Concise Selina Solutions for Class 9 Maths Chapter 8-Logarithms

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Exercise 8(A)

1. Express each of the following in logarithmic form:

(i) 5³ = 125

(ii) 3^-2 = 1/9

(iii) 10^-3 = 0.001

(iv) (81)^3/4 = 27

Solution:

We know that,

a^b = c ⇒ log_a c = b

(i) 5³ = 125

log₅ 125 = 3 

(ii) 3^-2 = 1/9

log₃ 1/9 = -2 

(iii) 10^-3 = 0.001

log₁₀ 0.001 = -3 

(iv) (81)^3/4 = 27

log₈₁ 27 = ¾ 

2. Express each of the following in exponential form:

(i) log₈ 0.125 = -1

(ii) log₁₀ 0.01 = -2

(iii) log_a A = x

(iv) log₁₀ 1 = 0

Solution:

We know that,

log_a c = b ⇒ a^b = c

(i) log₈ 0.125 = -1

8^-1 = 0.125

(ii) log₁₀ 0.01 = -2

8^-1 = 0.125

(iii) log_a A = x

a^x = A

(iv) log₁₀ 1 = 0

10⁰ = 1

3. Solve for x: log₁₀ x = -2.

Solution:

We have,

log₁₀ x = -2

10^-2 = x [As log_a c = b ⇒ a^b = c]

x = 10^-2

x = 1/10²

x = 1/100

Hence, x = 0.01

4. Find the logarithm of:

(i) 100 to the base 10

(ii) 0.1 to the base 10

(iii) 0.001 to the base 10

(iv) 32 to the base 4

(v) 0.125 to the base 2

(vi) 1/16 to the base 4

(vii) 27 to the base 9

(viii) 1/81 to the base 27

Solution;

(i) Let log₁₀ 100 = x

So, 10^x = 100

10^x = 10²

Then,

x = 2 [If a^m = aⁿ; then m = n]

Hence, log₁₀ 100 = 2

(ii) Let log₁₀ 0.1 = x

So, 10^x = 0.1

10^x = 1/10

10^x = 10^-1

Then,

x = -1 [If a^m = aⁿ; then m = n]

Hence, log₁₀ 0.1 = -1

(iii) Let log₁₀ 0.001 = x

So, 10^x = 0.001

10^x = 1/1000

10^x = 1/10³

10^x = 10^-3

Then,

x = -3 [If a^m = aⁿ; then m = n]

Hence, log₁₀ 0.001 = -3

(iv) Let log₄ 32 = x

So, 4^x = 32

4^x = 2 x 2 x 2 x 2 x 2

(2²)^x = 2⁵

2^2x = 2⁵

Then,

2x = 5 [If a^m = aⁿ; then m = n]

x = 5/2

Hence, log₄ 32 = 5/2

(v) Let log₂ 0.125 = x

So, 2^x = 0.125

2^x = 125/1000

2^x = 1/8

2^x = (½)³

2^x = 2^-3

Then,

x = -3 [If a^m = aⁿ; then m = n]

Hence, log₂ 0.125 = -3

(vi) Let log₄ 1/16 = x

So, 4^x = 1/16

4^x = (¼)^-2

4^x = 4^-2

Then,

x = -2 [If a^m = aⁿ; then m = n]

Hence, log₄ 1/16 = -2

(vii) Let log₉ 27 = x

So, 9^x = 27

9^x = 3 x 3 x 3

(3²)^x = 3³

3^2x = 3³

Then,

2x = 3 [If a^m = aⁿ; then m = n]

x = 3/2

Hence, log₉ 27 = 3/2

(viii) Let log₂₇ 1/81 = x

So, 27^x = 1/81

27^x = 1/9²

(3³)^x = 1/(3²)²

3^3x = 1/3⁴

3^3x = 3^-4

Then,

3x = -4 [If a^m = aⁿ; then m = n]

x = -4/3

Hence, log₂₇ 1/81 = -4/3

5. State, true or false:

(i) If log₁₀ x = a, then 10^x = a

(ii) If x^y = z, then y = log_z x

(iii) log₂ 8 = 3 and log₈ 2 = 1/3

Solution:

(i) We have,

log₁₀ x = a

So, 10^a = x

Thus, the statement 10^x = a is false

(ii) We have,

x^y = z

So, log_x z = y

Thus, the statement y = log_z x is false

(iii) We have,

log₂ 8 = 3

So, 2³ = 8 … (1)

Now consider the equation,

log₈ 2 = 1/3

8^1/3 = 2

(2³)^1/3 = 2 … (2)

Both equations (1) and (2) are correct

Thus, the given statements, log₂ 8 = 3 and log₈ 2 = 1/3 are true

6. Find x, if:

(i) log₃ x = 0

(ii) log_x 2 = -1

(iii) log₉243 = x

(iv) log₅ (x – 7) = 1

(v) log₄32 = x – 4

(vi) log₇ (2x² – 1) = 2

Solution:

(i) We have, log₃ x = 0

So, 3⁰ = x

1 = x

Hence, x = 1

(ii) we have, log_x 2 = -1

So, x^-1 = 2

1/x = 2

Hence, x = ½

(iii) We have, log₉ 243 = x

9^x = 243

(3²)^x = 3⁵

3^2x = 3⁵

On comparing the exponents, we get

2x = 5

x = 5/2 = 2½

(iv) We have, log₅ (x – 7) = 1

So, 5¹ = x – 7

5 = x – 7

x = 5 + 7

Hence, x = 12

(v) We have, log₄ 32 = x – 4

So, 4^{(x – 4)} = 32

(2²)^{(x – 4)} = 2⁵

2^{(2x – 8)} = 2⁵

On comparing the exponents, we get

2x – 8 = 5

2x = 5 + 8

Hence,

x = 13/2 = 6½

(vi) We have, log₇ (2x² – 1) = 2

So, (2x² – 1) = 7²

2x² – 1 = 49

2x² = 49 + 1

2x² = 50

x² = 25

Taking square root on both side, we get

x = ±5

Hence, x = 5 (Neglecting the negative value)

7. Evaluate:

(i) log₁₀ 0.01

(ii) log₂ (1 ÷ 8)

(iii) log₅ 1

(iv) log₅ 125

(v) log₁₆ 8

(vi) log_0.5 16

Solution:

(i) Let log₁₀ 0.01 = x

Then, 10^x = 0.01

10^x = 1/100 = 1/10²

So, 10^x = 10^-2

On comparing the exponents, we get

x = -2

Hence, log₁₀ 0.01 = -2

(ii) Let log₂ (1 ÷ 8) = x

Then, 2^x = 1/8

2^x = 1/2³

So, 2^x = 2^-3

On comparing the exponents, we get

x = -3

Hence, log₁₀ (1 ÷ 8) = -3

(iii) Let log₅ 1 = x

Then, 5^x = 1

5^x = 5⁰

On comparing the exponents, we get

x = 0

Hence, log₅ 1 = 0

(iv) Let log₅ 125 = x

Then, 5^x = 125

5^x = (5 x 5 x 5) = 5³

So, 5^x = 5³

On comparing the exponents, we get

x = 3

Hence, log₅ 125 = 3

(v) Let log₁₆ 8 = x

Then, 16^x = 8

(2⁴)^x = (2 x 2 x 2) = 2³

So, 2^4x = 2³

On comparing the exponents, we get

4x = 3

x = 3/4

Hence, log₁₆ 8 = 3/4

(vi) Let log_0.5 16 = x

Then, 0.5^x = 16

(5/10)^x = (2 x 2 x 2 x 2)

(1/2)^x = 2⁴

So, 2^-x = 2⁴

On comparing the exponents, we get

-x = 4

⇒ x = -4

Hence, log_0.5 16 = -4

8. If log_a m = n, express a^{n – 1} in terms in terms of a and m.

Solution:

We have, log_a m = n

So,

aⁿ = m

Dividing by a on both sides, we get

aⁿ/a = m/a

a^n-1 = m/a 

9. Given log₂ x = m and log₅ y = n

(i) Express 2^m-3 in terms of x

(ii) Express 5³ⁿ⁺² in terms of y 

Solution:

Given, log₂ x = m and log₅ y = n

So,

2^m = x and 5ⁿ = y

(i) Taking, 2^m = x

2^m/2³ = x/2³

2^m-3 = x/8

(ii) Taking, 5ⁿ = y

Cubing on both sides, we have

(5ⁿ)³ = y³

5³ⁿ = y³

Multiplying by 5² on both sides, we have

5³ⁿ x 5² = y³ x 5²

5³ⁿ⁺² = 25y³ 

10. If log₂ x = a and log₃ y = a, write 72^a in terms of x and y. 

Solution:

Given, log₂ x = a and log₃ y = a

So,

2^a = x and 3^a = y

Now, the prime factorization of 72 is

72 = 2 x 2 x 2 x 3 x 3 = 2³ x 3²

Hence,

(72)^a = (2³ x 3²)^a

= 2^3a x 3^2a

= (2^a)³ x (3^a)²

= x³y² [As 2^a = x and 3^a = y]

11. Solve for x: log (x – 1) + log (x + 1) = log₂ 1 

Solution:

We have,

log (x – 1) + log (x + 1) = log₂ 1

log (x – 1) + log (x + 1) = 0

log [(x – 1) (x + 1)] = 0

Then,

(x – 1) (x + 1) = 1 [As log 1 = 0]

x² – 1 = 1

x² = 1 + 1

x² = 2

x = ±√2

The value -√2 is not a possible, since log of a negative number is not defined.

Hence, x = √2 

12. If log (x² – 21) = 2, show that x = ± 11.

Solution:

Given, log (x² – 21) = 2

So,

x² – 21 = 10²

x² – 21 = 100

x² = 121

Taking square root on both sides, we get

x = ±11

Exercise 8(B)

1. Express in terms of log 2 and log 3:

(i) log 36 

(ii) log 144 

(iii) log 4.5

(iv) log 26/51 – log 91/119 

(v) log 75/16 – 2log 5/9 + log 32/243

Solution:

(i) log 36 = log (2 x 2 x 3 x 3)

= log (2² x 3²)

= log 2² + log 3² [Using log_a mn = log_a m + log_a n]

= 2log 2 + 2log 3 [Using log_a mⁿ = nlog_a m]

(ii) log 144 = log (2 x 2 x 2 x 2 x 3 x 3)

= log (2⁴ x 3²)

= log 2⁴ + log 3² [Using log_a mn = log_a m + log_a n]

= 4log 2 + 2log 3 [Using log_a mⁿ = nlog_a m]

(iii) log 4.5 = log 45/10

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

= log 3²/2

= log 3² – log 2 [Using log_a m/n = log_a m – log_a n]

= 2log 3 – log 2 [Using log_a mⁿ = nlog_a m]

(iv) log 26/51 – log 91/119 = log (26/51)/ (91/119)  [Using log_a m – log_a n = log_a m/n]

= log [(26/51) x (119/91)]

= log (2 x 13 x 7 x 117)/ (3 x 17 x 7 x 13)

= log 2/3

= log 2 – log 3 [Using log_a m/n = log_a m – log_a n]

(v) log 75/16 – 2log 5/9 + log 32/243

= log 75/16 – log (5/9)² + log 32/243 [Using nlog_a m = log_a mⁿ]

= log 75/16 – log 25/81 + log 32/243

= log [(75/16)/ (25/81)] + log 32/243 [Using log_a m – log_a n = log_a m/n]

= log (75 x 81)/ (16 x 25) + log 32/243

= log (3 x 81)/16 + log 32/243

= log 243/16 + log 32/243

= log (243/16) x (32/243) [Using log_a m + log_a n = log_a mn]

= log 32/16

= log 2

2. Express each of the following in a form free from logarithm:

(i) 2 log x – log y = 1

(ii) 2 log x + 3 log y = log a

(iii) a log x – b log y = 2 log 3

Solution:

(i) We have, 2 log x – log y = 1

Then,

log x² – log y = 1 [Using nlog_a m = log_a mⁿ]

log x²/y = 1 [Using log_a m – log_a n = log_a m/n]

Now, on removing log we have

x²/y = 10¹

⇒ x² = 10y

(ii) We have, 2 log x + 3 log y = log a

Then,

log x² + log y³ = log a [Using nlog_a m = log_a mⁿ]

log x²y³ = log a [Using log_a m + log_a n = log_a mn]

Now, on removing log we have

x²y³ = a

(iii) a log x – b log y = 2 log 3

Then,

log x^a – log y^b = log 3² [Using nlog_a m = log_a mⁿ]

log x^a/y^b = log 3² [Using log_a m – log_a n = log_a m/n]

Now, on removing log we have

x^a/y^b = 3²

⇒ x² = 9y^b

3. Evaluate each of the following without using tables:

(i) log 5 + log 8 – 2log 2

(ii) log₁₀ 8 + log₁₀ 25 + 2log₁₀ 3 – log₁₀ 18

(iii) log 4 + 1/3 log 125 – 1/5 log 32

Solution:

(i) We have, log 5 + log 8 – 2log 2

= log 5 + log 8 – log 2² [Using nlog_a m = log_a mⁿ]

= log 5 + log 8 – log 4

= log (5 x 8) – log 4 [Using log_a m + log_a n = log_a mn]

= log 40 – log 4

= log 40/4 [Using log_a m – log_a n = log_a m/n]

= log 10

= 1 

(ii) We have, log₁₀ 8 + log₁₀ 25 + 2log₁₀ 3 – log₁₀ 18

= log₁₀ 8 + log₁₀ 25 + log₁₀ 3² – log₁₀ 18 [Using nlog_a m = log_a mⁿ]

= log₁₀ 8 + log₁₀ 25 + log₁₀ 9 – log₁₀ 18

= log₁₀ (8 x 25 x 9) – log₁₀ 18 [Using log_a l + log_a m + log_a n = log_a lmn]

= log₁₀ 1800 – log₁₀ 18

= log₁₀ 1800/18 [Using log_a m – log_a n = log_a m/n]

= log₁₀ 100

= log₁₀ 10²

= 2log₁₀ 10 [Using log_a mⁿ = nlog_a m]

= 2 x 1

= 2

(iii) We have, log 4 + 1/3log 125 – 1/5log 32

= log 4 + log (125)^1/3 – log (32)^1/5 [Using nlog_a m = log_a mⁿ]

= log 4 + log (5³)^1/3 – log (2⁵)^1/5

= log 4 + log 5 – log 2

= log (4 x 5) – log 2 [Using log_a m + log_a n = log_a mn]

= log 20 – log 2

= log 20/2 [Using log_a m – log_a n = log_a m/n]

= log 10

= 1

4. Prove that:

2log 15/18 – log 25/162 + log 4/9 = log 2

Solution:

Taking L.H.S.,

= 2log 15/18 – log 25/162 + log 4/9

= log (15/18)² – log 25/162 + log 4/9 [Using nlog_a m = log_a mⁿ]

= log 225/324 – log 25/162 + log 4/9

= log [(225/324)/(25/162)] + log 4/9 [Using log_a m – log_a n = log_a m/n]

= log (225 x 162)/(324 x 25) + log 4/9

= log (9 x 1)/(2 x 1) + log 4/9

= log 9/2 + log 4/9 [Using log_a m + log_a n = log_a mn]

= log (9/2 x 4/9)

= log 2

= R.H.S.

5. Find x, if:

x – log 48 + 3 log 2 = 1/3 log 125 – log 3.

Solution:

We have,

x – log 48 + 3 log 2 = 1/3 log 125 – log 3

Solving for x, we have

x = log 48 – 3 log 2 + 1/3 log 125 – log 3

= log 48 – log 2³ + log 125^1/3 – log 3 [Using nlog_a m = log_a mⁿ]

= log 48 – log 8 + log (5³)^1/3 – log 3

= (log 48 – log 8) + (log 5 – log 3)

= log 48/8 + log 5/3 [Using log_a m – log_a n = log_a m/n]

= log (48/8 x 5/3) [Using log_a m + log_a n = log_a mn]

= log (2 x 5)

= log 10

= 1

Hence, x = 1

6. Express log₁₀ 2 + 1 in the form of log₁₀ x.

Solution:

Given, log₁₀ 2 + 1

= log₁₀ 2 + log₁₀ 10 [As, log₁₀ 10 = 1]

= log₁₀ (2 x 10) [Using log_a m + log_a n = log_a mn]

= log₁₀ 20

7. Solve for x:

(i) log₁₀ (x – 10) = 1

(ii) log (x² – 21) = 2

(iii) log (x – 2) + log (x + 2) = log 5

(iv) log (x + 5) + log (x – 5) = 4 log 2 + 2 log 3

Solution:

(i) We have, log₁₀ (x – 10) = 1

Then,

x – 10 = 10¹

x = 10 + 10

Hence, x = 20

(ii) We have, log (x² – 21) = 2

Then,

x² – 21 = 10²

x² – 21 = 100

x² = 100 + 21

x² = 121

Taking square root on both sides,

Hence, x = ±11

(iii) We have, log (x – 2) + log (x + 2) = log 5

Then,

log (x – 2)(x + 2) = log 5 [Using log_a m + log_a n = log_a mn]

log (x² – 2²) = log 5 [As (x – a)(x + a) = x² – a²]

log (x² – 4) = log 5

Removing log on both sides, we get

x² – 4 = 5

x² = 5 + 4

x² = 9

Taking square root on both sides,

x = ±3

(iv) We have, log (x + 5) + log (x – 5) = 4 log 2 + 2 log 3

Then,

log (x + 5) + log (x – 5) = log 2⁴ + log 3² [Using nlog_a m = log_a mⁿ]

log (x + 5)(x – 5) = log 16 + log 9 [Using log_a m + log_a n = log_a mn]

log (x² – 5²) = log (16 x 9) [As (x – a)(x + a) = x² – a²]

log (x² – 25) = log 144

Removing log on both sides, we have

x² – 25 = 144

x² = 144 + 25

x² = 169

Taking square root on both sides, we get

x = ±13

8. Solve for x:

(i) log 81/log 27 = x

(ii) log 128/log 32 = x

(iii) log 64/log 8 = log x

(iv) log 225/log 15 = log x

Solution:

(i) We have, log 81/log 27 = x

x = log 81/log 27

= log (3 x 3 x 3 x 3)/ log (3 x 3 x 3)

= log 3⁴/log 3³

= (4log 3)/(3log 3) [Using log_a mⁿ = nlog_a m]

= 4/3

Hence, x = 4/3

(ii) We have, log 128/log 32 = x

x = log 128/log 32

= log (2 x 2 x 2 x 2 x 2 x 2 x 2)/ log (2 x 2 x 2 x 2 x 2)

= log 2⁷/log 2⁵

= (7log 2)/(5log 2) [Using log_a mⁿ = nlog_a m]

= 7/5

Hence, x = 7/5

(iii) log 64/log 8 = log x

log x = log 64/log 8

= log (2 x 2 x 2 x 2 x 2 x 2)/ log (2 x 2 x 2)

= log 2⁶/log 2³

= (6log 2)/(3log 2) [Using log_a mⁿ = nlog_a m]

= 6/3

= 2

So, log x = 2

Hence, x = 10² = 100

(iv) We have, log 225/log 15 = log x

log x = log 225/log 15

= log (15 x 15)/ log 15

= log 15²/log 15

= (2log 15)/log 15 [Using log_a mⁿ = nlog_a m]

