Logarithms — Test Yourself

The value of log₃ 81 is:

1. 4

2. -4

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

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

Answer

Given,

⇒ log₃ 81

⇒ log₃ (3)⁴

⇒ 4.log₃ 3

⇒ 4 x 1

⇒ 4.

Hence, option 1 is the correct option.

The value of log₁₆ 2 is:

1. 4

2. -4

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

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

Answer

Let, log₁₆ 2 = x

⇒ 16^x = 2

⇒ (2⁴)^x = 2

⇒ 2^4x = 2¹

⇒ 4x = 1

⇒ x = $\dfrac{1}{4}$.

Hence, option 3 is the correct option.

If log(3x - 2) = 2, then the value of x is:

1. 34

2. 30

3. 17

4. none of these

Answer

Given, log(3x - 2) = 2

⇒ log₁₀(3x - 2) = 2

⇒ 3x - 2 = 10²

⇒ 3x - 2 = 100

⇒ 3x = 100 + 2

⇒ 3x = 102

⇒ x = $\dfrac{102}{3}$

⇒ x = 34.

Hence, option 1 is the correct option.

2 + $\dfrac{1}{2}$ log 9 - 2 log 5 is equal to:

1. -log 12

2. log 24

3. log 12

4. log $\dfrac{18}{25}$

Answer

Given,

⇒ 2 + $\dfrac{1}{2}$ log 9 - 2 log 5

⇒ 2log 10 + $\text{log 9}^\dfrac{1}{2}$ - log 5²

⇒ log 10² + log $\sqrt{9}$ - log 5²

⇒ log 100 + log 3 - log 25

⇒ log (100 x 3) - log 25

⇒ log 300 - log 25

⇒ log $\dfrac{300}{25}$

⇒ log 12.

Hence, option 3 is the correct option.

3 + log 10^-2 is equal to:

1. 5

2. 1

3. -5

4. -1

Answer

Given,

⇒ 3 + log 10^-2

⇒ 3 + (-2) log 10

⇒ 3 + (-2) × 1

⇒ 3 - 2

⇒ 1.

Hence, option 2 is the correct option.

Statement 1: log₂ (x² - 4) = 5 ⇒ x = 6

Statement 2: x² - 4 = 2⁵

⇒ x² = 36 and x = $\pm$ 6

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

Given,

⇒ log₂ (x² - 4) = 5

⇒ x² - 4 = 2⁵

⇒ x² - 4 = 32

⇒ x² = 32 + 4

⇒ x² = 36

⇒ x = $\sqrt{36}$

⇒ x = $\pm 6$

∴ Statement 1 is false, and statement 2 is true.

Hence, option 4 is the correct option.

Statement 1: log₃x = a, then 9^a = $\dfrac{1}{x^2}$

Statement 2: log₃x = a ⇒ 3^a = x

∴ 9^a = (3²)^a = (3^a)² = x²

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

Let, log₃x = a

⇒ 3^a = x

⇒ (3^a)² = x²

⇒ (3²)^a = x²

⇒ 9^a = x²

∴ Statement 1 is false, and statement 2 is true.

Hence, option 4 is the correct option.

Assertion (A): $\text{log}_{\sqrt{3}}x = 4$

⇒ x = 9

Reason (R): x = $(\sqrt{3})^4$ = 3² = 9

1. A is true, but R is false.

2. A is false, but R is true.

3. Both A and R are true and R is the correct reason for A.

4. Both A and R are true and R is the incorrect reason for A.

Answer

Given,

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

∴ Both A and R are true and R is the correct reason for A.

Hence, option 3 is the correct option.

Assertion (A): log 2 = a and log 3 = b

⇒ 1 + log 12 = 2a + b

Reason (R): 1 + log 12

= 1 + log 2 x 2 x 3

= 1 + 2log 2 + log 3

= 1 + 2a + b

1. A is true, but R is false.

2. A is false, but R is true.

3. Both A and R are true and R is the correct reason for A.

4. Both A and R are true and R is the incorrect reason for A.

Answer

Given, log 2 = a and log 3 = b

⇒ 1 + log 12

⇒ 1 + log (4 x 3)

⇒ 1 + log (2 x 2 x 3)

⇒ 1 + log 4 + log 3

⇒ 1 + log 2² + log 3

= 1 + 2log 2 + log 3

= 1 + 2a + b.

∴ A is false, but R is true.

Hence, option 2 is the correct option.

