Frank Solutions for Chapter 10 Logarithms Class 9 Mathematics ICSE
Exercise 10.1
1. Express each of the following in the logarithmic form:
(i) 33 = 27
(ii) 54 = 625
(iii) 90 = 1
(iv) (1/8) = 2-3
(v) 112 = 121
(vi) 3-2 = (1/9)
(vii) 10-4= 0.0001
(viii) 70 = 1
(ix) (1/3)4 = (1/81)
(x) 9– 4 = (1/6561)
Answer
The logarithmic forms of the given expressions are as follows:
(i) 33 = 27
log3 27 = 3
(ii) 54 = 625
log5 625 = 4
(iii) 90 = 1
log9 1 = 0
(iv) (1/8) = 2– 3
log2 (1/8) = – 3
(v) 112 = 121
log11 121 = 2
(vi) 3-2 = (1/9)
log3 (1/9) = – 2
(vii) 10-4 = 0.0001
log10 0.0001 = – 4
(viii) 70 = 1
log7 1 = 0
(ix) (1/3)4 = (1/81)
log1/3 (1/81) = 4
(x) 9-4 = (1/6561)
log9 (1/6561) = – 4
2. Express each of the following in the exponential form:
(i) log2 128 = 7
(ii) log3 81 = 4
(iii) log10 0.001 = – 3
(iv) log2 (1/32) = – 5
(v) logb a = c
(vi) log2 (1/2) = – 1
(vii) log5 a = 3
(viii) 
(xii) – 2 = log2 (0.25)
Answer
(i) log2 128 = 7
128 = 27
Hence, the exponential form of log2 128 = 7 is 27
(ii) log3 81 = 4
81 = 34
Hence, the exponential form of log3 81 = 4 is 34
(iii) log10 0.001 = – 3
0.001 = 10-3
Hence, the exponential form of log10 0.001 = –3 is 10-3
(iv) log2 (1/32) = – 5
(1/32) = 2– 5
Hence, the exponential form of log2 (1/32) = –5 is 2-5
(v) logb a = c
a = bc
Hence, the exponential form of logb a = c is bc
(vi) log2 (1/2) = – 1
(1/2) = 2– 1
Hence, the exponential form of log2 (1/2) = – 1 is 2-1
(vii) log5 a = 3
a = 53
Hence, the exponential form of log5 a = 3 is 53
(viii)
(ix)
(xi)
(i) logx 49 = 2
(ii) logx 125 = 3
(iii) logx 243 = 5
(iv) log8 x = (2 / 3)
(v) log7 x = 3
(vi) log4 x = – 4
(vii) log2 0.5 = x
(viii) log3 243 = x
(ix) log10 0.0001 = x
(x) log4 0.0625 = x
Answer
(i) logx 49 = 2
⇒ x2 = 49
⇒ x = 7
Therefore, the value of x is 7
(ii) logx 125 = 3
⇒ x3 = 125
⇒ x3 = 53
⇒ x = 5
Therefore, the value of x is 5
(iii) logx 243 = 5
⇒ x5 = 243
⇒ x5 = 35
⇒ x = 3
Therefore, the value of x is 3
(iv) log8 x = (2 / 3)
⇒ x = 82 / 3
Taking cube on both sides, we get,
x3 = 82
⇒ x3 = 64
⇒ x3 = 43
⇒ x = 4
Therefore, the value of x is 4
(v) log7 x = 3
⇒ x = 73
⇒ x = 343
Therefore, the value of x is 343
(vi) log4 x = – 4
⇒ x = 4– 4
⇒ x = (1/256)
Therefore, the value of x is (1/256)
(vii) log2 0.5 = x
⇒ 2x = 0.5
⇒ 2x = (1 / 2)
⇒ 2x = 2-1
⇒ x = –1
Therefore, the value of x is -1
(viii) log3 243 = x
⇒ 243 = 3x
⇒ 35 = 3x
⇒ x = 5
Therefore, the value of x is 5
(ix) log10 0.0001 = x
⇒ 0.0001 = 10x
⇒ 10x = 10– 4
⇒ x = – 4
Therefore, the value of x is -4
(x) log4 0.0625 = x
0.0625 = 4x
⇒ 4x = 4– 2
⇒ x = – 2
Therefore, the value of x is -2
4. Find the values of:
(i) log10 1000
(ii) log3 81
(iii) log5 3125
(iv) log2 128