= 2

So, log x = 2

Hence, x = 10² = 100

9. Given log x = m + n and log y = m – n, express the value of log 10x/y² in terms of m and n.

Solution:

Given, log x = m + n and log y = m – n

Now consider log 10x/y²,

log 10x/y² = log 10x – log y² [Using log_a m/n = log_a m – log_a n]

= log 10x – 2log y

= log 10 + log x – 2 log y

= 1 + (m + n) – 2 (m – n)

= 1 + m + n – 2m + 2n

= 1 + 3n – m

10. State, true or false:

(i) log 1 × log 1000 = 0

(ii) log x/log y = log x – log y

(iii) If log 25/log 5 = log x, then x = 2

(iv) log x × log y = log x + log y

Solution:

(i) We have, log 1 × log 1000 = 0

Now,

log 1 = 0 and

log 1000 = log 10³ = 3log 10 = 3 [Using log_a mⁿ = nlog_a m]

So,

log 1 × log 1000 = 0 x 3 = 0

Thus, the statement log 1 × log 1000 = 0 is true

(ii) We have, log x/log y = log x – log y

We know that,

log x/y = log x – log y

So,

log x/log y ≠ log x – log y

Thus, the statement log x/log y = log x – log y is false

(iii) We have, log 25/log 5 = log x

log (5 x 5)/log 5 = log x

log 5²/ log 5 = log x

2log 5/log 5 = log x [Using log_a mⁿ = nlog_a m]

2 = log x

So, x = 10²

x = 100

Thus, the statement x = 2 is false

(iv) We know, log x + log y = log xy

So,

log x + log y ≠ log x × log y

Thus, the statement log x + log y = log x × log y is false

11. If log₁₀ 2 = a and log₁₀ 3 = b; express each of the following in terms of ‘a’ and ‘b’:

(i) log 12

(ii) log 2.25

(iii) log 

(iv) log 5.4

(v) log 60

(iv) log 

Solution:

Given that log₁₀ 2 = a and log₁₀ 3 = b … (1)

(i) log 12 = log (2 x 2 x 3)

= log (2 x 2) + log 3 [Using log_a mn = log_a m + log_a n]

= log 2² + log 3

= 2log 2 + log 3 [Using log_a mⁿ = nlog_a m]

= 2a + b [From 1]

(ii) log 2.25 = log 225/100

= log (25 x 9)/(25 x 4)

= log 9/4

= log (3/2)²

= 2log 3/2 [Using log_a mⁿ = nlog_a m]

= 2(log 3 – log 2) [Using log_a m/n = log_a m – log_a n]

= 2(b – a) [From 1]

= 2b – 2a

(iii) log  = log 9/4

= log (3/2)²

= 2log 3/2 [Using log_a mⁿ = nlog_a m]

= 2(log 3 – log 2) [Using log_a m/n = log_a m – log_a n]

= 2(b – a) [From 1]

= 2b – 2a

(iv) log 5.4 = log 54/10

= log (2 x 3 x 3 x 3)/10

= log (2 x 3³) – log 10 [Using log_a m/n = log_a m – log_a n]

= log 2 + log 3³ – 1 [Using log_a mn = log_a m + log_a n and log 10 = 1]

= log 2 + 3log 3 – 1 [Using log_a mⁿ = nlog_a m]

= a + 3b – 1 [From 1]

(v) log 60 = log (10 x 3 x 2)

= log 10 + log 3 + log 2 [Using log_a lmn = log_a l + log_a m + log₁₀ n]

= 1 + b + a [From 1]

(vi) log = log 25/8

= log 5²/2³

= log 5² – log 2³ [Using log_a m/n = log_a m – log_a n]

= 2log 5 – 3log 2 [Using log_a mⁿ = nlog_a m]

= 2log 10/2 – 3log 2

= 2(log 10 – log 2) – 3log 2 [Using log_a m/n = log_a m – log_a n]

= 2log 10 – 2log 2 – 3log 2

= 2(1) – 2a – 3a [From 1]

= 2 – 5a

12. If log 2 = 0.3010 and log 3 = 0.4771; find the value of:

(i) log 12

(ii) log 1.2

(iii) log 3.6

(iv) log 15

(v) log 25

(vi) 2/3 log 8

Solution:

Given, log 2 = 0.3010 and log 3 = 0.4771

(i) log 12 = log (4 x 3)

= log 4 + log 3 [Using log_a mn = log_a m + log_a n]

= log 2² + log 3

= 2log 2 + log 3 [Using log_a mⁿ = nlog_a m]

= 2 x 0.3010 + 0.4771

= 1.0791

(ii) log 1.2 = log 12/10

= log 12 – log 10 [Using log_a m/n = log_a m – log_a n]

= log (4 x 3) – log 10

= log 4 + log 3 – log 10 [Using log_a mn = log_a m + log_a n]

= log 2² + log 3 – log 10

= 2log 2 + log 3 – log 10 [Using log_a mⁿ = nlog_a m]

= 2 x 0.3010 + 0.4771 – 1 [As log 10 = 1]

= 0.6020 + 0.4771 – 1

= 1.0791 – 1

= 0.0791

(iii) log 3.6 = log 36/10

= log 36 – log 10 [Using log_a m/n = log_a m – log_a n]

= log (2 x 2 x 3 x 3) – 1 [As log 10 = 1]

= log (2² x 3²) – 1

= log 2² + log 3² – 1 [Using log_a mn = log_a m + log_a n]

= 2log 2 + 2log 3 – 1 [Using log_a mⁿ = nlog_a m]

= 2 x 0.3010 + 2 x 0.4771 – 1

= 0.6020 + 0.9542 – 1

= 1.5562 – 1

= 0.5562

(iv) log 15 = log (15/10 x 10)

= log 15/10 + log 10 [Using log_a mn = log_a m + log_a n]

= log 3/2 + 1 [As log 10 = 1]

= log 3 – log 2 + 1 [Using log_a m/n = log_a m – log_a n]

= 0.4771 – 0.3010 + 1

= 1.1761

(v) log 25 = log (25/4 x 4)

= log 100/4

= log 100 – log 4 [Using log_a m/n = log_a m – log_a n]

= log 10² – log 2²

= 2log 10 – 2log 2 [Using log_a mⁿ = nlog_a m]

= (2 x 1) – (2 x 0.3010)

= 2 – 0.6020

= 1.398

13. Given 2 log₁₀ x + 1 = log₁₀ 250, find:

(i) x

(ii) log₁₀ 2x

Solution:

