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 =
p
x are inverse functions.
For example, 52
= 25 and
p
25 = 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
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Y:HAESEIGCSE01IG01_31625IGCSE01_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
4
1
2 = 2 , log4 2 = 1
2
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
= 10 000 f 7¡1
= 1
7 g 3¡3
= 1
27 h 27
1
3 = 3
i 5¡2
= 1
25 j 2¡ 1
2 = 1p
2
k 4
p
2 = 22:5
l 0:001 = 10¡3
2 Write an equivalent exponential statement for:
a log2 8 = 3 b log2 1 = 0 c log2
¡1
2
¢
= ¡1 d log2
p
2 = 1
2
e log2
³
1p
2
´
= ¡1
2 f logp
2 2 = 2 g logp
3 9 = 4 h log9 3 = 1
2
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
¡1
2
¢
j log3
¡1
9
¢
k log2(
p
2) l log2
¡p
8
¢
The logarithm of in base
is the exponent or power
of which gives .
81
3
3 81
Let log3 81 = x
626 Logarithms (Chapter 31)
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Y:HAESEIGCSE01IG01_31626IGCSE01_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 1
8
1
4
1
2 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 > 0g
² 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
4
2
-2
-4
y
O
y x¡=¡logx
627Logarithms (Chapter 31)
m log7
¡ 3
p
7
¢
n log2(4
p
2) o logp
2 2 p log2
³
1
4
p
2
´
q log10(0:01) r logp
2 4 s logp
3
¡1
3
¢
t log3
³
1
9
p
3
´
4 Rewrite as logarithmic equations:
a y = 4x
b y = 9x
c y = ax
d y = (
p
3)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
¡a
2
¢
g M = 1
2 log3 p h G = 5 logb m
i P = logp
b n
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
x
2
h w = log3(2x)
i z = 1
2 £ 3x
j y = 1
5 £ 4x
k D = 1
10 £ 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 = 3
x
2
So, f¡1
(x) = 3
x
2
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) = 1
2 (5x
) g f(x) = 3 log2 x h f(x) = 5 log3 x
i f(x) = logp
2 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.
x5-5
5
-5
y
OO
y¡=¡2x
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
¡x
4
¢
= 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 = ap
aq
= ap+q
x
y
=
ap
aq
= ap¡q
xn
= (ap
)n
= anp
) loga(xy) = p + q = loga x + loga y ffrom *g
loga
µ
x
y
¶
= p ¡ q = loga x ¡ loga y
loga(xn
) = np = n loga x
loga (xy) = loga x + loga y
loga
µ
x
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
¡ 64
125
¢
a log3 40
= log3(5 £ 8)
= log3 5 + log3 8
= p + q
b log3 25
= log3 52
= 2 log3 5
= 2p
c log3
¡ 64
125
¢
= log3
µ
82
53
¶
= log3 82
¡ log3 53
= 2 log3 8 ¡ 3 log3 5
= 2q ¡ 3p
RULES FOR LOGARITHMS [3.10]C
Simplify: a log2 7 ¡ 1
2 log2 3 + log2 5 b 3 ¡ log2 5
a log2 7 ¡ 1
2 log2 3 + log2 5
= log2 7 + log2 5 ¡ log2 3
1
2
= log2(7 £ 5) ¡ log2
p
3
= log2
³
35p
3
´
b 3 ¡ log2 5
= log2 23
¡ log2 5
= log2
¡8
5
¢
= 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
¡3
7
¢
c log2 49 d log2 27
e log2
¡7
9
¢
f log2(63) g log2
¡56
9
¢
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 = 1
2 log2 v ¡ 2 log2 u
i log6 y = 2 ¡ 1
3 log6 u j log3 y = 1
2 log3 u + log3 v + 1
4 Without using a calculator, simplify:
a
log2 16
log2 4
b
logp 16
logp 4
c
log5 25
log5
¡1
5
¢ d
logm 25
logm
¡1
5
¢
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(100000) = 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:HAESEIGCSE01IG01_31630IGCSE01_31.CDR Friday, 31 October 2008 9:46:05 AM PETER

