Logarithms — Multiple Choice Questions

If $\text{log}_{\sqrt{3}} \space 27 = x$, then the value of x is

1. 3

2. 4

3. 6

4. 9

Answer

Given,

$\Rightarrow x = \text{log}_{\sqrt{3}} \space 27 \\[1em] = \dfrac{\text{log}_{10} \space 27}{\text{log}_{10} \space \sqrt{3}} \\[1em] = \dfrac{\text{log}_{10} \space 3^3}{\text{log}_{10} \space 3^{\dfrac{1}{2}}} \\[1em] = \dfrac{3\text{log}_{10} \space 3}{\dfrac{1}{2}\text{log}_{10} \space 3} \\[1em] = \dfrac{3}{\dfrac{1}{2}} \\[1em] = 3 \times 2 = 6.$

Hence, Option 3 is the correct option.

If log₅ (0.04) = x, then the value of x is

1. 2

2. 4

3. -4

4. -2

Answer

Given,

log₅ (0.04) = x

⇒ 5^x = 0.04

⇒ 5^x = $\dfrac{4}{100} = \dfrac{1}{25}$

⇒ 5^x = $\dfrac{1}{5^2}$

⇒ 5^x = 5^-2

⇒ x = -2.

Hence, Option 4 is the correct option.

If log_0.5 64 = x, then the value of x is

1. -4

2. -6

3. 4

4. 6

Answer

Given,

log_0.5 64 = x

$\Rightarrow (0.5)^x = 64 \\[1em] \Rightarrow \Big(\dfrac{5}{10}\Big)^x = 64 \\[1em] \Rightarrow \Big(\dfrac{1}{2}\Big)^x = (2)^6 \\[1em] \Rightarrow (2^{-1})^x = (2)^6 \\[1em] \Rightarrow 2^{-x} = (2)^6 \\[1em] \Rightarrow -x = 6 \\[1em] \Rightarrow x = -6.$

Hence, Option 2 is the correct option.

If $\text{log}_{\sqrt[3]{5}} \space x$ = -3, then the value of x is

1. $\dfrac{1}{5}$

2. $-\dfrac{1}{5}$

3. -1

4. 5

Answer

Given,

$\Rightarrow \text{log}_{\sqrt[3]{5}} \space x = -3 \\[1em] \Rightarrow x = (\sqrt[3]{5})^{-3} \\[1em] \Rightarrow x = [(5)^{\dfrac{1}{3}}]^{-3} \\[1em] \Rightarrow x = (5)^{\dfrac{1}{3} \times (-3)} \\[1em] \Rightarrow x = (5)^{-1} = \dfrac{1}{5}.$

Hence, Option 1 is the correct option.

If log (3x + 1) = 2, then the value of x is

1. $\dfrac{1}{3}$

2. 99

3. 33

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

Answer

Given,

log (3x + 1) = 2

⇒ 10² = 3x + 1

⇒ 100 = 3x + 1

⇒ 3x = 100 - 1

⇒ 3x = 99

⇒ x = 33.

Hence, Option 3 is the correct option.

The value of 2 + log₁₀ (0.01) is

1. 4

2. 3

3. 1

4. 0

Answer

Given,

2 + log₁₀ (0.01)

⇒ 2 + log₁₀ (10^-2)

⇒ 2 + (-2)log₁₀ 10

⇒ 2 + (-2)1

⇒ 2 + (-2)

⇒ 0

Hence, Option 4 is the correct option.

The value of $\dfrac{\text{log} \space 8 - \text{log} \space 2}{\text{log} \space 32}$ is

1. $\dfrac{2}{5}$

2. $\dfrac{1}{4}$

3. $-\dfrac{2}{5}$

4. $\dfrac{1}{3}$

Answer

Given,

$\Rightarrow \dfrac{\text{log} \space 8 - \text{log} \space 2}{\text{log} \space 32} \\[1em] \Rightarrow \dfrac{\text{log} \space 2^3 - \text{log} \space 2}{\text{log} \space 2^5} \\[1em] \Rightarrow \dfrac{3\text{log} \space 2 - \text{log} \space 2}{5\text{log} \space 2} \\[1em] \Rightarrow \dfrac{2 \text{log} \space 2}{5\text{log} \space 2} \\[1em] \Rightarrow \dfrac{2}{5}.$

Hence, Option 1 is the correct option.

Consider the following two statements :

Statement 1: log₅ 150 = log₅ 25 + log₅ 125

Statement 2: log_a (b + c) = log_a b + log_a c

Which of the following is valid?

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and Statement 2 is false.

4. Statement 1 is false, and Statement 2 is true.

Answer

According to statement 1; log₅ 150 = log₅ 25 + log₅ 125

Solving R.H.S.,

⇒ log₅ 25 + log₅ 125

⇒ log₅ (25 x 125)

⇒ log₅ 3125.

As, log₅ 150 ≠ log₅ 3125

∴ Statement 1 is false.

According to statement 2; log_a (b + c) = log_a b + log_a c

Solving R.H.S.

⇒ log_a b + log_a c

⇒ log_a (b x c)

⇒ log_a bc.

As, log_a (b + c) ≠ log_a bc

∴ Statement 2 is false.

∴ Both statements are false.

Hence, option 2 is the correct option.