(i) Given equation, 2log₁₀ x + 1 = log₁₀ 250

log₁₀ x² + log₁₀ 10 = log₁₀ 250 [Using nlog_a m = log_a mⁿ and log₁₀ 10 = 1]

log₁₀ 10x² = log₁₀ 250 [Using log_a m + log_a n = log_a mn]

Removing log on both sides, we have

10x² = 250

x² = 25

x = ±5

As x cannot be a negative value, x = -5 is not possible

Hence, x = 5

(ii) Now, from (i) we have x = 5

So,

log₁₀ 2x = log₁₀ 2(5)

= log₁₀ 10

= 1

14. Given 3log x + ½ log y = 2, express y in term of x.

Solution:

We have, 3log x + ½ log y = 2

log x³ + log y^1/2 = 2 [Using log_a m + log_a n = log_a mn]

log x³y^1/2 = 2

Removing logarithm, we get

x³y^1/2 = 10²

y^1/2 = 100/x³

On squaring on both sides, we get

y = 10000/x⁶

y = 10000x^-6

15. If x = (100)^a, y = (10000)^b and z = (10)^c, find log 10√y/x²z³ in terms of a, b and c.

Solution:

We have,

x = (100)^a, y = (10000)^b and z = (10)^c

So,

log x = a log 100, log y = b log 10000 and log z = c log 10

⇒ log x = a log 10², log y = b log 10⁴ and log z = c log 10

⇒ log x = 2a log 10, log y = 4b log 10 and log z = c log 10

⇒ log x = 2a, log y = 4b and log z = c … (i)

Now,

log 10√y/x²z³ = log 10√y – log x²z³ [Using log_a m/n = log_a m – log_a n]

= (log 10 + log √y) – (log x² + log z³) [Using log_a mn = log_a m + log_a n]

= 1 + log y^1/2 – log x² – log z³

= 1 + ½ log y – 2 log x – 3 log z [Using log_a mⁿ = nlog_a m]

= 1 + ½(4b) – 2(2a) – 3c … [Using (i)]

= 1 + 2b – 4a – 3c

16. If 3(log 5 – log 3) – (log 5 – 2 log 6) = 2 – log x, find x.

Solution:

We have,

3(log 5 – log 3) – (log 5 – 2 log 6) = 2 – log x

3log 5 – 3log 3 – log 5 + 2log 6 = 2 – log x

3log 5 – 3log 3 – log 5 + 2log (3 x 2) = 2 – log x

2log 5 – 3log 3 + 2(log 3 + log 2) = 2 – log x [Using log_a mn = log_a m + log_a n]

2log 5 – log 3 + 2log 3 + 2log 2 = 2 – log x

2log 5 – log 3 + 2log 2 = 2 – log x

2log 5 – log 3 + 2log 2 + log x = 2

log 5² – log 3 + log 2² + log x = 2 [Using nlog_a m = log_a mⁿ]

log 25 – log 3 + log 4 + log x = 2

log (25 × 4 × x)/3 = 2 [Using log_a m + log_a n = log_a mn & log_a m – log_a n = log_a m/n]

log 100x/3 = 2

On removing logarithm,

100x/3 = 10²

100x/3 = 100

Dividing by 100 on both sides, we have

x/3 = 1

Hence, x = 3

Exercise 8(C)

1. If log₁₀ 8 = 0.90; find the value of:

(i) log₁₀ 4

(ii) log √32

(iii) log 0.125

Solution:

Given, log₁₀ 8 = 0.90

log₁₀ (2 x 2 x 2) = 0.90

log₁₀ 2³ = 0.90

3 log₁₀ 2 = 0.90

log₁₀ 2 = 0.90/3

log₁₀ 2 = 0.30 … (1)

(i) log 4 = log₁₀ (2 x 2)

= log₁₀ 2²

= 2 log₁₀ 2

= 2 x 0.60 … [From (1)]

= 1.20

(ii) log √32 = log₁₀ 32^1/2

= ½ log₁₀ (2 x 2 x 2 x 2 x 2)

= ½ log₁₀ 2⁵

= ½ x 5 log₁₀ 2

= ½ x 5 x 0.30 … [From (1)]

= 0.75

(iii) log 0.125 = log 125/1000

= log₁₀ 1/8

= log₁₀ 1/2³

= log₁₀ 2^-3

= -3 log₁₀ 2

= -3 x 0.30 … [From (1)]

= -0.90

2. If log 27 = 1.431, find the value of:

(i) log 9 (ii) log 300

Solution:

Given, log 27 = 1.431

So, log 3³ = 1.431

3log 3 = 1.431

log 3 = 1.431/3

= 0.477 … (1)

(i) log 9 = log 3²

= 2log 3

= 2 x 0.477 … [From (1)]

= 0.954

(ii) log 300 = log (3 x 100)

= log 3 + log 100

= log 3 + log 10²

= log 3 + 2log 10

= log 3 + 2 [As log 10 = 1]

= 0.477 + 2

= 2.477

3. If log₁₀ a = b, find 10^{3b – 2} in terms of a.

Solution:

Given, log₁₀ a = b

Now,

Let 10^{3b – 2} = x

Applying log on both sides,

log 10^{3b – 2} = log x

(3b – 2)log 10 = log x

3b – 2 = log x

3log₁₀ a – 2 = log x

3log₁₀ a – 2log 10 = log x

log₁₀ a³ – log 10² = log x

log₁₀ a³ – log 100 = log x

log₁₀ a³/100 = log x

On removing logarithm, we get

a³/100 = x

Hence, 10^{3b – 2} = a³/100

4. If log₅ x = y, find 5^{2y+ 3} in terms of x.

Solution:

Given, log₅ x = y

So, 5^y = x

Squaring on both sides, we get

(5^y)² = x²

5^2y = x²

5^2y × 5³ = x² × 5³

Hence,

5^2y+3 = 125x²

5. Given: log₃ m = x and log₃ n = y.

(i) Express 3^{2x – 3} in terms of m.

(ii) Write down 3^{1 – 2y + 3x} in terms of m and n.

(iii) If 2 log₃ A = 5x – 3y; find A in terms of m and n.

