Frank Solutions for Class 9 Maths Chapter 10 Logarithms

Frank Solutions for Class 9 Maths Chapter 10 Logarithms help students to understand the concepts covered in this chapter easily. The solutions are created by subject experts at BYJU’S with the aim of clearing students’ doubts instantly, at their convenience. Students who refer to these solutions during exam preparation will improve their problem-solving skills.

Chapter 10 mainly deals with the study of Logarithms. Students can cross-check their answers by referring to these solutions while solving exercise problems. Regular practice of these solutions will build confidence in students who find difficulty in solving textbook problems. In order to prepare well for the annual exams, students can download Frank Solutions for Class 9 Maths Chapter 10 Logarithms PDF from the link given below and practise regularly.

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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)

Solution:

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)

log_{1 / 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) log_b a = c

(vi) log₂ (1 / 2) = – 1

(vii) log₅ a = 3

(viii)

(ix)

(x)

(xi)

(xii) – 2 = log₂ (0.25)

Solution:

(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) log_b a = c

a = b^c

Hence, the exponential form of log_b a = c is b^c

(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)

27 =

Hence, the exponential form of
is

(ix)

Hence, the exponential form of
is 25^{1 / 4}

(x)

p = a^q

Hence, the exponential form of
is a^q

(xi)

Hence, the exponential form of
is

(xii) -2 =

We get,

2^-2 = 0.25

3. Find x in each of the following when:

(i) log_x 49 = 2

(ii) log_x 125 = 3

(iii) log_x 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

Solution:

(i) log_x 49 = 2

x² = 49

x = 7

Therefore, the value of x is 7

(ii) log_x 125 = 3

x³ = 125

x³ = 5³

x = 5

Therefore, the value of x is 5

(iii) log_x 243 = 5

x⁵ = 243

x⁵ = 3⁵

x = 3

Therefore, the value of x is 3

(iv) log₈ x = (2 / 3)

x = 8^{2 / 3}

Taking the 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

2^x = 0.5

2^x = (1 / 2)

2^x = 2^-1

x = – 1

Therefore, the value of x is -1

(viii) log₃ 243 = x

243 = 3^x

3⁵ = 3^x

x = 5

Therefore, the value of x is 5

(ix) log₁₀ 0.0001 = x

0.0001 = 10^x

10^x = 10^{– 4}

x = – 4

Therefore, the value of x is -4

(x) log₄ 0.0625 = x

0.0625 = 4^x

4^x = 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) log_{1 / 5} 125

(vi) log₁₀ 0.0001

(vii) log₅ 125

(viii) log₈ 2

(ix) log_{1 / 2} 16

(x) log_0.01 10

(xi) log₃ 81

(xii) log₅ (1 / 25)

(xiii) log₂ 8

(xiv) log_a a³

(xv) log_0.1 10

(xvi)

Solution:

(i) log₁₀ 1000

Let log₁₀ 1000 = x

10^x = 1000

10^x = 10³

We get,

x = 3

Hence, the value of x is 3

(ii) log₃ 81

Let log₃ 81 = x

3^x = 81

3^x = 3⁴

We get,

x = 4

Hence, the value of x is 4

(iii) log₅ 3125

Let log₅ 3125 = x

5^x = 3125

5^x = 5⁵

We get,

x = 5

Hence, the value of x is 5

(iv) log₂ 128

Let log₂ 128 = x

2^x = 128

2^x = 2⁷

We get,

x = 7

Hence, the value of x is 7

(v) log_{1 / 5} 125

Let log_{1 / 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 = 10^x

10^x = 10^{– 4}

We get,

x = – 4

Hence, the value of x is -4

(vii) log₅ 125

Let log₅ 125 = x

125 = 5^x

5^x = 5³

We get,

x = 3

Hence, the value of x is 3

(viii) log₈ 2

Let log₈ 2 = x

2 = 8^x

This can be written as,

(2³)^x = 2

2^3x = 2¹

3x = 1

We get,

x = (1 / 3)

Hence, the value of x is (1 / 3)

(ix) log_{1 / 2} 16

Let log_{1 / 2} 16 = x

16 = (1 / 2)^x

2^{– x} = 2⁴

– x = 4

We get,

x = -4

Hence, the value of x is -4

(x) log_0.01 10

Let log_0.01 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

3^x = 81

3^x = 3⁴

We get,

x = 4

Hence, the value of x is 4

(xii) log₅ (1 / 25)

Let log₅ (1 / 25) = x

5^x = (1 / 25)

5^x = 5^-2

We get,

x = -2

Hence, the value of x is -2

(xiii) log₂ 8

Let log₂ 8 = x

2^x = 8

2^x = 2³

We get,

x = 3

Hence, the value of x is 3

(xiv) log_a a³

Let log_a a³ = x

a^x = a³

We get,

x = 3

Hence, the value of x is 3

(xv) log_0.1 10

Let log_0.1 10 = x

(0.1)^x = 10

(10^-1)^x = 10¹

-x = 1

We get,

x = -1

Hence, the value of x is -1

(xvi)

Let
= x

⁼ 3√3

3^{x / 2} = 3^{1 + 1 / 2}

3^{x / 2} = 3^{3 / 2}

(x / 2) = (3 / 2)

We get,

x = 3

Hence, the value of x is 3

5. If log₁₀ x = a, express the following in terms of x:

(i) 10^2a

(ii) 10^{a + 3}

(iii) 10^{– a}

(iv) 10^{2a – 3}

Solution:

(i) 10^2a

log₁₀ x = a

x = 10^a

Hence,

10^2a = (10^a)²

10^2a = x²

(ii) 10^{a + 3}

log₁₀ x = a

x = 10^a

Hence,

10^{a + 3} = 10^a. 10³

10^{a + 3} = x.1000

10^{a + 3} = 1000x

(iii) 10^-a

log₁₀ x = a

x = 10^a

Hence,

10^-a = x^-1

10^-a = (1 / x)

