Logarithms — Exercise 8(C)

If log₂ (log₃ x) = 4, the value of x is :

1. 3¹⁶

2. 16³

3. 3 × 16

4. 16 ÷ 3

Answer

Given,

⇒ log₂ (log₃ x) = 4

⇒ log₃ x = 2⁴

⇒ log₃ x = 16

⇒ x = 3¹⁶.

Hence, Option 1 is the correct option.

If log (5x - 4) - log (x + 1) = log 4, the value of x is :

1. 6

2. 8

3. 4

4. 12

Answer

Given,

⇒ log (5x - 4) - log (x + 1) = log 4

⇒ log $\dfrac{5x - 4}{x + 1}$ = log 4

⇒ $\dfrac{5x - 4}{x + 1} = 4$

⇒ 5x - 4 = 4(x + 1)

⇒ 5x - 4 = 4x + 4

⇒ 5x - 4x = 4 + 4

⇒ x = 8.

Hence, Option 2 is the correct option.

If log x - log (2x - 1) = 1, the value of x is :

1. $\dfrac{19}{10}$

2. 19 × 10

3. $\dfrac{10}{19}$

4. $\dfrac{1}{19 \times 10}$

Answer

Given,

$\Rightarrow \text{log } x - \text{log } (2x - 1) = 1 \\[1em] \Rightarrow \text{log } \dfrac{x}{2x - 1} = \text{log } 10 \\[1em] \Rightarrow \dfrac{x}{2x - 1} = 10 \\[1em] \Rightarrow x = 10(2x - 1) \\[1em] \Rightarrow x = 20x - 10 \\[1em] \Rightarrow 20x - x = 10 \\[1em] \Rightarrow 19x = 10 \\[1em] \Rightarrow x = \dfrac{10}{19}.$

Hence, Option 3 is the correct option.

If x = log $\dfrac{3}{5}, \text{ y = log } \dfrac{5}{4}\text{ and z = 2 log } \dfrac{\sqrt{3}}{2}$, the value of x + y - z is :

1. 1

2. -1

3. 2

4. 0

Answer

Solving,

$\Rightarrow x + y - z = \text{log } \dfrac{3}{5} + \text{log } \dfrac{5}{4} - \text{ 2 log } \dfrac{\sqrt{3}}{2} \\[1em] = \text{log } \dfrac{3}{5} + \text{log } \dfrac{5}{4} - \text{log } \Big(\dfrac{\sqrt{3}}{2}\Big)^2 \\[1em] = \text{log } \dfrac{3}{5} + \text{log } \dfrac{5}{4} - \text{log } \dfrac{3}{4} \\[1em] = \text{log } \dfrac{\dfrac{3}{5} \times \dfrac{5}{4}}{\dfrac{3}{4}} \\[1em] = \text{log } \dfrac{\dfrac{3}{4}}{\dfrac{3}{4}} \\[1em] = \text{log } 1 \\[1em] = 0.$

Hence, Option 4 is the correct option.

If log v + log 3 = log π + log 4 + 3 log r, the value of v in terms of r and other constants is :

1. $\dfrac{4}{3}πr^3$

2. 4πr²

3. πr³

4. $\dfrac{4}{3}πr^2$

Answer

Given,

⇒ log v + log 3 = log π + log 4 + 3 log r

⇒ log 3v = log π + log 4 + log r³

⇒ log 3v = log 4πr³

⇒ 3v = 4πr³

⇒ v = $\dfrac{4}{3}πr^3$

Hence, Option 1 is the correct option.

If log y + 2 log x = 2, the value of y in terms of x is :

1. x² ÷ 100

2. 100 ÷ x²

3. x ÷ 10

4. 10 ÷ x

Answer

Given,

⇒ log y + 2 log x = 2

⇒ log y + log x² = 2

⇒ log yx² = 2

⇒ yx² = 10²

⇒ yx² = 100

⇒ y = 100 ÷ x².

Hence, Option 2 is the correct option.