(v) log1/5 125
(vi) log10 0.0001
(vii) log5 125
(viii) log8 2
(ix) log1/2 16
(x) log0.01 10
(xi) log3 81
(xii) log5 (1/25)
(xiii) log2 8
(xiv) loga a3
(xv) log0.1 10
(xvi)
Answer
(i) log10 1000
Let log10 1000 = x
10x = 1000
⇒ 10x = 103
We get,
x = 3
Hence, the value of x is 3
(ii) log3 81
Let log3 81 = x
⇒ 3x = 81
⇒ 3x= 34
We get,
x = 4
Hence, the value of x is 4
(iii) log5 3125
Let log5 3125 = x
5x = 3125
⇒ 5x = 55
We get,
x = 5
Hence, the value of x is 5
(iv) log2 128
Let log2 128 = x
⇒ 2x= 128
⇒ 2x = 27
We get,
x = 7
Hence, the value of x is 7
(v) log1/5 125
Let log1/5 125 = x
⇒ (1/5)x = 125
⇒ 5– x = 53
⇒ – x = 3
We get,
x = – 3
Hence, the value of x is -3
(vi) log10 0.0001
Let log10 0.0001 = x
⇒ 0.0001 = 10x
⇒ 10x = 10– 4
We get,
x = – 4
Hence, the value of x is -4
(vii) log5 125
Let log5 125 = x
⇒ 125 = 5x
⇒ 5x = 53
We get,
x = 3
Hence, the value of x is 3
(viii) log8 2
Let log8 2 = x
⇒ 2 = 8x
This can be written as,
(23)x = 2
⇒ 23x = 21
⇒ 3x = 1
We get,
x = (1/3)
Hence, the value of x is (1/3)
(ix) log1/2 16
Let log1/2 16 = x
⇒ 16 = (1/2)x
⇒ 2– x = 24
⇒ – x = 4
We get,
x = -4
Hence, the value of x is -4
(x) log0.01 10
Let log0.01 10 = x
⇒ (0.01)x = 10
⇒ (10-2)x = 101
⇒ 10-2x = 101
⇒ -2x = 1
We get,
x = (- 1/2)
Hence, the value of x is (-1/2)
(xi) log3 81
Let log3 81 = x
⇒ 3x = 81
⇒ 3x = 34
We get,
x = 4
Hence, the value of x is 4
(xii) log5 (1/25)
Let log5 (1/25) = x
⇒ 5x = (1/25)
⇒ 5x = 5-2
We get,
x = -2
Hence, the value of x is -2
(xiii) log2 8
Let log2 8 = x
⇒ 2x = 8
⇒ 2x = 23
We get,
x = 3
Hence, the value of x is 3
(xiv) loga a3
Let loga a3 = x
⇒ ax = a3
We get,
x = 3
Hence, the value of x is 3
(xv) log0.1 10
Let log0.1 10 = x
⇒ (0.1)x = 10
⇒ (10-1)x = 101
⇒ -x = 1
We get,
x = -1
Hence, the value of x is -1
(xvi)
5. If log10x = a, express the following in terms of x:
(i) 102a
(ii) 10a + 3
(iii) 10– a
(iv) 102a – 3
Answer
(i) 102a
log10 x = a
⇒ x = 10a
Hence,
102a = (10a)2
⇒ 102a = x2
(ii) 10a + 3
⇒ log10 x = a
⇒ x = 10a
Hence,
10a + 3 = 10a. 103
⇒ 10a + 3 = x.1000
⇒ 10a + 3 = 1000x
(iii) 10-a
⇒ log10 x = a
⇒ x = 10a
Hence,
10-a = x-1
⇒ 10-a = (1/x)
(iv) 102a – 3
⇒ log10 x = a
⇒ x = 10a
Hence,
102a – 3 = 102a.10-3
⇒ 102a – 3 = (10a)2 10– 3
⇒ 102a – 3 = (x2/1000)
6. If log10m = n, express the following in terms of m:
(i) 10n – 1
(ii) 102n – 1
(iii) 10– 3n
Answer
(i) 10n – 1
⇒ log10 m = n
⇒ m = 10n
Therefore,
10n – 1 = 10n.10– 1
⇒ 10n – 1 = (m/10) [m = 10n]
(ii) 102n + 1
⇒ log10 m = n
⇒ m = 10n
Therefore,
102n + 1 = 102n. 101
⇒ 102n + 1 = (10n)2 . 10
⇒ 102n + 1 = (m)2 . 10 [m = 10n]
⇒ 102n + 1 = 10m2
(iii) 10-3n
⇒ log10 m = n
⇒ m = 10n
Therefore,
10-3n = (10n)– 3
⇒ 10-3n = (m)-3 [m = 10n]
