# 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)

The logarithmic forms of the given expressions are as follows:

(i) 33 = 27

log3 27 = 3

(ii) 54 = 625

log625 = 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

(xii) – 2 = log2 (0.25)

(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)

(x) (xi)

(xii)

3. Find x in each of the following when:

(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

(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

(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

(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

(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

(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.

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

(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

(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:

(iv) log (26/51) – log (91/119)

(v) log (225/16) – 2 log (5/9) + log (2/3)5

(i)

(ii)
(iii)

(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

(iv) V = (4/3) πr3

(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)

(iv) V = (4 / 3) πr3

On taking log on both the sides, we get,

log V = log {(4 / 3)πr3}

⇒ log V = log 4 + log π + log r3 – log 3

⇒ log V = log 22 + log π + 3 log r – log 3

⇒ log V = 2 log 2 – log 3 + log π + 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

(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)

(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)

(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 {(x– 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

(i)
(ii)
(iii)
(iv)
(v) (vi)

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

(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

(i) (ii) (iii)

(iv)

(v)

5. Express log103 + 1 in terms of log 10 x.

6. State, true or false:

(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.

(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

(a)

(b)
(c)
(d)
(e)

8. If log x = p + q and log y = p – q, find the value of log 10x/y2 in terms of p and q.

9. If log a = p and log b = q, express a3/b2 in terms of p and q.

10. If log x = A + B and log y = A – B, express the value of log x2/10y in terms of A and B.

11. If log x = a and log y = b, write down

(i) 10a-1 in terms of x.

(ii) 102b in terms of y.

(i)

(ii)

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

(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

(a)

(b)

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

(i)

(ii)

(iii)

(iv)

16. If 2 log y – log x – 3 = 0, express x in terms of y.

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

(i)

(ii)

18. If log 4 = 0.6020, find the value of each of the following:

(i) log 8

(ii) log 2.5

(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

(i)

(ii)

20. If log 27 = 1.431, find the value of each of the following:

(i) log 9

(ii) log 300

(i)

(ii)

21. If x2 + y2 = 6xy, prove that log(x – y)/2 = 1/2 (log x + log y)

22. If x2 + y2 = 7xy, prove that log (x+ y/3) = 1/2(log x + log y)

23. Find x and y, if (log x)/(log 5) = (log 36)/(log 6) = (log 64)/(log y)

24. If log x2 – log √y = 1, express y in terms of x. Hence find y when x = 2.

25. If 2 log x + 1 = log 360, find

(i) x

(ii) log (2x – 2)

(iii) log (3x2 – 8)

(i)

(ii)

(iii)

26. If x + log 4 + 2 log 5 + 3 log 3 + 2 log 2 = log 108, find the value of x.

(a)

(b)

(c)

29. If a = log 3/5, b = log 5/4 and c = log √3/4 prove that 5a+b-c = 1

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

(i)

(ii)
(iii)
(iv)
(v)

31. Prove that log(1 + 2 + 3) = log 1 + log 2 + log 3 . Is it true for any three numbers x, y, z?

32. Prove that (log a)2 – (log b)2 = log (a/b).log (ab)

33. If a log b + b log a – 1 = 0, then prove that ba.bb = 10

34. If log (a + 1) = log (4a – 3) – log 3; find a.

35. Prove that log 10 125 = 3(1 – log 10 2)

36. Prove that (log p x)/(log pq x) q

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

(a)

(b)

38. If a = log.p2/qr, b = log .q2/rp, c = log.r2/pq, find the value of a + b + c.