Solution:

Given, log₃ m = x and log₃ n = y

So, 3^x = m and 3^y = n … (1)

(i) Taking the given expression, 3^{2x – 3}

3^{2x – 3} = 3^2x . 3^-3

= 3^2x . 1/3³

= (3^x)²/3³

= m²/3³ … [Using (1)]

= m²/27

Hence, 3^{2x – 3} = m²/27

(ii) Taking the given expression, 3^{1 – 2y + 3x} 

3^{1 – 2y + 3x} = 3¹ . 3^-2y . 3^3x 

= 3 . (3^y)^-2 . (3^x)³ 

= 3 . n^-2 . m³ … [Using (1)]

= 3m³/n²

Hence, 3^{1 – 2y + 3x} = 3m³/n²

(iii) Taking the given equation,

2 log₃ A = 5x – 3y

log₃ A² = 5x – 3y

log₃ A² = 5log₃ m – 3log₃ n … [Using (1)]

log₃ A² = log₃ m⁵ – log₃ n³

log₃ A² = log₃ m⁵/n³

Removing logarithm on both sides, we get

A² = m⁵/n³

Hence, by taking square root on both sides

A = √(m⁵/n³) = m^5/2/n^3/2

6. Simplify:

(i) log (a)³ – log a

(ii) log (a)³ ÷ log a

Solution:

(i) We have, log (a)³ – log a

= 3log a – log a

= 2log a

(ii) We have, log (a)³ ÷ log a

= 3log a/ log a

= 3

7. If log (a + b) = log a + log b, find a in terms of b.

Solution:

We have, log (a + b) = log a + log b

Then,

log (a + b) = log ab

So, on removing logarithm, we have

a + b = ab

a – ab = -b

a(1 – b) = -b

a = -b/(1 – b)

Hence,

a = b/(b – 1)

8. Prove that:

(i) (log a)² – (log b)² = log (a/b). log (ab)

(ii) If a log b + b log a – 1 = 0, then b^a. a^b = 10

Solution:

(i) Taking L.H.S. we have,

= (log a)² – (log b)²

= (log a + log b) (log a – log b) [As x² – y² = (x + y)(x – y)]

= (log ab). (log a/b)

= R.H.S.

– Hence proved

(ii) We have, a log b + b log a – 1 = 0

So,

log b^a + log a^b – 1 = 0

log b^a + log a^b = 1

log b^aa^b = 1

On removing logarithm, we get

b^aa^b = 10

– Hence proved

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

(ii) If 2 log y – log x – 3 = 0, express x in terms of y.

(iii) Prove that: log₁₀ 125 = 3(1 – log₁₀2).

Solution:

(i) Given, log (a + 1) = log (4a – 3) – log 3

So,

log (a + 1) = log (4a – 3)/3

On removing logarithm on both sides, we have

a + 1 = (4a – 3)/3

3(a + 1) = 4a – 3

3a + 3 = 4a – 3

Hence, a = 6

(ii) Given, 2log y – log x – 3 = 0

So,

log y² – log x = 3

log y²/x = 3

On removing logarithm, we have

y²/x = 10³ = 1000

Hence, x = y²/1000

(iii) Considering the L.H.S., we have

log₁₀ 125 = log₁₀ (5 x 5 x 5)

= log₁₀ 5³

= 3log₁₀ 5

= 3log₁₀ 10/2

= 3(log₁₀ 10 – log₁₀ 2)

= 3(1 – log₁₀ 2) [Since, log₁₀ 10 = 1]

= R.H.S. 

– Hence proved

10. Given log x = 2m – n, log y = n – 2m and log z = 3m – 2n. Find in terms of m and n, the value of log x²y³/z⁴.

Solution:

We have, log x = 2m – n, log y = n – 2m and log z = 3m – 2n

Now, considering

log x²y³/z⁴ = log x²y³ – log z⁴

= (log x² + log y³) – log z⁴

= 2log x + 3log y – 4log z

= 2(2m – n) + 3(n – 2m) – 4(3m – 2n)

= 4m – 2n + 3n – 6m – 12m + 8n

= -14m + 9n

11. Given log_x 25 – log_x 5 = 2 – log_x 1/125; find x.

Solution:

We have, log_x 25 – log_x 5 = 2 – log_x 1/125

log_x (5 x 5) – log_x 5 = 2 – log_x 1/(5 x 5 x 5)

log_x 5² – log_x 5 = 2 – log_x 1/5³

2log_x 5 – log_x 5 = 2 – log_x 1/5³

log_x 5 = 2 – 3log_x 1/5

log_x 5 = 2 + 3log_x (1/5)^-1

log_x 5 = 2 + 3log_x 5

2 = log_x 5 – 3log_x 5

2 = -2log_x 5

-1 = log_x 5

Removing logarithm, we get

x^-1 = 5

Hence, x = 1/5

Exercise 8(D)

1. If 3/2 log a + 2/3 log b – 1 = 0, find the value of a⁹.b⁴

Solution:

Given equation,

3/2 log a + 2/3 log b – 1 = 0

log a^3/2 + log b^2/3 – 1 = 0

log a^3/2 × b^2/3 – 1 = 0

log a^3/2.b^2/3 = 1

Removing logarithm, we have

a^3/2.b^2/3 = 10

On manipulating,

(a^3/2.b^2/3)⁶ = 10⁶

Hence,

a⁹.b⁴ = 10⁶

2. If x = 1 + log 2 – log 5, y = 2log3 and z = log a – log 5; find the value of a if x + y = 2z.

Solution:

Given, x = 1 + log 2 – log 5, y = 2log3 and z = log a – log 5

Now, considering the given equation x + y = 2z

(1 + log 2 – log 5) + (2log 3) = 2(log a – log 5)

1 + log 2 – log 5 + 2log 3 = 2log a – 2log 5

1 + log 2 – log 5 + 2log 3 + 2log 5 = 2log a

log 10 + log 2 + log 3² + log 5 = log a²

log 10 + log 2 + log 9 + log 5 = log a²

log (10 x 2 x 9 x 5) = log a²

log 900 = log a²

On removing logarithm on both sides, we have

900 = a²

Taking square root, we get

a = ±30

Since, a cannot be a negative value,

Hence, a = 30

3. If x = log 0.6; y = log 1.25 and z = log 3 – 2log 2, find the values of:

(i) x + y – z        (ii) 5^{x + y – z}

Solution:

Given,

x = log 0.6, y = log 1.25 and z = log 3 – 2log 2

So, z = log 3 – log 2²

= log 3 – log 4

= log ¾

= log 0.75 … (1)

(i) Considering,

x + y – z = log 0.6 + log 1.25 – log 0.75 … [From (1)]

= log (0.6 x 1.25)/0.75

= log 0.75/0.75

= log 1

= 0 … (2)

(ii) Now, considering

5^x+y-z = 5⁰ … [From (2)]

= 1

4. If a² = log x, b³ = log y and 3a² – 2b³ = 6 log z, express y in terms of x and z.

Solution:

We have, a² = log x and b³ = log y

Now, considering the equation

3a² – 2b³ = 6log z

3log x – 2log y = 6log z

log x³ – log y² = log z⁶

log x³/y² = log z⁶

On removing logarithm on both sides, we get

x³/y² = z⁶

So,

y² = x³/z⁶

Taking square root on both sides, we get

y = √(x³/z⁶)

Hence, y = x^3/2/z³

5. If log (a – b)/2 = ½ (log a + log b), show that: a² + b² = 6ab.

Solution:

We have, log (a – b)/2 = ½ (log a + log b)

log (a – b)/2 = ½ log a + ½ log b

= log a^1/2 + log b^1/2

= log √a + log √b

= log √(ab)

Now, removing logarithm on both sides, we get

(a – b)/2 = √(ab)

Squaring on both sides, we get

[(a – b)/2]² = [√(ab)]²

(a – b)²/4 = ab

(a – b)² = 4ab

a² + b² – 2ab = 4ab

a² + b² = 4ab + 2ab

a² + b² = 6ab

– Hence proved

6. If a² + b² = 23ab, show that: log (a + b)/5 = ½ (log a + log b).

Solution:

Given, a² + b² = 23ab

Adding 2ab on both sides,

a² + b² + 2ab = 23ab + 2ab

(a + b)² = 25ab

(a + b)²/25 = ab

[(a + b)/5]² = ab

Taking logarithm on both sides, we have

log [(a + b)/5]² = log ab

2log (a + b)/5 = log ab

2log (a + b)/5 = log a + log b

Thus,

log (a + b)/5 = ½ (log a + log b)

7. If m = log 20 and n = log 25, find the value of x, so that: 2log (x – 4) = 2m – n.

Solution:

Given, m = log 20 and n = log 25

Now, considering the given expression

2log (x – 4) = 2m – n

2log (x – 4) = 2log 20 – log 25

log (x – 4)² = log 20² – log 25

log (x – 4)² = log 400 – log 25

log (x – 4)² = log 400/25

Removing logarithm on both sides,

(x – 4)² = 400/25

x² – 8x + 16 = 16

x² – 8x = 0

x(x – 8) = 0

So,

x = 0 or x = 8

If x = 0, then log (x – 4) doesn’t exist

Hence, x = 8

8. Solve for x and y; if x > 0 and y > 0; log xy = log x/y + 2log 2 = 2.

Solution:

We have,

log xy = log x/y + 2log 2 = 2

Considering the equation,

log xy = 2

log xy = 2log 10

log xy = log 10²

log xy = log 100

On removing logarithm,

xy = 100 … (1)

Now, consider the equation

log x/y + 2log 2 = 2

log x/y + log 2² = 2

log x/y + log 4 = 2

log 4x/y = 2

Removing logarithm, we get

4x/y = 10²

4x/y = 100

x/y = 25

(xy)/y² = 25

100/y² = 25 … [From (1)]

y² = 100/25

y² = 4

y = 2 [Since, y > 0]

From log xy = 2

Substituting the value of y, we get

log 2x = 2

On removing logarithm,

2x = 10²

2x = 100

x = 100/2

x = 50

Thus, the values x and y are 50 and 2 respectively

9. Find x, if:

(i) log_x 625 = -4 

(ii) log_x (5x – 6) = 2

Solution:

(i) We have, log_x 625 = -4

On removing logarithm,

x^-4 = 625

(1/x)⁴ = 5⁴

Taking the fourth root on both sides,

1/x = 5

Hence, x = 1/5

(ii) We have, log_x (5x – 6) = 2

On removing logarithm,

x² = 5x – 6

x² – 5x + 6 = 0

x² – 3x – 2x + 6 = 0

x(x – 3) – 2(x – 2) = 0

(x – 2)(x – 3) = 0

Hence,

x = 2 or 3

10. If p = log 20 and q = log 25, find the value of x, if 2log (x + 1) = 2p – q. 

Solution:

Given, p = log 20 and q = log 25

Considering the equation,

2log (x + 1) = 2p – q

log (x + 1)² = 2p – q

log (x + 1)² = 2log 20 – log 25

log (x + 1)² = log 20² – log 25

log (x + 1)² = log 400 – log 25

log (x + 1)² = log 400/25

Removing logarithm on both sides, we have

(x + 1)² = 400/25 = 16

(x + 1)² = (4)²

Taking square root on both sides, we have

x + 1 = 4

x = 4 -1

Hence, x = 3  

11. If log₂ (x + y) = log₃ (x – y) = log 25/log 0.2, find the value of x and y.

Solution:

Considering the relation, log₂ (x + y) = log 25/log 0.2

log₂ (x + y) = log_0.2 25

= log_2/10 5²

= 2log_1/5 5

= 2log₅^-1 5

= -2log₅ 5

= -2 x 1

= -2

So, we have

log₂ (x + y) = -2

Removing logarithm, we get

x + y = 2^-2

x + y = 1/2²

x + y = ¼ … (i)

Now, considering the relation log₃ (x – y) = log 25/log 0.2

log₃ (x – y) = = log_0.2 25

= log_2/10 5²

= 2log_1/5 5

= 2log₅^-1 5

= -2log₅ 5

= -2 x 1

= -2

So, we have

log₃ (x – y) = -2

Removing logarithm, we get

x – y = 3^-2

x – y = 1/3²

x – y = 1/9 … (ii)

On adding (i) and (ii), we get

x + y = ¼

x – y = 1/9

—————-

2x = ¼ + 1/9

2x = (9 + 4)/36

2x = 13/36

x = 13/(36 x 2)

= 13/72

Now, substituting the value of x in (i), we get

13/72 + y = ¼

y = ¼ – 13/72

= (18 – 13)/72

= 5/72

Hence, the values of x and are is 13/72 and 5/72 respectively

12. Given: log x/log y = 3/2 and log xy = 5; find the values of x and y. 

Solution:

Given, log x/log y = 3/2 … (i) and log xy = 5 … (ii)

So,

log xy = log x + log y = 5

And, we have log y = (2log x)/3 … [From (i)]

Now,

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

3log x + 2log x = 5 x 3

5log x = 15

log x = 15/5

log x = 3

Removing logarithm, we get

x = 10³ = 1000

Substituting value of x in (ii), we get

log xy = 5

Removing logarithm, we get

xy = 10⁵

(10³). y = 10⁵

y = 10⁵/10³

y = 10²

y = 100

13. Given log₁₀ x = 2a and log₁₀ y = b/2

(i) Write 10^a in terms of x

(ii) Write 10^{2b + 1} in terms of y 

(iii) If log₁₀ p = 3a – 2b, express p in terms of x and y.