⇒ 10-3n = (1/m3)
7. If log10x = p, express the following in terms of x:
(i) 10p
(ii) 10p + 1
(iii) 102p – 3
(iv) 102 – p
Answer
(i) 10p
⇒ log10 x = p
We get,
x = 10p
(ii) 10p + 1
⇒ log10 x = p
⇒ x = 10p
Therefore,
10p + 1 = 10p.101
⇒ 10p + 1 = (x). 10 [x = 10p]
We get,
10p + 1 = 10x
(iii) 102p – 3
⇒ log10 x = p
⇒ x = 10p
Therefore,
102p – 3 = 102p. 10-3
⇒ 102p – 3 = (10p)2. 10-3
⇒ 102p – 3 = (x)2.10-3 [x = 10p]
102p – 3 = (x2/1000)
(iv) 102 – p
⇒ log10 x = p
⇒ x = 10p
Therefore,
102 – p = 102.10-p
⇒ 102 – p = 100.x-1
⇒ 102 – p = (100/x)
8. If log10x = a, log10y = b and log10 z = 2a – 3b, express z in terms of x and y.
Answer
log10 x = a
⇒ x = 10a
⇒ log10 y = b
⇒ y = 10b
log10 z = 2a – 3b
⇒ z = 102a – 3b
Therefore,
z = 102a – 3b
⇒ z = (10a)2.(10b)-3
⇒ z = (x)2.(y)-3
⇒ z = (x2/y3)
9. Express the following in terms of log 2 and log 3:
(i) log 36
(ii) log 54
(iii) log 144
(iv) log 216
(v) log 648
(vi) log 128
Answer
(i) log 36
⇒ log 36 = log (2 × 2 × 3 × 3)
⇒ log 36 = log (22 × 32)
⇒ log 36 = log 22 + log 32
We get,
log 36 = 2 log 2 + 2 log 3
(ii) log 54
⇒ log 54 = log (2 × 3 × 3 × 3)
⇒ log 54 = log (2 × 33)
⇒ log 54 = log 2 + log 33
We get,
log 54 = log 2 + 3 log 3
(iii) log 144
⇒ log 144 = log (24 × 32)
⇒ log 144 = log 24 + log 32
We get,
log 144 = 4 log 2 + 2 log 3
(iv) log 216
⇒ log 216 = log (23 × 33)
⇒ log 216 = log 23 + log 33
We get,
log 216 = 3 log 2 + 3 log 3
(v) log 648
⇒ log 648 = log (23 × 34)
⇒ log 648 = log 23 + log 34
We get,
log 648 = 3 log 2 + 4 log 3
(vi) log 128
⇒ log 128 = log (3 × 22)8
⇒ log 128 = 8 log (3 × 22)
⇒ log 128 = 8 {log 3 + log 22}
We get,
log 128 = 8 {log 3 + 2 log 2}
10. Express the following in terms of log 5 and / or log 2:
(i) log 20
(ii) log 80
(iii) log 125
(iv) log 160
(v) log 500
(vi) log 250
Answer
(i) log 20
⇒ log 20 = log (22 × 5)
⇒ log 20 = log 22 + log 5
We get,
log 20 = 2 log 2 + log 5
(ii) log 80
⇒ log 80 = log (24 × 5)
⇒ log 80 = log 24 + log 5
We get,
log 80 = 4 log 2 + log 5
(iii) log 125
⇒ log 125 = log 53
We get,
log 125 = 3 log 5
(iv) log 160
⇒ log 160 = log (25 × 5)
⇒ log 160 = log 25 + log 5
We get,
log 160 = 5 log 2 + log 5
(v) log 500
⇒ log 500 = log (22 × 53)
⇒ log 500 = log 22 + log 53
We get,
log 500 = 2 log 2 + 3 log 5
(vi) log 250 = log (53 × 2)
⇒ log 250 = log 53 + log 2
We get,
log 250 = 3 log 5 + log 2
11. Express the following in terms of log 2 and log 3:
(i)
(iv) log (26/51) – log (91/119)
(v) log (225/16) – 2 log (5/9) + log (2/3)5
Answer
(i)
(iv) log (26/51) – log (91/119)
= log {(2×13)/(3×17)} – log {(7×13)/(7×17)}
= log {(2×13)/(3×17)} – log (13/17)
= (log 13 + log 2 – log 3 – log 17) – (log 13 – log 17)
= log 13 + log 2 – log 3 – log 17 – log 13 + log 17
