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

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

(viii) 

(ix)
(x)
(xi)

(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)

(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

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)

(ii)
(iii)

(iv) log (26/51) – log (91/119)

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

Answer

(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

(iii)

(iv) V = (4/3) πr3

(v)

Answer

(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

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 {(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

Answer
(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

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: 

(i)
(ii)
(iii)

(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)

(iii)
(iv)
(v)
(vi)
(vii)
(viii)

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

Answer:


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.

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)

(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.

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) 

(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

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)

(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

Answer

(i)

(ii)

(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)

(b)

28. Find the value of:

(a) (log √8)/8

(b)

(c)

Answer

(a) 

(b)

(c)

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)

(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?

Answer


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

Answer


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

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

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