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

Answer

Given,

⇒ log₂ x = a

⇒ x = 2^a ........(1)

⇒ log₃ y = a

⇒ y = 3^a ........(2)

Simplifying (72)^a, we get :

⇒ 72^a

⇒ (2³ × 3²)^a

⇒ (2³)^a × (3²)^a

⇒ (2^a)³ × (3^a)²

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

⇒ x³.y²

Hence, 72^a = x³.y²

Solve for x :

log (x - 1) + log (x + 1) = log₂ 1

Answer

Given,

⇒ log (x - 1) + log (x + 1) = log₂ 1

⇒ log (x - 1)(x + 1) = 0

⇒ log (x² + x - x - 1) = 0

⇒ log (x² - 1) = 0

⇒ x² - 1 = 10⁰

⇒ x² - 1 = 1

⇒ x² = 1 + 1

⇒ x² = 2

⇒ x = $\sqrt{2}$.

Hence, x = $\sqrt{2}$.

If log (x² - 21) = 2, show that x = $\pm 11$.

Answer

Given,

⇒ log (x² - 21) = 2

⇒ x² - 21 = 10²

⇒ x² - 21 = 100

⇒ x² = 100 + 21

⇒ x² = 121

⇒ x = $\sqrt{121}$

⇒ x = $\pm 11$.

Hence, proved that x = $\pm 11$.

If x = (100)^a, y = (10000)^b and z = (10)^c, find log $\dfrac{10\sqrt{y}}{x^2z^3}$ in terms of a, b and c.

Answer

Given,

⇒ x = (100)^a

⇒ x = (10²)^a

⇒ x = 10^2a ..........(1)

Given,

⇒ y = (10000)^b

⇒ y = (10⁴)^b

⇒ y = 10^4b ..........(2)

Given,

⇒ z = 10^c ..........(3)

Substituting value of x, y and z from equations (1), (2) and (3) respectively in log $\dfrac{10\sqrt{y}}{x^2z^3}$, we get :

$\Rightarrow \text{log } \dfrac{10\sqrt{y}}{x^2z^3} \\[1em] \Rightarrow \text{log } \dfrac{10 \times \sqrt{10^{4b}}}{(10^{2a})^2 \times (10^c)^3} \\[1em] \Rightarrow \text{log } \dfrac{10 \times \sqrt{(10^{b})^4}}{(10^{4a}) \times (10^{3c})} \\[1em] \Rightarrow \text{log } \dfrac{10 \times 10^{2b}}{10^{4a + 3c}} \\[1em] \Rightarrow \text{log } \dfrac{10^{2b + 1}}{10^{4a + 3c}} \\[1em] \Rightarrow \text{log } 10^{2b + 1} - \text{log } 10^{4a + 3c} \\[1em] \Rightarrow (2b + 1) \times \text{log 10} - (4a + 3c) \times \text{log 10} \\[1em] \Rightarrow (2b + 1) \times 1 - (4a + 3c) \times 1 \\[1em] \Rightarrow (2b + 1) - (4a + 3c) \\[1em] \Rightarrow 2b + 1 - 4a - 3c.$

Hence, log $\dfrac{10\sqrt{y}}{x^2z^3}$ = 2b + 1 - 4a - 3c.

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

Answer

Given,

⇒ 3(log 5 - log 3) - (log 5 - 2 log 6) = 2 - log x

⇒ 3log 5 - 3log 3 - log 5 + 2 log 6 = 2 - log x

⇒ 2log 5 - 3log 3 + 2log 6 = 2 - log x

⇒ 2log 5 - log 3³ + log 6² = 2 - log x

⇒ log 5² - log 27 + log 36 + log x = 2

⇒ log 25 + log 36 + log x - log 27 = 2

⇒ log $\dfrac{25 \times 36 \times x}{27}$ = 2

⇒ $\dfrac{100x}{3} = 10^2$

⇒ 100x = 3 × 10²

⇒ 100x = 300

⇒ x = $\dfrac{300}{100}$ = 3.

Hence, x = 3.

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 $\dfrac{x^2y^3}{z^4}$.