We get,
= log 2 – log 3
(v) log (225/16) – 2 log (5/9) + log (2/3)5
= log (225/16) – 2 log (5/9) + 5 log (2/3)
= log 225 – log 16 – 2 {log 5 – log 9} + 5 {log 2 – log 3}
= log (52×32) – log 24 – 2 {log 5 – log 32} + 5 {log 2 – log 3}
= log 52 + log 32 – 4 log 2 – 2 {log 5 – 2 log 3} + 5 {log 2 – log 3}
= 2 log 5 + 2 log 3 – 4 log 2 – 2 log 5 + 4 log 3 + 5 log 2 – 5 log 3
We get,
= log 2 + log 3
12. Write the logarithmic equation for:
(i) F = {G (m1m2)/d2}
(ii) E = (1/2) mv2
(iii)
(iv) V = (4/3) πr3
(v)
(i) F = {G (m1m2) / d2}
Taking log on both the sides, we get,
log F = log [{G (m1m2)} d2]
⇒ log F = log (Gm1m2) – log d2
We get,
log F = log G + log m1 + log m2 – 2 log d
(ii) E = (1 / 2) mv2
Taking log on both the sides, we get,
log E = log {(1 / 2) mv2}
⇒ log E = log (1 / 2) + log m + log v2
We get,
log E = log 1 – log 2 + log m + 2 log v
(iii)
On taking log on both the sides, we get,
log V = log {(4 / 3)Ï€r3}
⇒ log V = log 4 + log Ï€ + log r3 – log 3
⇒ log V = log 22 + log Ï€ + 3 log r – log 3
⇒ log V = 2 log 2 – log 3 + log Ï€ + 3 log r
(v)
13. Express the following as a single logarithm:
(i) log 18 + log 25 – log 30
(ii) log 144 – log 72 + log 150 – log 50
(iii) 2 log 3 – (1/2) log 16 + log 12
(iv) 2 + (1/2) log 9 – 2 log 5
(v) 2 log (9/5) – 3 log (3/5) + log (16/20)
(vi) 2 log (15/18) – log (25/162) + log (4 / 9)
(vii) 2 log (16/25) – 3 log (8/5) + log 90
(viii) (1/2) log 25 – 2 log 3 + log 36
(ix) log (81/8) – 2 log (3/5) + 3 log (2/5) + log (25/9)
(x) 3 log (5/8) + 2 log (8/15) – (1/2) log (25/81) + 3
Answer
(i) log 18 + log 25 – log 30
This can be written as,
= log (2×32) + log 52 – log (2×3×5)
= log 2 + log 32 + 2 log 5 – {log 2 + log 3 + log 5}
= log 2 + 2 log 3 + 2 log 5 – log 2 – log 3 – log 5
= log 3+ log 5
= log (3×5)
We get,
= log 15
(ii) log 144 – log 72 + log 150 – log 50
This can be written as,
= log (24×32) – log (23×32) + log (2×3×52) – log (2×52)
= log 24 + log 32 – {log 23 + log 32) + log 2 + log 3 + log 52 – {log 2 + log 52}
= 4 log 2 + 2 log 3 – 3 log 2 – 2 log 3 + log 2 + log 3 + 2 log 5 – log 2 – 2 log 5
We get,
= log 2 + log 3
= log (2 × 3)
We get,
= log 6
(iii) 2 log 3 – (1/2) log 16 + log 12
= 2 log 3 – (1/2) log 24 + log (22 × 3)
= 2 log 3 – (1/2) × 4 log 2 + log 22 + log 3
We get,
= 2 log 3 – 2 log 2 + 2 log 2 + log 3
= 3 log 3
= log 33
We get,
= log 27
(iv) 2 + (1/2) log 9 – 2 log 5
This can be written as,
= 2 + (1/2) log 32 – 2 log 5
= 2 log 10 + (1/2)×2 log 3 – 2 log 5
= log 102 + log 3 – log 52
= log 100 + log 3 – log 25
= log {(100 × 3)/25}
We get,
= log 12
(v) 2 log (9/5) – 3 log (3/5) + log (16/20)
= 2 log 9 – 2 log 5 – 3 log 3 + 3 log 5 + log 16 – log 20