(iv) 10^{2a – 3}

log₁₀ x = a

x = 10^a

Hence,

10^{2a – 3} = 10^2a.10^-3

10^{2a – 3} = (10^a)² 10^{– 3}

10^{2a – 3} = (x² / 1000)

6. If log₁₀ m = n, express the following in terms of m:

(i) 10^{n – 1}

(ii) 10^{2n – 1}

(iii) 10^{– 3n}

Solution:

(i) 10^{n – 1}

log₁₀ m = n

m = 10ⁿ

Therefore,

10^{n – 1} = 10ⁿ.10^{– 1}

10^{n – 1} = (m / 10) [m = 10ⁿ]

(ii) 10^{2n + 1}

log₁₀ m = n

m = 10ⁿ

Therefore,

10^{2n + 1} = 10²ⁿ. 10¹

10^{2n + 1} = (10ⁿ)² . 10

10^{2n + 1} = (m)² . 10 [m = 10ⁿ]

10^{2n + 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 log₁₀ x = p, express the following in terms of x:

(i) 10^p

(ii) 10^{p + 1}

(iii) 10^{2p – 3}

(iv) 10^{2 – p}

Solution:

(i) 10^p

log₁₀ x = p

We get,

x = 10^p

(ii) 10^{p + 1}

log₁₀ x = p

x = 10^p

Therefore,

10^{p + 1} = 10^p.10¹

10^{p + 1} = (x). 10 [x = 10^p]

We get,

10^{p + 1} = 10x

(iii) 10^{2p – 3}

log₁₀ x = p

x = 10^p

Therefore,

10^{2p – 3} = 10^2p. 10^-3

10^{2p – 3} = (10^p)². 10^-3

10^{2p – 3} = (x)².10^-3 [x = 10^p]

10^{2p – 3} = (x² / 1000)

(iv) 10^{2 – p}

log₁₀ x = p

x = 10^p

Therefore,

10^{2 – p} = 10².10^-p

10^{2 – p} = 100.x^-1

10^{2 – p} = (100 / x)

8. If log₁₀ x = a, log₁₀ y = b and log₁₀ z = 2a – 3b, express z in terms of x and y.

Solution:

log₁₀ x = a

x = 10^a

log₁₀ y = b

y = 10^b

log₁₀ z = 2a – 3b

z = 10^{2a – 3b}

Therefore,

z = 10^{2a – 3b}

z = (10^a)².(10^b)^-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

Solution:

(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

Solution:

(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

Solution:

(i)

= log (144)^{1 / 3}

= (1 / 3) log 144

= (1 / 3) log (2⁴ × 3²)

= (1 / 3) log 2⁴ + (1 / 3) log 3²

We get,

= (4 / 3) log 2 + (2 / 3) log 3

(ii)

= log (216)^{1 / 5}

= (1 / 5) log 216

= (1 / 5) log (2³ × 3³)

= (1 / 5) log 2³ + (1 / 5) log 3³

We get,

= (3 / 5) log 2 + (3 / 5) log 3

(iii)

= log (648)^{1 / 4}

= (1 / 4) log 648

= (1 / 4) log (2³ × 3⁴)

= (1 / 4) log 2³ + (1 / 4) log 3⁴

= (3 / 4) log 2 + (4 / 4) log 3

We get,

= (3 / 4) log 2 + 1 log 3

= (3 / 4) log 2 + log 3

(iv) log (26 / 51) – log (91 / 119)

= log {(2 × 13) / (3 × 17)} – log {(7 x 13) / (7 x 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)

Solution:

(i) F = {G (m₁m₂) / d²}

Taking log on both 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 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)

n = (M.g / m.l)^{1 / 2}

On taking log on both sides, we get,

log n = log (M.g / m.l)^{1 / 2}

log n = (1 / 2) log (M.g / m. l)

log n = (1 / 2) {log (M.g) – log (m.l)}

We get,

log n = (1 / 2) {log M + log g – log m – log l}

(iv) V = (4 / 3) πr³

On taking log on both 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)

V = 1 / Dl (T / πr)^{1 / 2}

On taking log on both sides, we get,

log V = log {1 / Dl (T / πr)^{1 / 2}}

log V = log (1 / Dl) + log (T / πr)^{1 / 2}

log V = (log 1 – log D – log l) + (1 / 2) log (T / πr)

log V = (0 – log D – log l) + (1 / 2) {(log T – log π – log r)}

We get,

log V = (1 / 2) (log T – log π – log r) – log D – log l

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

Solution:

(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 log 2 + 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)} – (1 / 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)

Solution:

(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 x 3)} – 3 log (5⁴ / 3⁵) + log {2 / (3 x 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 x 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)

Solution:

(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, the 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

The negative value is rejected

Hence,

x = 8

(vi) log₈ (x² – 1) – log₈ (3x + 9) = 0

log₈ {(x² – 1) / (3x + 9)} = log₈ 1

(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

The 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 / log_x 2) + (1 / log_x 2²) + (1 / log_x 2⁴) = (21 / 4)

(1 / log_x 2) + (1 / 2 log_x 2) + (1 / 4 log_x 2) = (21 / 4)

Taking (1 / log_x 2) as common, we get,

(1 / log_x 2) {1 + (1 / 2) + (1 / 4)} = (21 / 4)

We get,

(1 / log_x 2) (7 / 4) = (21 / 4)

log_x 2= (7 / 4) × (4 / 21)

log_x 2 = (1 / 3)

So,

x^{1 / 3} = 2

We get,

x = 2³

x = 8