Answer

Given,

1st equation :

⇒ log x = 2m - n

⇒ x = 10^{2m - n} .......(1)

2nd equation :

⇒ log y = n - 2m

⇒ y = 10^{n - 2m} .......(2)

3rd equation :

⇒ log z = 3m - 2n

⇒ z = 10^{3m - 2n} .......(3)

Substituting value of x, y and z in log $\dfrac{x^2y^3}{z^4}$, we get :

$\Rightarrow \text{log } \dfrac{(10^{2m - n})^2(10^{n - 2m})^3}{(10^{3m - 2n})^4} \\[1em] \Rightarrow \text{log } \dfrac{10^{2(2m - n)}.10^{3(n - 2m)}}{10^{4(3m - 2n)}} \\[1em] \Rightarrow \text{log } \dfrac{10^{4m - 2n}.10^{3n - 6m}}{10^{12m - 8n}} \\[1em] \Rightarrow \text{log } 10^{4m - 2n + 3n - 6m - (12m - 8n)} \\[1em] \Rightarrow \text{log } 10^{n - 2m - 12m + 8n} \\[1em] \Rightarrow \text{log } 10^{9n - 14m} \\[1em] \Rightarrow (9n - 14m) \text{ log 10} \\[1em] \Rightarrow (9n - 14m) \times 1 \\[1em] \Rightarrow 9n - 14m.$

Hence, log $\dfrac{x^2y^3}{z^4}$ = 9n - 14m.

Given log_x 25 - log_x 5 = 2 - log_x $\dfrac{1}{125}$; find x.

Answer

Given,

⇒ log_x 25 - log_x 5 = 2 - log_x $\dfrac{1}{125}$

⇒ log_x 25 - log_x 5 + log_x $\dfrac{1}{125}$ = 2

⇒ log_x $\dfrac{25 \times \dfrac{1}{125}}{5}$ = 2

⇒ log_x $\dfrac{25}{625}$ = 2

⇒ x² = $\dfrac{25}{625}$

⇒ $x^2 = \dfrac{1}{25}$

⇒ x = $\sqrt{\dfrac{1}{25}} = \dfrac{1}{5}$.

Hence, x = $\dfrac{1}{5}$.

Solve for x, if :

log_x 49 - log_x 7 + log_x $\dfrac{1}{343}$ + 2 = 0

Answer

Given,

$\Rightarrow \text{log}_x \space {49} + \text{log}_x \space {\dfrac{1}{343}} - \text{log}_x7 = -2 \\[1em] \Rightarrow \text{log}_x \space {\dfrac{49 \times \dfrac{1}{343}}{7}} = -2 \\[1em] \Rightarrow \text{log}_x \space {\dfrac{49}{7 \times 343}} = -2 \\[1em] \Rightarrow \text{log}_x \space {\dfrac{1}{49}} = -2 \\[1em] \Rightarrow x^{-2} = \dfrac{1}{49} \\[1em] \Rightarrow x^{-2} = \dfrac{1}{7^2} \\[1em] \Rightarrow x^{-2} = 7^{-2} \\[1em] \Rightarrow x = 7.$

Hence, x = 7.

If a² = log x, b³ = log y and $\dfrac{a^2}{2} - \dfrac{b^3}{3}$ = log c, find c in terms of x and y.

Answer

Given,

a² = log x and b³ = log y

Substituting value of a² and b³ in $\dfrac{a^2}{2} - \dfrac{b^3}{3}$ = log c, we get :

$\Rightarrow \dfrac{a^2}{2} - \dfrac{b^3}{3} = \text{log c} \\[1em] \Rightarrow \dfrac{\text{log x}}{2} - \dfrac{\text{log y}}{3} = \text{log c} \\[1em] \Rightarrow \dfrac{\text{3 log x - 2 log y}}{6} = \text{log c} \\[1em] \Rightarrow \text{log x}^3 - \text{log y}^2 = \text{6 log c} \\[1em] \Rightarrow \text{log x}^3 - \text{log y}^2 = \text{log c}^6 \\[1em] \Rightarrow \text{log c}^6 = log \dfrac{x^3}{y^2} \\[1em] \Rightarrow c^6 = \dfrac{x^3}{y^2} \\[1em] \Rightarrow c = \sqrt[6]{\dfrac{x^3}{y^2}}.$

Hence, c = $\sqrt[6]{\dfrac{x^3}{y^2}}.$

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

(i) x - y - z

(ii) 13^{x - y - z}

Answer

Given,

x = log₁₀ 12, y = log₄ 2 × log₁₀ 9 and z = log₁₀ 0.4

(i) Substituting value of x, y and z in equation x - y - z, we get :

⇒ x - y - z = log₁₀ 12 - log₄ 2 × log₁₀ 9 - log₁₀ 0.4

= log₁₀ 12 - log(_2²) 2 × log₁₀ 9 - log₁₀ 0.4

= log₁₀ 12 - $\dfrac{1}{2}$ log ₂ 2 × log₁₀ 9 - log₁₀ 0.4

= log₁₀ 12 - $\dfrac{1}{2}$ × 1 × log₁₀ 9 - log₁₀ 0.4

= log₁₀ 12 - $\dfrac{1}{2}$ × log₁₀ 9 - log₁₀ 0.4

= log₁₀12 - log₁₀ $9^{\dfrac{1}{2}}$ - log₁₀ 0.4

= log₁₀12 - log₁₀3 - log₁₀0.4

= log₁₀12 - (log₁₀3 + log₁₀0.4)

= log₁₀ $\dfrac{12}{3 \times 0.4}$

= log₁₀ $\dfrac{12}{1.2}$

= log₁₀ 10

= 1.