This can be written as,
= 2 log (32) – 2 log 5 – 3 log 3 + 3 log 5 + log (42) – log (5×4)
= 4 log 3 – 2 log 5 – 3 log 3 + 3 log 5 + 2 log 4 – log 5 – log 4
= (4 – 3) log 3 + (-2 -1 + 3) log 5 + log 4
= log 3 + log 4
= log (3×4)
We get,
= log 12
(vi) 2 log (15/18) – log (25/162) + log (4/9)
= 2 log {5/(2×3)} – log {52/(2×34)} + log (22/32)
= 2 log 5 – 2 log 2 – 2 log 3 – {log 52 – log 2 – log 34} + log 22 – log 32
= 2 log 5 – 2 log 2 – 2 log 3 – 2 log 5 + log 2 + 4 log 3 + 2 log 2 – 2 log 3
We get,
= log 2
(vii) 2 log (16/25) – 3 log (8/5) + log 90
This can be written as,
= 2 log (24/52) – 3 log (23/5) + log (2 × 5 × 32)
= 2 log 24 – 2 log 52 – 3 {log 23 – log 5) + log 2 + log 5 + log 32
= 4×2 log 2 – 2×2 log 5 – 3×3 log2 + 3 log 5 + log 2 + log 5 + 2 log 3
= 8 log 2 – 4 log 5 – 9 log 2 + 3 log 5 + log 2 + log 5 + 2 log 3
= 2 log 3
= log 32
We get,
= log 9
(viii) (1/2) log 25 – 2 log 3 + log 36
= (1/2) log 52 – 2 log 3 + log (22×32)
= (1/2)×2 log 5 – 2 log 3 + log 22 + log 32
= log 5 – log 32 + 2 log 2 + log 32
= log 5 + 2 log 2
= log 5 + log 22
= log 5 + log 4
= log (5×4)
We get,
= log 20
(ix) log (81/8) – 2 log (3/5) + 3 log (2/5) + log (25/9)
= log (34/23) – 2 log (3/5) + 3 log (2/5) + log (52/32)
= log 34 – log 23 – 2 log 3 + 2 log 5 + 3 log 2 – 3 log 5 + log 52 – log 32
= 4 log 3 – 3 log 2 – 2 log 3 + 2 log 5 + 3 log 2 – 3 log 5 + 2 log 5 – 2 log 3
We get,
= log 5
(x) 3 log (5/8) + 2 log (8/15) – (1/2) log (25/81) + 3
This can be written as,
= 3 log (5/23) + 2 log {23/(3×5)} – ( /2) log (52/34) + 3 log 10
= 3 log 5 – 3 log 23 + 2 log 23 – 2 log 3 – 2 log 5 – (1/2) log 52 + (1/2) log 34 + 3 log (2×5)
= 3 log 5 – 3×3 log 2 + 2×3 log 2 – 2 log 3 – 2 log 5 – (1/2)×2 log 5 + (1/2)×4 log 3 + 3 log 2 + 3 log 5
= 3 log 5 – 9 log 2 + 6 log 2 – 2 log 3 – 2 log 5 – log 5 + 2 log 3 + 3 log 2 + 3 log 5
= 3 log 5
= log 53
We get,
= log 125
14. Simplify the following:
(i) 2 log 5 + log 8 – (1/2) log 4
(ii) 2 log 7 + 3 log 5 – log (49/8)
(iii) 3 log (32/27) + 5 log (125/24) – 3 log (625/ 243) + log (2/75)
(iv) 12 log (3/2) + 7 log (125/27) – 5 log (25/36) – 7 log 25 + log (16/3)
Answer
(i) 2 log 5 + log 8 – (1/2) log 4
= 2 log 5 + log 23 – (1/2) log 22
= 2 log 5 + 3 log 2 – (1/2)×2 log 2
= 2 log 5 + 3 log 2 – log 2
= 2 log 5 + 2 log 2
= 2 (log 5 + log 2)
= 2 log (5 × 2)
We get,
= 2 log 10
= 2 × 1
= 2
(ii) 2 log 7 + 3 log 5 – log (49 / 8)
= 2 log 7 + 3 log 5 – log 49 + log 8
= 2 log 7 + 3 log 5 – log 72 + log 23
= 2 log 7 + 3 log 5 – 2 log 7 + 3 log 2
= 3 log 5 + 3 log 2
= 3 (log 5 + log 2)
= 3 log (5×2)
= 3 log 10
We get,
= 3 × 1
= 3
(iii) 3 log (32/27) + 5 log (125/24) – 3 log (625/ 243) + log (2/75)
