Logarithms — Exercise 8(A)

The value of $\text{log}_{\sqrt{2}} \space 8$ is :

1. 6

2. 4

3. 3

4. 8

Answer

Let value of $\text{log}_{\sqrt{2}} \space 8$ be x.

$\therefore \text{log}_{\sqrt{2}} \space 8 = x \\[1em] \Rightarrow (\sqrt{2})^x = 8 \\[1em] \Rightarrow (\sqrt{2})^x = (\sqrt{2})^6 \\[1em] \Rightarrow x = 6.$

Hence, Option 1 is the correct option.

If log₄ x = 2.5, the value of x is :

1. 12.5

2. 32

3. 10

4. 20

Answer

Given,

⇒ log₄ x = 2.5

⇒ x = 4^2.5

⇒ x = 4^{2 + 0.5}

⇒ x = 4².4^0.5

⇒ x = $16.(4)^{\dfrac{1}{2}}$

⇒ x = $16.(2^2)^{\dfrac{1}{2}}$

⇒ x = 16 × 2 = 32.

Hence, Option 2 is the correct option.

If $\text{log}_{\sqrt{3}} \space x = 4$, the value of x is :

1. 12

2. 6

3. 9

4. 24

Answer

Given,

$\Rightarrow \text{log}_{\sqrt{3}} \space x = 4 \\[1em] \Rightarrow x = (\sqrt{3})^4 \\[1em] \Rightarrow x = 9.$

Hence, Option 3 is the correct option.

If log_x 64 = 1.5, the value of x is :

1. 48

2. 32

3. 64

4. 16

Answer

Given,

⇒ log_x 64 = 1.5

⇒ 64 = x^1.5

⇒ 64 = (x)^{0.5 + 0.5 + 0.5}

⇒ 64 = x^0.5.x^0.5.x^0.5

⇒ 64 = $\sqrt{x} \times \sqrt{x} \times \sqrt{x}$

⇒ $4^3 = (\sqrt{x})^3$

⇒ $\sqrt{x} = 4$

Squaring both sides, we get :

⇒ $(\sqrt{x})^2 = 4^2$

⇒ x = 16.

Hence, Option 4 is the correct option.

If log₂ (x² - 4) = 5, the value of x is :

1. ±6

2. 6

3. -6

4. ±12

Answer

Given,

⇒ log₂ (x² - 4) = 5

⇒ x² - 4 = 2⁵

⇒ x² - 4 = 32

⇒ x² = 32 + 4

⇒ x² = 36

⇒ x = $\sqrt{36}$

⇒ x = $\pm 6$.

Hence, Option 1 is the correct option.

If log₁₀ x = a, the value of 10^{a - 1} in terms of x is :

1. 10x

2. $\dfrac{x}{10}$

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

4. $\dfrac{1}{10x}$

Answer

Given,

⇒ log₁₀ x = a

⇒ x = 10^a .......(1)

We need to find the value of:

⇒ 10^{a - 1}

⇒ 10^a.10^-1

Substituting value of 10^a from equation (1), in above equation, we get :

⇒ x.10^-1

⇒ $\dfrac{x}{10}$.

Hence, Option 2 is the correct option.

Express each of the following in logarithmic form :

(i) 5³ = 125

(ii) 3^-2 = $\dfrac{1}{9}$

(iii) 10^-3 = 0.001

(iv) $(81)^{\dfrac{3}{4}} = 27$

Answer

(i) Given,

⇒ 5³ = 125

⇒ log₅125 = 3.

Hence, required logarithmic form is log₅ 125 = 3.

(ii) Given,

$\Rightarrow 3^{-2} = \dfrac{1}{9} \\[1em] \Rightarrow \text{log}_{3} \space {\dfrac{1}{9}} = -2.$

Hence, required logarithmic form is $\text{log}_{3} \space {\dfrac{1}{9}} = -2.$

(iii) Given,

⇒ 10^-3 = 0.001

⇒ log₁₀ 0.001 = -3.

Hence, required logarithmic form is log₁₀ 0.001 = -3.

(iv) Given,

$\Rightarrow (81)^{\dfrac{3}{4}} = 27 \\[1em] \Rightarrow \text{log}_{81} \space 27 = \dfrac{3}{4}.$

Hence, required logarithmic form is $\text{log}_{81} \space 27 = \dfrac{3}{4}.$

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

Answer

(i) Given,

⇒ log₈ 0.125 = -1

⇒ 8^-1 = 0.125

Hence, required exponential form is 8^-1 = 0.125

(ii) Given,

⇒ log₁₀ 0.01 = -2

⇒ 10^-2 = 0.01

Hence, required exponential form is 10^-2 = 0.01

(iii) Given,

⇒ log_a A = x

⇒ a^x = A.

Hence, required exponential form is a^x = A.

(iv) Given,

⇒ log₁₀ 1 = 0

⇒ 10⁰ = 1.

Hence, required exponential form is 10⁰ = 1.

Solve for x : log₁₀ x = -2.

