81
Logarithms
Introduction
Logarithms were developed for making complicated calculations simple. However, with the
advent of computers and hand calculators, doing calculations with the use of logarithms is
no longer necessary. But still, logarithmic and exponential equations and functions are very
common in mathematics.
8.1 Logarithms
To learn the concept of logarithm, consider the equality 23 = 8, another way of writing this is
log2 8 = 3.
It is read as “logarithm (abbreviated ‘log’) of 8 to the base 2 is equal to 3”. Thus, 23 = 8 is
equivalent to log2 8 = 3. In general, we have:
Definition. If a is any positive real number (except 1), n is any rational number and an = b, then
n is called logarithm of b to the base a. It is written as log a b (read as log of b to the base a). Thus,
an = b if and only if log a b = n.
a n = b is called the exponential form and log a b = n is called the logarithmic form.
For example:
(i) 32 = 9, ∴
log3 9 = 2.
4
(ii) 5 = 625, ∴ log5 625 = 4.
0
(iii) 7 = 1, ∴
log7 1 = 0.
(iv) 2–3 =
(v) (10)–2 =
Remarks
1
23
=
1
,
8
1
= ·01, 100
∴
1
8
log2 = – 3.
∴ log10 (·01) = – 2.
❐Since a is any positive real number (except 1), a n is always a positive real number for
every rational number n i.e. b is always a positive real number, therefore, logarithm of only
positive real numbers are defined.
❐Since a0 = 1, log a 1 = 0 and a1 = a, log a a = 1.
Thus, remember that
(i) log a 1 = 0
(ii) log a a = 1
where a is any positive real number (except 1).
❐If log a x = log a y = n (say), then x = a n and y = a n, so x = y.
Thus, remember that
log a x = log a y ⇒ x = y.
❐Logarithms to the base 10 are called common logarithms.
❐If no base is given, the base is always taken as 10.
For example, log 2 = log10 2.
Illustrative Examples
Example 1. Convert the following to logarithmic form:
(i) (10) 4 = 10000
(ii) 3–5 = x
(iii) (0·3)3 = 0·027.
Solution. (i) (10)4 = 10000 ⇒ log10 10000 = 4.
(ii) 3–5 = x ⇒ log3 x = – 5.
(iii) (0·3)3 = 0·027 ⇒ log0·3 (0·027) = 3.
Example 2. Convert the following to exponential form:
(i) log3 81 = 4
(ii) log8 32 =
5
3
(iii) log10 (0·1) = – 1.
Solution. (i) log3 81 = 4 ⇒ 34 = 81.
5
⇒ (8)5/3 = 32.
3
(ii) log 8 32 =
(iii) log10 (0·1) = – 1 ⇒ (10) – 1 = 0·1.
Example 3. Find the value of the following (by converting to exponential form):
(i) log2 16
(ii) log16 2
(iii) log3 1
3
(iv) log
2
8 (v) log5 (0·008).
Solution. (i) Let log2 16 = x ⇒ 2x = 16 ⇒ 2x = (2)4 ⇒ x = 4, ∴
log2 16 = 4.
(ii) Let log16 2 = x ⇒ 16 x = 2 ⇒ (2 4)x = 2
⇒
2 4x = 21 ⇒ 4x = 1 ⇒ x = 1 ,
∴
log 16 2
4
= 1.
4
(iii) Let log3
1
1
= x ⇒ 3x =
⇒ 3x = (3)–1 ⇒ x = – 1,
3
3
∴
1
= – 1.
3
log3 (iv) Let log
2
8 = x ⇒ ( 2 )x = 8 ⇒ (21/2)x = 23
⇒
2x/2 = 23 ⇒ x = 3 ⇒ x = 6,
∴
log
2
2
8 = 6.
(v) Let log5 (0·008) = x ⇒ 5 x = 0·008
8
1
⇒ 5 x =
⇒ 5 x = (5) – 3 ⇒ x = – 3,
1000
125
⇒
5 x =
∴
log5 (0·008) = – 3.
logarithms
1139
Example 4. Find the value of x in each of the following:
(i) log2 x = 5
(ii) log4 x = 2·5
(iii) log64 x =
2
3
(iv) log
3
x = 4.
Solution. (i) log2 x = 5 ⇒ x = 25 ⇒ x = 32.
(ii) log4 x = 2·5 ⇒ x = 42·5 ⇒ x = (22)5/2
⇒
x= 2
(iii) log64 x =
2×
5
2
⇒ x = 25 ⇒ x = 32.
2
⇒ x = (64)2/3 ⇒ x = (43)2/3
3
3×
2
3
⇒ x = 42 ⇒ x = 16.