= 3 log (25/33) + 5 log {53/(23×3)} – 3 log (54/35) + log {2/(3×52)}
= 3 log 25 – 3 log 33 + 5 log 53 – 5 log 23 – 5 log 3 – 3 log 54 + 3 log 35 + log 2 – log 3 -log 52
= 15 log 2 – 9 log 3 + 15 log 5 – 15 log 2 – 5 log 3 – 12 log 5 + 15 log 3 + log 2 – log 3 – 2 log 5
= log 2 + log 5
(iv) 12 log (3/2) + 7 log (125/27) – 5 log (25/36) – 7 log 25 + log (16/3)
= 12 log (3/2) + 7 log (53/33) – 5 log {52/(22×32)} – 7 log 52 + log (24/3)
= 12 log 3 – 12 log 2 + 7 log 53 – 7 log 33 – 5 log 52 + 5 log 22 + 5 log 32 – 7 log 52 + log 24 – log 3
= 12 log 3 – 12 log 2 + 21 log 5 – 21 log 3 – 10 log 5 + 10 log 2 + 10 log 3 – 14 log 5 + 4 log 2 – log 3
We get,
= 2log 2 – 3 log 5
15. Solve the following:
(i) log (3 – x) – log (x – 3) = 1
(ii) log (x2 + 36) – 2 log x = 1
(iii) log 7 + log (3x – 2) = log (x + 3) + 1
(iv) log (x + 1) + log (x – 1) = log 11 + 2 log 3
(v) log4 x + log4 (x – 6) = 2
(vi) log8 (x2 – 1) – log8 (3x + 9) = 0
(vii) log (x + 1) + log (x – 1) = log 48
(viii) log2 x + log4 x + log16 x = (21/4)
Answer
(i) log (3 – x) – log (x – 3) = 1
This can be written as,
log {(3 – x)/(x – 3)} = 1
⇒ log {(3 – x)/(x – 3)} = log 10
⇒ (3 – x)/(x – 3) = 10
On calculating further, we get,
(3 – x) = 10 (x – 3)
⇒ (3 – x) = 10 x – 30
⇒ 11x = 33
We get,
x = 3
(ii) log (x2 + 36) – 2 log x = 1
This can be written as,
log (x2 + 36) – log x2 = 1
⇒ log {(x2 + 36)/x2} = 1
⇒ log {(x2 + 36)/x2} = log 10
⇒ {(x2 + 36)/x2} = 10
On further calculation, we get,
x2 + 36 = 10x2
⇒ 9x2 = 36
⇒ x2 = 4
We get,
x = 2
(iii) log 7 + log (3x – 2) = log (x + 3) + 1
log 7 + log (3x – 2) – log (x + 3) = 1
This can be written as,
log {7.(3x – 2)/(x + 3)} = log 10
⇒ {7. (3x – 2)/(x + 3)} = 10
On further calculation, we get,
21x – 14 = 10 (x + 3)
⇒ 21x – 10x = 30 + 14
⇒ 11x = 44
⇒ x = (44/11)
We get,
x = 4
(iv) log (x + 1) + log (x – 1) = log 11 + 2 log 3
This can be written as,
log {(x + 1) (x – 1)} = log 11 + log 32
⇒ log (x2 – 1) = log (11. 9)
⇒ log (x2 – 1) = log 99
⇒ x2 – 1 = 99
⇒ x2 = 99 + 1
⇒ x2 = 100
Hence,
x = 10 or -10
Here, negative value is rejected
Therefore,
x = 10
(v) log4 x + log4 (x – 6) = 2
⇒ log4 {x (x – 6)} = 2 log4 4
⇒ log4 {x2 – 6x} = log4 42
⇒ x2 – 6x = 16
⇒ x2 – 6x – 16 = 0
⇒ x2 – 8x + 2x – 16 = 0
⇒ x (x – 8) + 2 (x – 8) = 0
⇒ (x – 8) (x + 2) = 0
We get,
x = 8 or -2
Negative value is rejected
Hence,
x = 8
(vi) log8 (x2 – 1) – log8 (3x + 9) = 0
⇒ log8 {(x2 – 1)/(3x + 9)} = log81
⇒ (x2 – 1)/(3x + 9) = 1
⇒ x2 – 1 = 3x + 9
On calculating further, we get,
x2 – 3x – 10 = 0
⇒ x2 – 5x + 2x – 10 = 0
⇒ x (x – 5) + 2 (x – 5) = 0
⇒ (x – 5) (x + 2) = 0
⇒ x = 5 or x = -2
negative value is rejected,
Hence,
x = 5
(vii) log (x + 1) + log (x – 1) = log 48
