c log2 10

14 Find y in terms of u and v if:

a log2 y = 4 log2 u b log5 y = ¡2 log5 v c log3 y = 12 log3 u+ log3 v

15 Find log3 15 correct to 4 decimal places.

16 Use a graphics calculator to solve, correct to 4 significant figures:

a 2x = 4¡ 3x b log x = 3¡x



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Review set 31B#endboxedheading

1 a On the same set of axes, sketch the graphs of y = 3x and y = log3 x.

b State the domain and range of each function.

2 Find the value of:

a log2p2 b log2

1p8

c logp3 27 d log9 27

3 Write the following in terms of logarithms:

a y = 4x b y = a¡n

4 Write the following as exponential equations:

a y = log2 d b M = 12 loga k

5 Make x the subject of:

a y = log3 x b T = logb(3x) c 3t = 5£ 2x+1

6 Find the inverse function f¡1(x) of:

a f(x) = 6x b f(x) = 12 log5 x

7 Solve for x, giving your answers correct to 4 significant figures:

a 3x = 3000 b (1:13)x = 2 c 2(2x) = 10

8

a What was the value of the banknote in 1970?

b What was the value of the banknote in 2005?

c


a log2 5 + log2 3 b log3 8¡ log3 2 c 2 log 5¡ 1 d 2 log2 5¡ 1

10 Write as a logarithmic equation in base 10:

a D =100

n2b G2 = c3d

11 Write as an equation without logarithms:

a logM = 2x+ 1 b logG = 12 log d¡ 1

12 If log3 7 = a and log3 4 = b, find in terms of a and b:

a log3¡47

¢b log3 28 c log3

¡73

¢13 Find y in terms of c and d if:

a log2 y = 2 log2 c b log3 y = 13 log3 c¡ 2 log3 d

14 Find log7 200 correct to 3 decimal places.

15 Use a graphics calculator to solve, correct to 4 significant figures:

a 3x = 0:6x+ 2 b log(2x) = (x¡ 1)(x¡ 4)


The value of a rare banknote has been modelled by V = 400£ 20:15t US dollars, where t is the

time in years since 1970.

When is the banknote expected to have a value of $100 000?


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Challenge#endboxedheading

1 Where is the error in the following argument?12 > 1

4

) log(12 ) > log(14)

) log(12 ) > log(12)2

) log(12 ) > 2 log(12 )

) 1 > 2 fdividing both sides by log(12 )g2 Solve for x:

a 4x ¡ 2x+3 + 15 = 0 Hint: Let 2x = m, say.

b log x = 5 log 2¡ log(x+ 4).

3 a Find the solution of 2x = 3 to the full extent of your calculator’s display.

b The solution of this equation is not a rational number, so it is irrational.

Consequently its decimal expansion is infinitely long and neither terminates nor recurs. Copy

and complete the following argument which proves that the solution of 2x = 3 is irrational

without looking at the decimal expansion.

Proof:

Assume that the solution of 2x = 3 is rational.

(The opposite of what we are trying to prove.)

) there exist positive integers p and q such that x =p

q, q 6= 0

Thus 2pq

= 3

) 2p = ::::::

and this is impossible as the LHS is ...... and the RHS is ...... no matter what values p and qmay take.

Clearly, we have a contradiction and so the original assumption is incorrect.

Consequently, the solution of 2x = 3 is ......

4 Prove that:

a the solution of 3x = 4 is irrational

b the exact value of log2 5 is irrational.



0 05 5

25

25

75

75

50

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95

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100 0 05 5

25

25

75

75

50

50

95

95

100

100

Y:\HAESE\IGCSE01\IG01_31\638IGCSE01_31.cdr Tuesday, 4 November 2008 12:05:20 PM PETER

Chapter 31 logarithms

Education

Transcript of Chapter 31 logarithms