⇒
x= 4
(iv) log
3
⇒
x = 32 ⇒ x = 9.
x = 4 ⇒ x = ( 3 )4 ⇒ x = (31/2)4
Example 5. Solve for x:
(i) logx 243 = – 5
(ii) logx 16 = 2
(iii) log9 27 = 2 x + 3
(iv) log (3 x – 2) = 2
(v) logx 64 = 3 2
(vi) log2 (x2 – 4) = 5.
Solution. (i) Given logx 243 = – 5 ⇒ x – 5 = 243
⇒
⇒
n

 1 
 a− n =   
 a 


5
 1
x – 5 = 35 ⇒   = 35
x
1
1
=3 ⇒x= .
x
3
(ii)Given logx 16 = 2 ⇒ x2 = 16 ⇒ x = ± 4.
But the base of a logarithm cannot be negative, so x = – 4 is rejected.
∴
The solution of the given equation is x = 4.
(iii)Given log9 27 = 2 x + 3 ⇒ 9 2 x + 3 = 27 ⇒ (32) 2 x + 3 = 33
⇒
32(2x + 3) = 33 ⇒ 2 (2 x + 3) = 3 ⇒ 4 x + 6 = 3
⇒
4 x = – 3 ⇒ x = – 3 .
4
(iv)Given log (3 x – 2) = 2 ⇒ log10 (3 x – 2) = 2
10 2
⇒
3 x – 2 =
⇒
3 x = 102 ⇒ x = 34.
[If no base is given, we take it as 10.]
⇒ 3 x – 2 = 100
(v)Given logx 64 = 3 ⇒ x3/2 = 64
2
(64)2/3
⇒
x=
⇒
x = 16.
= (26)2/3 = 26 × 2/3 = 24
(vi)Given log2 (x2 – 4) = 5 ⇒ x2 – 4 = 25
⇒
x2 – 4 = 32 ⇒ x2 = 36 ⇒ x = ± 6.
Example 6. Given log10 x = a, log10 y = b,
(i) write down 10 a – 1 in terms of x.
(ii) write down 102b in terms of y.
(iii) if log10 P = 2 a – b, express P in terms of x and y.
Solution. Given log10 x = a ⇒ 10a = x, and log10 y = b ⇒ 10b = y.
(i) 10 a – 1 = 10 a . 10 – 1 = 10 a . 1140
1
x
=
.
10
10
Understanding ICSE mathematics – Ix
(ii) 10 2 b = (10 b) 2 = y 2.
(iii) log10 P = 2 a – b ⇒ 10 2 a – b = P,
∴
2
P = 10 2 a . 10 – b = (10a)2 . 1 = x .
b
y
10
Example 7. If log3 x = a, find 81a – 1 in terms of x.
Solution. Given log3 x = a ⇒ 3a = x
…(i)
∴ 81 a – 1 = (34)a – 1 = 34a – 4 = 34a × 3– 4
=
( 3 a )4
34
=
x4
81
[Using (i)]
Example 8. If log2 x = a and log3 y = a, find 122a – 1 in terms of x and y.
Solution. Given log2 x = a and log3 y = a
⇒ 2a = x and 3a = y
∴ 122a – 1=
(22
=
24a – 2
=
×
…(i)
3)2a – 1
×
=
32a – 1
(2 a )4 × (3 a )2
22 × 31
=
22(2a – 1)
=
24a
×
32a – 1
× 2–2 × 32a × 3–1
x4y2
12
[Using (i)]
Exercise 8.1
1. Convert the following to logarithmic form:
(i) 52 = 25
(ii) a5 = 64
(iii) 7x = 100
(v) 61 = 6
1
(vi) 3–2 = (vii) 10–2 = 0·01
9
(ii) log3 81 = 4
2
(iv) log8 4 = 3
(v) log8 32 =
5
3
(iii) log3 1
= – 1
3
(vi) log10 (0·001) = – 3
 1
(vii) log2 0·25 = – 2
(viii) loga   = – 1.
a
3.By converting to exponential form, find the values of:
(i) log2 16
(ii) log5 125
(iii) log4 8
(v) log10 (·01)
(vi) log7 1 (vii) log.5 256
(viii) (81)3/4 = 27.
2. Convert the following into exponential form:
(i) log2 32 = 5
(iv) 90 = 1
7
(iv) log9 27
(viii) log2 0·25.
4.Solve the following equations for x:
(i) log3 x = 2
(ii) logx 25 = 2
(iii) log10 x = – 2
(iv) log4 x = 1 2
(v) logx 11 = 1
(vi) logx
(vii) log81 x = 3 (viii) log9 x = 2·5
2
(x) log
5
(xi) logx 0·001 = – 3
x = 2
(xiii) log4 (2 x + 3) = 3 (xiv) log 3 2 x = 3
(xvi) log x = – 1
(xvii) log (2x – 3) = 1
2
1
= – 1
4
(ix) log4 x = – 1·5
(xii) log
3
(x + 1) = 2
(xv) log2 (x2 – 1) = 3
(xviii) log x = – 2, 0,
1
.
3
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