Solution:

Given, log₁₀ x = 2a and log₁₀ y = b/2

(i) Taking log₁₀ x = 2a

Removing logarithm on both sides,

x = 10^2a

Taking square root on both sides, we get

x^1/2 = 10^2a/2

Hence, 10^a = x^1/2

(ii) Taking log₁₀ y = b/2

Removing logarithm on both sides,

y = 10^b/2

On manipulating,

y⁴ = 10^{b/2 x 4}

y⁴ = 10^2b

10y⁴ = 10^2b x 10

Hence, 10^{2b + 1} = 10y⁴

(iii) We have, 10^a = x^1/2

and y = 10^b/2

Considering the equation, log₁₀ p = 3a – 2b

log₁₀ p = 3a – 2b

Removing logarithm, we get

p = 10^{3a – 2b}

p = 10^3a/10^2b

p = (10^a)³/(10^b/2)⁴

p = (x^1/2)³/(y)⁴

Hence, p = x^3/2/y⁴

14. Solve:

log₅ (x + 1) – 1 = 1 + log₅ (x – 1).

Solution:

Considering the given equation,

log₅ (x + 1) – 1 = 1 + log₅ (x – 1)

log₅ (x + 1) – log₅ (x – 1) = 1 + 1

log₅ (x + 1)/(x – 1) = 2

Removing logarithm, we have

(x + 1)/(x – 1) = 5²

(x + 1)/(x – 1) = 25

(x + 1) = 25(x – 1)

x + 1 = 25x – 25

25x – x = 25 + 1

24x = 26

x = 26/24

Hence, x = 13/12

15. Solve for x, if:

log_x 49 – log_x 7 + log_x 1/343 + 2 = 0 

Solution:

We have,

log_x 49 – log_x 7 + log_x 1/343 + 2 = 0 

log_x 49/(7 x 343) + 2 = 0

log_x 1/49 = -2

log_x 1/7² = -2

log_x 7^-2 = -2

-2log_x 7 = -2

So,

log_x 7 = 1

Removing logarithm, we get

x = 7

16. If a² = log x, b³ = log y and a²/2 – b³/3 = log c, find c in terms of x and y.  

Solution:

Given,

a² = log x, b³ = log y

Considering the given equation,

a²/2 – b³/3 = log c

(log x)/2 – (log y)/3 = log c

½ log x – 1/3 log y = log c

log x^1/2 – log y^1/3 = log c

log x^1/2/y^1/3 = log c

On removing logarithm, we get

x^1/2/y^1/3 = c

Hence, c = x^1/2/y^1/3 is the required relation 

17. Given: x = log₁₀ 12, y = log₄ 2 x log₁₀ 9 and z = log₁₀ 0.4, find

(i) x – y – z

(ii) 13^{x – y – z} 

Solution:

(i) Considering, x – y – z

= log₁₀ 12 – (log₄ 2 x log₁₀ 9) – log₁₀ 0.4

= log₁₀ 12 – (log₄ 2 x log₁₀ 9) – log₁₀ 0.4

= log₁₀ (4 x 3) – (log₁₀ 2/ log₁₀ 4 x log₁₀ 9) – log₁₀ 0.4

= log₁₀ 4 + log₁₀ 3 – (log₁₀ 2 x log₁₀ 3²)/ log₁₀ 2² – log₁₀ 4/10

= log₁₀ 4 + log₁₀ 3 – (log₁₀ 2 x 2log₁₀ 3)/ 2log₁₀ 2 – (log₁₀ 4 – log₁₀ 10)

= log₁₀ 4 + log₁₀ 3 – log₁₀ 3 – log₁₀ 4 + log₁₀ 10

= log₁₀ 4 + log₁₀ 3 – log₁₀ 3 – log₁₀ 4 + 1

= 1

(ii) Now,

13^{x – y – z} = 13¹ = 13 

18. Solve for x, log_x 15√5 = 2 – log_x 3√5 

Solution:

Considering the given equation,

log_x 15√5 = 2 – log_x 3√5 

log_x 15√5 + log_x 3√5 = 2

log_x (15√5 x 3√5) = 2

log_x (45 x 5) = 2

log_x 225 = 2

Removing logarithm, we get

x² = 225

Taking square root on both sides,

x = 15 

19. Evaluate:

(i) log_b a x log_c b x log_a c

(ii) log₃ 8 ÷ log₉ 16

(iii) log₅ 8/(log₂₅ 16 x log₁₀₀ 10) 

Solution:

Using log_b a = 1/log_a b and log_x a/log_x b = log_b a, we have

(i) log_b a x log_c b x log_a c

= 1

(ii) log₃ 8 ÷ log₉ 16

= log₃ 8/ log₉ 16

= 3 x ½

= 3/2

(iii) log₅ 8/(log₂₅ 16 x log₁₀₀ 10)

= 3 x ½ x 2

= 3  

20. Show that:

log_a m ÷ log_ab m = 1 + log_a b 

Solution:

Considering the L.H.S.,

log_a m ÷ log_ab m = log_a m/log_ab m

= log_m ab/log_m a [As log_b a = 1/log_a b]

= log_a ab [As log_x a/log_x b = log_b a]

= log_a a + log_a b

= 1 + log_a b 

21. If log_√27 x = 2 2/3, find x.

Solution:

We have,

log_√27 x = 2 2/3

log_√27 x = 8/3

Removing logarithm, we get

x = √27^8/3

= 27^{1/2 x 8/3}

= 27^4/3

= 3^{3 x 4/3}

= 3⁴

Hence, x = 81

22. Evaluate:

1/(log_a bc + 1) + 1/(log_b ca + 1) + 1/(log_c ab + 1)

Solution:

We have,

Selina Solutions for Class 9 Maths Chapter 8- Logarithms

The Chapter 8, Logarithms, contains 4 exercises and the Solutions given here contains the answers for all the questions present in these exercises. Let us have a look at some of the topics that are being discussed in this chapter.

8.1 Introduction

8.2 Interchanging

8.3 Laws of logarithms with use

8.4 Expansion of expressions with the help of laws of logarithms

8.5 More about logarithms

Selina Solutions for Class 9 Maths Chapter 8- Logarithms

The Chapter 8 of class 9 teaches the students the method of calculating using logarithm.Logarithms are used to make long and complicated calculations easy. Logarithm is related to exponents or indices. I.e., while a^b=c is called the exponential form, log_ac=b is called the logarithmic form. Read and learn the Chapter 8 of Selina textbook to get to know more about Logarithms. Learn the Selina Solutions for Class 9 effectively to attain excellent result in the examination.