This can be written as,
log {(x + 1) (x – 1)} = log 48
⇒ log (x2 – 1) = log 48
⇒ x2 – 1 = 48
⇒ x2 = 48 + 1
⇒ x2 = 49
⇒ x = 7 or -7
neglecting the negative value
Therefore,
x = 7
(viii) log2 x + log4 x + log16 x = (21 / 4)
⇒ (1/logx2) + (1/logx22) + (1/logx24) = (21/4)
⇒ (1/logx2) + (1/2 logx2) + (1/4 logx2) = (21/4)
Taking (1/logx2) as common, we get,
(1/logx2) {1 + (1/2) + (1/4)} = (21/4)
We get,
(1/logx 2) (7/4) = (21/4)
⇒ logx 2 = (7/4) × (4/21)
⇒ logx 2 = (1/3)
So,
x1/3 = 2
We get,
x = 23
⇒ x = 8
Exercise 10.2
1. Express the following in terms of log 2 and log 3:
(i) log 36
(ii) log 54
(iii) log 144
(iv) log 216
(v) log 216
(vi) log128
2. Express the following in terms of log 5 and / or log 2:
(i) log 20
(ii) log 80
(iii) log 125
(iv) log 160
(v) log 500
(vi) log 250
Answer
(i) log 20
= log 20
= log (22×5)
= log 22 + log 5
= 2 log 2 + log 5
(ii) log 80
log 80
= log (24 ×5)
= log 24 + log 5
= 4 log 2 + log 5
(iii) log 125
= log 125
= log 53
= 3 log 5
(iv) log 160
log 160
= log (25 ×5)
= log 25 + log 5
= 5 log 2 + log 5
(v) log 500
log 500
= log (25 × 53)
= log 22 + log 53
= 2 log 2 + 3 log 5
(vi) log 250
= log (53 × 2)
= log 53 + log 2
= 3 log 5 + log 2
3. Express the following in terms of log 2 and log 3:
(vi) log 26/51 – log 91/119
(v) log 225/6 – 2log 5/9 + log (2/3)5
Answer
(i)
(ii)
(iii)
(iv)
(v)
4. Solve for x:
(i) log (x + 5) = 1
(ii) log 27/log 243 = x
(iii) log 81/log 9 = x
(iv) log 121/log 11 = log x
(v) log 125/log 5 = log x
(vi) log 128/log 32 = x
(vii) log 1331/log 11 = log x
(viii) log 289/log 17 = log x
Answer
(i)
(ii)
5. Express log103 + 1 in terms of log 10 x.
Answer:
(a) log (x + y) = log xy
(b) log 4 × log 1 = 0
(c) log b a = - log a b
(d) If log 49/log 7 = log y, then y = 100.
Answer
(a) False, since log xy = log x + log y
(b) True, since log 1 = 0 and anything multiplied by 0 is 0.
(c) False, since log b a = 1/(log a b).
(d) True,
7. (a) If log 16 = a, log 9 = b and log 5 = c, evaluate the following in terms of a, b, c: log 12
(b) If log 16 = a, log 9 = b and log 5 = c, evaluate the following in terms of a, b, c: log 75
(c) If log 16 = a, log 9 = b and log 5 = c, evaluate the following in terms of a, b, c: log 720
(d) If log 16 = a, log 9 = b and log 5 = c, evaluate the following in terms of a, b, c: log 2.25
(e) If log 16 = a, log 9 = b and log 5 = c, evaluate the following in terms of a, b, c: log 2.1/4
Answer
(a)
Answer
9. If log a = p and log b = q, express a3/b2 in terms of p and q.
Answer
10. If log x = A + B and log y = A – B, express the value of log x2/10y in terms of A and B.
Answer
11. If log x = a and log y = b, write down
(i) 10a-1 in terms of x.