Hence, x - y - z = 1.

(ii) Substituting value of x - y - z in 13^{x - y - z}, we get :

⇒ 13¹ = 13.

Hence, 13^{x - y - z} = 13.

Solve for x, log_x $15\sqrt{5} = 2 - \text{log}_x \space 3\sqrt{5}$

Answer

Given,

$\Rightarrow \text{log}_x \space 15\sqrt{5} = 2 - \text{log}_x \space 3\sqrt{5} \\[1em] \Rightarrow \text{log}_x \space 15\sqrt{5} + \text{log}_x \space 3\sqrt{5} = 2 \\[1em] \Rightarrow \text{log}_x \space (15\sqrt{5} \times 3\sqrt{5}) = 2 \\[1em] \Rightarrow \text{log}_x \space 225 = 2 \\[1em] \Rightarrow x^2 = 225 \\[1em] \Rightarrow x = \sqrt{225} = 15.$

Hence, x = 15.

Evaluate :

log_b a × log_c b × log_a c

Answer

Simplifying the expression,

$\Rightarrow \text{log}_b a \times \text{log}_c b \times \text{log}_a c \\[1em] \Rightarrow \dfrac{\text{log a}}{\text{log b}} \times \dfrac{\text{log b}}{\text{log c}} \times \dfrac{\text{log c}}{\text{log a}} \\[1em] \Rightarrow \dfrac{\text{log a.log b.log c}}{\text{log b.log c.log a}} \\[1em] \Rightarrow 1.$

Hence, log_b a × log_c b × log_a c = 1.

Evaluate :

log₃ 8 ÷ log₉ 16

Answer

Simplifying the expression,

$\Rightarrow \text{log}_3 \space 8 ÷ \text{log}_9 \space 16 \\[1em] \Rightarrow \dfrac{\text{log 8}}{\text{log 3}} ÷ \dfrac{\text{log 16}}{\text{log 9}} \\[1em] \Rightarrow \dfrac{\text{log 8}}{\text{log 3}} \times \dfrac{\text{log 9}}{\text{log 16}} \\[1em] \Rightarrow \dfrac{\text{log 2}^3}{\text{log 3}} \times \dfrac{\text{log 3}^2}{\text{log 2}^4} \\[1em] \Rightarrow \dfrac{\text{3 log 2}}{\text{log 3}} \times \dfrac{\text{2 log 3}}{\text{4 log 2}} \\[1em] \Rightarrow \dfrac{\text{6 log 2.log 3}}{\text{4 log 2.log 3}} \\[1em] \Rightarrow \dfrac{6}{4} \\[1em] \Rightarrow \dfrac{3}{2} \\[1em] \Rightarrow 1\dfrac{1}{2}.$

Hence, log₃ 8 ÷ log₉ 16 = $1\dfrac{1}{2}$.

Evaluate :

$\dfrac{\text{log}_5 \space 8}{\text{log }_{25} \space 16 \times \text{log }_{100} \space 10}$

Answer

Simplifying the expression,

$\Rightarrow \dfrac{\text{log}_5 \space 8}{\text{log }_{25} \space 16 \times \text{log }_{100} \space 10} \\[1em] \Rightarrow \dfrac{\text{log}_5 \space 8}{\text{log }_{5^2} \space 16 \times \text{log}_{10^2} \space 10} \\[1em] \Rightarrow \dfrac{\text{log}_5 \space 8}{\dfrac{1}{2} \times \text{log }_{5} \space 16 \times \dfrac{1}{2} \times \text{log}_{10} \space 10} \\[1em] \Rightarrow \dfrac{\text{log}_5 \space 8}{\dfrac{1}{4} \times \text{log}_{5} \space 16} \\[1em] \Rightarrow \dfrac{4\text{log}_5 \space 8}{\text{log}_{5} \space 16} \\[1em] \Rightarrow \dfrac{4 \times \dfrac{\text{log 8}}{\text{log 5}}}{\dfrac{\text{log 16}}{\text{log 5}}} \\[1em] \Rightarrow \dfrac{\text{4 log 8. log 5}}{\text{log 16. log 5}} \\[1em] \Rightarrow \dfrac{\text{4 log 8}}{\text{log 16}} \\[1em] \Rightarrow \dfrac{\text{4 log 2}^3}{\text{log 2}^4} \\[1em] \Rightarrow \dfrac{4 \times 3 \times \text{log 2}}{4 \times \text{log 2}} \\[1em] \Rightarrow \dfrac{12}{4} \\[1em] \Rightarrow 3.$

Hence, $\dfrac{\text{log}_5 \space 8}{\text{log }_{25} \space 16 \times \text{log }_{100} \space 10}$ = 3.