2 Copy and complete: Number Number as a power of 10 log of number
p
10
3
p
10
p
1000
1p
10
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) = log x + log y
log
µ
x
y
¶
= log x ¡ log y
log(xn
) = n log x
Write as a single logarithm:
a log 2 + log 7 b log 6 ¡ log 3 c 2 + log 9 d
log 49
log
¡1
7
¢
a log 2 + log 7
= log(2 £ 7)
= log 14
b log 6 ¡ log 3
= log
¡6
3
¢
= log 2
c 2 + log 9
= log 102
+ log 9
= log(100 £ 9)
= log 900
d
log 49
log
¡1
7
¢
=
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 100:301
log 20 ¼ 1:301
) 20 ¼ 101:301
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Y:HAESEIGCSE01IG01_31631IGCSE01_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 1
2 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
¡1
8
¢
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:
a
log 8
log 2
b
log 9
log 3
c
log 4
log 8
d
log 5
log
¡1
5
¢
e
log(0:5)
log 2
f
log 8
log(0:25)
g
log 2b
log 8
h
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
¡1
7
¢
= ¡ log 7
d log
¡1
4
¢
= ¡2 log 2 e log
p
5 = 1
2 log 5 f log 3
p
2 = 1
3 log 2
g log
³
1p
3
´
= ¡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 3
4 log 2 + 1
4 log 3 + 1
2 .
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 = a3
b2
b y =
m
p
n
a y = a3
b2
) log y = log(a3
b2
)
) log y = log a3
+ log b2
) log y = 3 log a + 2 log b
b y =
m
p
n
) log y = log
µ
m
n
1
2
¶
) log y = log m ¡ log n
1
2
) log y = log m ¡ 1
2 log n
Write as powers of 10 using a = 10log a
:
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Y:HAESEIGCSE01IG01_31632IGCSE01_31.CDR Thursday, 30 October 2008 11:59:02 AM PETER

Write these equations without logarithms:
a log D = 2x + 1 b log N ¼ 1:301 ¡ 2x
a log D = 2x + 1
) D = 102x+1
or D = (100)x
£ 10
b log N ¼ 1:301 ¡ 2x
) N ¼ 101:301¡2x
) N ¼
101:301
102x
¼
20
102x
Write these equations without logarithms:
a log C = log a + 3 log b b log G = 2 log d ¡ 1
a log C = log a + 3 log b
= log a + log b3
= log(ab3
)
) C = ab3
b log G = 2 log d ¡ 1
= log d2
¡ log 101
= log
µ
d2
10
¶
) G =
d2
10
EXERCISE 31D.2
1 Write the following as logarithmic equations in base 10:
a y = ab2
b y =
a2
b
c y = d
p
p
d M = a2
b5
e P =
p
ab f Q =
p
m
n
g R = abc2
h T = 5
r
d
c
i M =
ab3
p
c
2 Write these equations without logarithms:
a log Q = x + 2 b log J = 2x ¡ 1 c log M = 2 ¡ x
d log P ¼ 0:301 + x e log R ¼ x + 1:477 f log K = 1
2 x + 1
3 Write these equations without logarithms:
a log M = log a + log b b log N = log d ¡ log e
c log F = 2 log x d log T = 1
2 log p
e log D = ¡ log g f log S = ¡2 log b
g log A = log B ¡ 2 log C h 2 log p + log q = log s
i ¡ log d + 3 log m = log n ¡ 2 log p j log m ¡ 1
2 log n = 2 log P
k log N = 1 + log t l log P = 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 LOGARITHMIC
EQUATIONS [3.10]
E
) log2 11 =
log 11
log 2
¼ 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 a
g
) log x = ¡1 log 3
) log x = log(3¡1
)
) log x = log
¡1
3
¢
) 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:
a log2 16 b log3
¡1
3
¢
c log2
p
32 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 = 1
3 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
b2
b M = 3
q
a
b
12 Write as an equation without logarithms:
a log T = ¡x + 3 b log N = 2 log c ¡ log d
13 If log2 3 = a and log2 5 = b, find in terms of a and b:
a log2 15 b log2
¡
12
3
¢
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 = 1
2 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 log2
p
2 b log2
1p
8
c logp
3 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 = 1
2 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) = 1
2 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
n2
b G2
= c3
d
11 Write as an equation without logarithms:
a log M = 2x + 1 b log G = 1
2 log d ¡ 1
12 If log3 7 = a and log3 4 = b, find in terms of a and b:
a log3
¡4
7
¢
b log3 28 c log3
¡7
3
¢
13 Find y in terms of c and d if:
a log2 y = 2 log2 c b log3 y = 1
3 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?
1
2 > 1
4
) log(1
2 ) > log(1
4 )
) log(1
2 ) > log(1
2 )2
) log(1
2 ) > 2 log(1
2 )
) 1 > 2 fdividing both sides by log(1
2 )g
2 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 2
p
q
= 3
) 2p
= ::::::
and this is impossible as the LHS is ...... and the RHS is ...... no matter what values p and q
may 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.
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Y:HAESEIGCSE01IG01_31638IGCSE01_31.cdr Tuesday, 4 November 2008 12:05:20 PM PETER

Chapter 31 logarithms

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Chapter 31 logarithms