Answer

Given,

⇒ log₁₀ x = -2

⇒ x = 10^-2 = $\dfrac{1}{100}$ = 0.01

Hence, x = 0.01

Find the logarithm of 100 to the base 10.

Answer

Let,

⇒ log₁₀ 100 = x

⇒ 100 = 10^x

⇒ 10² = 10^x

⇒ x = 2.

Hence, required value = 2.

Find the logarithm of 0.1 to the base 10.

Answer

Let,

⇒ log₁₀ 0.1 = x

⇒ (10)^x = 0.1

⇒ 10^x = $\dfrac{1}{10}$

⇒ 10^x = (10^-1)

⇒ x = -1.

Hence, required value = -1.

Find the logarithm of 0.001 to the base 10.

Answer

Let,

⇒ log₁₀ (0.001) = x

⇒ (10)^x = 0.001

⇒ $10^x = \dfrac{1}{1000}$

⇒ 10^x = $\dfrac{1}{10^3}$

⇒ 10^x = 10^-3

⇒ x = -3.

Hence, required value = -3.

Find the logarithm of 32 to the base 4.

Answer

Let,

⇒ log₄ 32 = x

⇒ 32 = 4^x

⇒ (2)⁵ = (2²)^x

⇒ (2)⁵ = (2)^2x

⇒ 2x = 5

⇒ x = $\dfrac{5}{2}$.

Hence, required value = $\dfrac{5}{2}$.

Find the logarithm of 0.125 to the base 2.

Answer

Let,

⇒ log₂ (0.125) = x

⇒ 0.125 = 2^x

⇒ $\dfrac{125}{1000} = 2^x$

⇒ $\dfrac{1}{8} = 2^x$

⇒ $\dfrac{1}{2^3} = 2^x$

⇒ 2^-3 = 2^x

⇒ x = -3.

Hence, required value = -3.

Find the logarithm of $\dfrac{1}{16}$ to the base 4.

Answer

Let,

$\Rightarrow \text{log}_{(4)} \space \dfrac{1}{16} = x \\[1em] \Rightarrow \dfrac{1}{16} = 4^x\\[1em] \Rightarrow \dfrac{1}{4^2} = 4^x \\[1em] \Rightarrow 4^x = 4^{-2} \\[1em] \Rightarrow x = -2.$

Hence, required value = -2.

Find the logarithm of 27 to the base 9.

Answer

Let,

⇒ log₉ 27 = x

⇒ 27 = 9^x

⇒ 3³ = (3²)^x

⇒ 3³ = 3^2x

⇒ 2x = 3

⇒ x = $\dfrac{3}{2}$.

Hence, required value = $\dfrac{3}{2}$.

Find the logarithm of $\dfrac{1}{81}$ to the base 27.

Answer

Let,

$\Rightarrow \text{log}_{27} \space \Big(\dfrac{1}{81}\Big) = x \\[1em] \Rightarrow \dfrac{1}{81} = 27^x \\[1em] \Rightarrow \Big(\dfrac{1}{3^4}\Big) = (3^3)^x \\[1em] \Rightarrow 3^{-4} = 3^{3x} \\[1em] \Rightarrow 3x = -4 \\[1em] \Rightarrow x = -\dfrac{4}{3}.$

Hence, required value = $-\dfrac{4}{3}$.

State, true or false :

If log₁₀ x = a, then 10^x = a

Answer

Given,

⇒ log₁₀ x = a

⇒ 10^a = x

Hence, the statement "If log₁₀ x = a, then 10^x = a" is false.

State, true or false :

If x^y = z, then y = log_z x

Answer

Given,

⇒ x^y = z

⇒ log_x z = y.

Hence, the statement "If x^y = z, then y = log_z x" is false.

State, true or false :

log₂ 8 = 3 and log₈ 2 = $\dfrac{1}{3}$.

Answer

Given,

⇒ log₂ 8 = 3

⇒ 2³ = 8

⇒ 8 = 8, which is true.

Given,

⇒ log₈ 2 = $\dfrac{1}{3}$

⇒ $8^{\dfrac{1}{3}}$ = 2

⇒ $(2^3)^{\dfrac{1}{3}}$ = 2

⇒ 2 = 2, which is true.

Hence, the statement "log₂ 8 = 3 and log₈ 2 = $\dfrac{1}{3}$" is True.

Find x, if log₃ x = 0

Answer

Given,

⇒ log₃ x = 0

⇒ x = 3⁰

⇒ x = 1.

Hence, x = 1.

Find x, if log_x 2 = -1

Answer

Given,

⇒ log_x 2 = -1

⇒ (x)^-1 = 2

⇒ $\dfrac{1}{x} = 2$

⇒ x = $\dfrac{1}{2}$.

Hence, x = $\dfrac{1}{2}$.

Find x, if log₉ 243 = x

Answer

Given,

⇒ log₉ 243 = x

⇒ 243 = 9^x

⇒ 3⁵ = (3²)^x

⇒ 3⁵ = 3^2x

⇒ 2x = 5

⇒ x = $\dfrac{5}{2} = 2\dfrac{1}{2}$.

