Logarithms 31 Contents: 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 = x 2 and y = p x are inverse functions. For example, 5 2 = 25 and p 25 = 5. If y = a x then we say “x is the logarithm of y in base a”, and write this as x = log a y. LOGARITHMS IN BASE a [3.10] A Opening problem 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 IGCSE01 Y:\HAESE\IGCSE01\IG01_31\625IGCSE01_31.CDR Tuesday, 18 November 2008 11:10:27 AM PETER
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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)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:\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.
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 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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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 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.
628 Logarithms (Chapter 31)
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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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, 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
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
µ82
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 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.
630 Logarithms (Chapter 31)
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
Example 7 Self Tutor
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
631Logarithms (Chapter 31)
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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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