If log₁₀ 8 = 0.90; find the value of :

(i) log₁₀ 4

(ii) log $\sqrt{32}$

(iii) log 0.125

Answer

Given,

⇒ log₁₀ 8 = 0.90

⇒ log₁₀ 2³ = 0.90

⇒ 3log₁₀ 2 = 0.90

⇒ log₁₀ 2 = $\dfrac{0.90}{3}$

⇒ log₁₀ 2 = 0.30 .........(1)

(i) Given,

⇒ log₁₀ 4

⇒ log₁₀ 2²

⇒ 2 log₁₀ 2

Substituting value of log₁₀ 2 from equation (1) in above equation, we get :

⇒ 2 × 0.30

⇒ 0.60

Hence, log₁₀ 4 = 0.60

(ii) Given,

$\Rightarrow \text{log } \sqrt{32} \\[1em] \Rightarrow \text{log } 4\sqrt{2} \\[1em] \Rightarrow \text{log } 4 + \text{log } \sqrt{2} \\[1em] \Rightarrow \text{log } 2^2 + \text{log } 2^{\dfrac{1}{2}} \\[1em] \Rightarrow 2 \times \text{log 2} + \dfrac{1}{2} \times \text{log 2}$

Substituting value of log 2 from equation (1), in above equation, we get :

$\Rightarrow 2 \times 0.30 + \dfrac{1}{2} \times 0.30 \\[1em] \Rightarrow 0.60 + 0.15 \\[1em] \Rightarrow 0.75$

Hence, log $\sqrt{32}$ = 0.75

(iii) Given,

$\Rightarrow \text{log } 0.125 \\[1em] \Rightarrow \text{log } \dfrac{125}{1000} \\[1em] \Rightarrow \text{log } \dfrac{1}{8} \\[1em] \Rightarrow \text{log } \dfrac{1}{2^3} \\[1em] \Rightarrow \text{log } 2^{-3} \\[1em] \Rightarrow -3 \times \text{log 2} \\[1em] \Rightarrow -3 \times 0.30 \\[1em] \Rightarrow -0.90$

Hence, log 0.125 = -0.90

If log 27 = 1.431, find the value of :

(i) log 9

(ii) log 300

Answer

Given,

⇒ log 27 = 1.431

⇒ log 3³ = 1.431

⇒ 3 log 3 = 1.431

⇒ log 3 = $\dfrac{1.431}{3}$ = 0.477 .......(1)

(i) Solving,

⇒ log 9

⇒ log 3²

⇒ 2 log 3

Substituting value of log 3 from equation (1) in above equation, we get :

⇒ 2 × 0.477

⇒ 0.954

Hence, log 9 = 0.954

(ii) Solving,

⇒ log 300

⇒ log 3 × 100

⇒ log 3 + log 100

⇒ 0.477 + log 10²

⇒ 0.477 + 2 log 10

⇒ 0.477 + 2 × 1

⇒ 0.477 + 2

⇒ 2.477

Hence, log 300 = 2.477

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

Answer

Given,

⇒ log₁₀ a = b

⇒ 10^b = a

Cubing both sides, we get :

⇒ (10^b)³ = a³

⇒ 10^3b = a³

Dividing both sides by 10², we get :

$\Rightarrow \dfrac{10^{3b}}{10^2} = \dfrac{a^3}{10^2} \\[1em] \Rightarrow 10^{3b - 2} = \dfrac{a^3}{100}.$

Hence, 10^{3b - 2} = $\dfrac{a^3}{100}.$

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

Answer

Given,

⇒ log₅ x = y

⇒ x = 5^y ......(1)

Solving, equation 5^{2y + 3}, we get :

⇒ 5^2y.5³

⇒ (5^y)² × 125

Substituting value of 5^y from equation (1) in above equation, we get :

⇒ 125x².

Hence, 5^{2y + 3} = 125x².

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.

Answer

Given,

1st equation :

⇒ log₃ m = x

⇒ m = 3^x ......(1)

2nd equation :

⇒ log₃ n = y

⇒ n = 3^y ......(2)

(i) Given,

3^{2x - 3}

⇒ 3^2x.3^-3

⇒ (3^x)².3^-3

Substituting value of 3^x from equation (1) in above equation, we get :

⇒ m².3^-3

⇒ $\dfrac{m^2}{3^3}$

⇒ $\dfrac{m^2}{27}$.

Hence, 3^{2x - 3} = $\dfrac{m^2}{27}$.

(ii) Simplifying the expression,

⇒ 3^{1 - 2y + 3x}

⇒ 3¹.3^-2y.3^3x

⇒ 3.(3^y)^-2.(3^x)³

Substituting value of 3^x and 3^y from equation (1) and (2) in above equation, we get :

⇒ 3.n^-2.m³

⇒ $\dfrac{3m^3}{n^2}$.