Hence, x = $2\dfrac{1}{2}$.

Find x, if log₅ (x - 7) = 1

Answer

Given,

⇒ log₅ (x - 7) = 1

⇒ x - 7 = 5¹

⇒ x - 7 = 5

⇒ x = 5 + 7 = 12.

Hence, x = 12.

Find x, if log₄ 32 = x - 4

Answer

Given,

⇒ log₄ 32 = x - 4

⇒ 32 = 4^{x - 4}

⇒ 2⁵ = (2²)^{x - 4}

⇒ 2⁵ = 2^{2(x - 4)}

⇒ 2⁵ = 2^{2x - 8}

⇒ 5 = 2x - 8

⇒ 2x = 8 + 5

⇒ 2x = 13

⇒ x = $\dfrac{13}{2} = 6\dfrac{1}{2}$.

Hence, x = $6\dfrac{1}{2}$.

Find x, if log₇ (2x² - 1) = 2

Answer

Given,

⇒ log₇ (2x² - 1) = 2

⇒ 2x² - 1 = 7²

⇒ 2x² - 1 = 49

⇒ 2x² = 49 + 1

⇒ 2x² = 50

⇒ x² = $\dfrac{50}{2}$

⇒ x² = 25

⇒ x = $\sqrt{25} = \pm 5$.

Hence, x = $\pm 5$.

Evaluate log₁₀ 0.01

Answer

Let,

$\Rightarrow \text{log}_{10} \space (0.01) = x \\[1em] \Rightarrow 0.01 = 10^x \\[1em] \Rightarrow \dfrac{1}{100} = 10^x \\[1em] \Rightarrow \dfrac{1}{10^2} = 10^x \\[1em] \Rightarrow 10^{-2} = 10^x \\[1em] \Rightarrow x = -2.$

Hence, log₁₀ 0.01 = -2.

Evaluate log₂ (1 ÷ 8)

Answer

Let,

$\Rightarrow \text{log}_{2} \space (1 ÷ 8) = x \\[1em] \Rightarrow 1 ÷ 8 = 2^x \\[1em] \Rightarrow \dfrac{1}{8} = 2^x \\[1em] \Rightarrow \dfrac{1}{2^3} = 2^x \\[1em] \Rightarrow 2^{-3} = 2^x \\[1em] \Rightarrow x = -3.$

Hence, log₂ (1 ÷ 8) = -3.

Evaluate log₅ 1

Answer

Let,

⇒ log₅ 1 = x

⇒ 1 = 5^x

⇒ 5⁰ = 5^x

⇒ x = 0.

Hence, log₅ 1 = 0.

Evaluate log₅ 125

Answer

Let,

⇒ log₅ 125 = x

⇒ 125 = 5^x

⇒ 5³ = 5^x

⇒ x = 3.

Hence, log₅ 125 = 3.

Evaluate log₁₆ 8

Answer

Let,

⇒ log₁₆ 8 = x

⇒ 8 = 16^x

⇒ 2³ = (2⁴)^x

⇒ 2³ = 2^4x

⇒ 4x = 3

⇒ x = $\dfrac{3}{4}$.

Hence, log₁₆ 8 = $\dfrac{3}{4}$.

Evaluate log_0.5 16

Answer

Let,

$\Rightarrow \text{log}_{0.5} \space 16 = x \\[1em] \Rightarrow 16 = (0.5)^x \\[1em] \Rightarrow 16 = \Big(\dfrac{5}{10}\Big)^x \\[1em] \Rightarrow 2^4 = \Big(\dfrac{1}{2}\Big)^x \\[1em] \Rightarrow 2^4 = (2^{-1})^x \\[1em] \Rightarrow 2^4 = 2^{-x} \\[1em] \Rightarrow -x = 4 \\[1em] \Rightarrow x = -4.$

Hence, log_0.5 16 = -4.

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

Answer

Given,

⇒ log_a m = n

⇒ m = aⁿ

We need to find the value of:

a^{n - 1}

⇒ aⁿ.a^-1

⇒ m.a^-1

⇒ $\dfrac{m}{a}$.

Hence, a^{n - 1} = $\dfrac{m}{a}$.

Given log₂ x = m and log₅ y = n.

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

(ii) Express 5^{3n + 2} in terms of y.

Answer

Given,

⇒ log₂ x = m and log₅ y = n

⇒ x = 2^m ......(1)

and,

⇒ y = 5ⁿ .......(2)

(i) Given,

⇒ 2^{m - 3}

⇒ 2^m.2^-3

⇒ $\dfrac{2^m}{2^3}$

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

⇒ $\dfrac{x}{8}$.

Hence, 2^{m - 3} = $\dfrac{x}{8}$.

(ii) Given,

⇒ 5^{3n + 2}

⇒ (5ⁿ)³.5²

Substituting value of 5ⁿ from equation (2) in above equation, we get :

⇒ 25y³

Hence, 5^{3n + 2} = 25y³.