(ii) 102b in terms of y.
Answer
(i)
12. If log 3 m = x and log 3 n = y, write down
(i) 32x-3 in terms of m
(ii) 31-2y+3x in terms of m and n
Answer
(i)
(ii)
14. (a) If log 10 25 = x and log 10 27 = y; evaluate without using logarithmic tables, in terms of x and y: log10 5
(b) If log 10 25 = x and log 10 27 = y; evaluate without using logarithmic tables, in terms of x and y: log10 3
Answer
(a)
15. If log 2 = 0.3010, log 3 = 0.4771 and log 5 = 0.6990, find the values of :
(i) log 18
(ii) log 45
(iii) log 540
(iv) log
Answer
(i)
(iii)
(iv)
16. If 2 log y – log x – 3 = 0, express x in terms of y.
Answer
17. If log 2 = x and log 3 = y, find the value of each of the following in terms of x and y:
(i) log 60
(ii) log 1.2
Answer
(i)
(ii)
18. If log 4 = 0.6020, find the value of each of the following:
(i) log 8
(ii) log 2.5
Answer
(i) log 8
log 4 = 0.6020
⇒ log 22 = 0.6020
⇒ 2 log 2 = 0.6020
⇒ log 2 = 0.6020/2
= 0.3010
∴ log 8 = log log 23 = 3 log 2 = 3 × 0.3010
= 0.9030
(ii) log 2.5
log 2.5 = log (10/4)
= log 10 – log 4
= 1 = 0.6020
= 0.3980
19. If log 8 = 0.90, find the value of each of the following:;
(i) log 4
(ii) log
Answer
(i)
(ii)
20. If log 27 = 1.431, find the value of each of the following:
(i) log 9
(ii) log 300
Answer
(i)
(ii)
21. If x2 + y2 = 6xy, prove that log(x – y)/2 = 1/2 (log x + log y)
Answer
22. If x2 + y2 = 7xy, prove that log (x+ y/3) = 1/2(log x + log y)
Answer
23. Find x and y, if (log x)/(log 5) = (log 36)/(log 6) = (log 64)/(log y)
Answer
24. If log x2 – log √y = 1, express y in terms of x. Hence find y when x = 2.
Answer
25. If 2 log x + 1 = log 360, find
(i) x
(ii) log (2x – 2)
(iii) log (3x2 – 8)
Answer
(i)
(ii)
(iii)
26. If x + log 4 + 2 log 5 + 3 log 3 + 2 log 2 = log 108, find the value of x.
Answer
27. Simplify:
(a) log a2 + log a-1
(b) log b ÷ log b2
Answer
(a)
28. Find the value of:
(a) (log √8)/8
(b)
29. If a = log 3/5, b = log 5/4 and c = log √3/4 prove that 5a+b-c = 1
Answer
30. Express each of the following in a form free from logarithm:
(i) 3 log x – 2 log y = 2
(ii) 2 log x + 3 log y = log a
(iii) m log x – n log y = 2 log 5
(iv) 2 log x + ½ log y = 1
(v) 5 log m – 1 = 3 log n
Answer
(i)
31. Prove that log(1 + 2 + 3) = log 1 + log 2 + log 3 . Is it true for any three numbers x, y, z?
Answer
32. Prove that (log a)2 – (log b)2 = log (a/b).log (ab)
Answer
Answer
34. If log (a + 1) = log (4a – 3) – log 3; find a.
Answer
35. Prove that log 10 125 = 3(1 – log 10 2)
Answer
36. Prove that (log p x)/(log pq x) q
Answer
37. Prove that :
(a) 1/log 2 30 + 1/log 3 30 + 1/log 5 30 = 1
(b) 1/log 8 36 + 1/log 9 36 + 1/log 18 36 = 2
Answer
(a)
(b)
38. If a = log.p2/qr, b = log .q2/rp, c = log.r2/pq, find the value of a + b + c.
Answer
39. If a = log 20 b = log 25 and 2 log (p – 4) = 2a – b, find the value of ‘p’.
Answer