Hence, 3^{1 - 2y + 3x} = $\dfrac{3m^3}{n^2}$.

(iii) Given,

⇒ 2log₃ A = 5x - 3y

⇒ log₃ A² = 5x - 3y

⇒ A² = 3^{5x - 3y}

⇒ A² = 3^5x.3^-3y

⇒ A² = (3^x)⁵.(3^y)^-3

Substituting value of 3^x and 3^y from equation (1) and (2) in above equation, we get :

⇒ A² = m⁵.n^-3

⇒ A² = $\dfrac{m^5}{n^3}$

⇒ A = $\sqrt{\dfrac{m^5}{n^3}}$.

Hence, A = $\sqrt{\dfrac{m^5}{n^3}}$.

Simplify :

log (a)³ - log a

Answer

Simplifying the expression,

⇒ log (a)³ - log a

⇒ 3log a - log a

⇒ 2log a

⇒ log a²

⇒ 2 log a.

Hence, log (a)³ - log a = 2 log a.

Simplify :

log (a)³ ÷ log a

Answer

Simplifying the expression,

⇒ log (a)³ ÷ log a

⇒ 3log a ÷ log a

⇒ $\dfrac{\text{3 log a}}{\text{log a}}$

⇒ 3.

Hence, log (a)³ ÷ log a = 3.

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

Answer

Given,

⇒ log (a + b) = log a + log b

⇒ log (a + b) = log ab

⇒ a + b = ab

⇒ ab - a = b

⇒ a(b - 1) = b

⇒ a = $\dfrac{b}{b - 1}$.

Hence, a = $\dfrac{b}{b - 1}$.

Prove that :

(log a)² - (log b)² = log $\Big(\dfrac{a}{b}\Big)$. log (ab)

Answer

To prove:

(log a)² - (log b)² = log $\Big(\dfrac{a}{b}\Big)$. log (ab)

Solving L.H.S. of the above equation, we get :

⇒ (log a)² - (log b)²

⇒ (log a + log b)(log a - log b)

⇒ log ab. log $\Big(\dfrac{a}{b}\Big)$

Hence, proved that (log a)² - (log b)² = log $\Big(\dfrac{a}{b}\Big)$. log (ab)

Prove that :

If a log b + b log a - 1 = 0, then b^a.a^b = 10

Answer

Given,

⇒ a log b + b log a - 1 = 0

⇒ log b^a + log a^b = 1

⇒ log b^a + log a^b = log 10

⇒ log b^a.a^b = log 10

⇒ b^a.a^b = 10.

Hence, proved that b^a.a^b = 10.

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

Answer

Given,

⇒ log (a + 1) = log (4a - 3) - log 3

⇒ log (a + 1) = log $\dfrac{4a - 3}{3}$

⇒ (a + 1) = $\dfrac{4a - 3}{3}$

⇒ 4a - 3 = 3(a + 1)

⇒ 4a - 3 = 3a + 3

⇒ 4a - 3a = 3 + 3

⇒ a = 6.

Hence, a = 6.

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

Answer

Given,

⇒ 2 log y - log x - 3 = 0

⇒ 2 log y - log x = 3

⇒ log y² - log x = 3

⇒ log $\dfrac{y^2}{x}$ = 3

⇒ $\dfrac{y^2}{x} = 10^3$

⇒ y² = x.10³

⇒ x = $\dfrac{y^2}{10^3} = \dfrac{y^2}{1000}$.

Hence, X = $\dfrac{y^2}{1000}$.

Prove that :

log₁₀ 125 = 3(1 - log₁₀ 2)

Answer

To prove:

log₁₀ 125 = 3(1 - log₁₀ 2)

Solving R.H.S. of the equation, we get :

⇒ 3(1 - log₁₀ 2)

⇒ 3(log₁₀ 10 - log₁₀ 2)

⇒ 3(log₁₀ $\dfrac{10}{2}$)

⇒ 3 log₁₀ 5

⇒ log₁₀ 5³

⇒ log₁₀ 125

Since, L.H.S. = R.H.S.

Hence, proved that log₁₀ 125 = 3(1 - log₁₀ 2).