Show that :

log_a m ÷ log_ab m = 1 + log_a b

Answer

Simplifying L.H.S. of the given equation, we get :

$\Rightarrow \text{log}_a \space m ÷ \text{log}_{ab} \space m \\[1em] \Rightarrow \dfrac{\text{log m}}{\text{log a}} ÷ \dfrac{\text{log m}}{\text{log ab}}\\[1em] \Rightarrow \dfrac{\text{log m}}{\text{log a}} \times \dfrac{\text{log ab}}{\text{log m}}\\[1em] \Rightarrow \dfrac{\text{log ab}}{\text{log a}} \\[1em] \Rightarrow \text{log}_{a} \space ab \\[1em] \Rightarrow \text{log}_{a} \space a + \text{log}_{a} \space b \\[1em] \Rightarrow 1 + \text{log}_{a} \space b.$

Hence, proved that log_a m ÷ log_ab m = 1 + log_a b.

If $\text{log}_{\sqrt{27}} \space x = 2\dfrac{2}{3}$, find x.

Answer

Given,

$\Rightarrow \text{log}_{\sqrt{27}} \space x = 2\dfrac{2}{3} \\[1em] \Rightarrow \text{log}_{\sqrt{27}} \space x = \dfrac{8}{3} \\[1em] \Rightarrow \text{log}_{\sqrt{3^3}} \space x = \dfrac{8}{3} \\[1em] \Rightarrow \text{log}_{3^{\frac{3}{2}}} \space x = \dfrac{8}{3} \\[1em] \Rightarrow \dfrac{1}{\dfrac{3}{2}} \text{ log}_{3} \space x = \dfrac{8}{3} \\[1em] \Rightarrow \dfrac{2}{3}\text{ log}_{3} \space x = \dfrac{8}{3} \\[1em] \Rightarrow \text{ log}_{3} \space x = \dfrac{8}{3} \times \dfrac{3}{2} \\[1em] \Rightarrow \text{ log}_{3} \space x = 4 \\[1em] \Rightarrow x = 3^4 = 81.$

Hence, x = 81.

$\dfrac{1}{\text{log}_{a} \space bc + 1} + \dfrac{1}{\text{log}_{b} \space ca + 1} + \dfrac{1}{\text{log}_{c} \space ab + 1}$

Answer

Evaluating,

$\Rightarrow \dfrac{1}{\text{log}_{a} \space bc + 1} + \dfrac{1}{\text{log}_{b} \space ca + 1} + \dfrac{1}{\text{log}_{c} \space ab + 1} \\[1em] \Rightarrow \dfrac{1}{\dfrac{\text{log bc}}{\text{log a}} + 1} + \dfrac{1}{\dfrac{\text{log ca}}{\text{log b}} + 1} + \dfrac{1}{\dfrac{\text{log ab}}{\text{log c}} + 1} \\[1em] \Rightarrow \dfrac{1}{\dfrac{\text{log bc + log a}}{\text{log a}}} + \dfrac{1}{\dfrac{\text{log ca + log b}}{\text{log b}}} + \dfrac{1}{\dfrac{\text{log ab + log c}}{\text{log c}}} \\[1em] \Rightarrow \dfrac{\text{log a}}{\text{log bc + log a}} + \dfrac{\text{log b}}{\text{log ca + log b}} + \dfrac{\text{log c}}{\text{log ab + log c}} \\[1em] \Rightarrow \dfrac{\text{log a}}{\text{log b + log c + log a}} + \dfrac{\text{log b}}{\text{log c + log a + log b}} + \dfrac{\text{log c}}{\text{log a + log b + log c}} \\[1em] \Rightarrow \dfrac{\text{log a + log b + log c}}{\text{log a + log b + log c}} \\[1em] \Rightarrow 1.$

Hence, $\dfrac{1}{\text{log}_{a} \space bc + 1} + \dfrac{1}{\text{log}_{b} \space ca + 1} + \dfrac{1}{\text{log}_{c} \space ab + 1}$ = 1.