Logarithms — Exercise 8(D)

The value of (log 8 - log 2) ÷ log 32 is :

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

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

3. 2

4. 4

Answer

Simplifying the expression :

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

Hence, Option 1 is the correct option.

If log₃ (x + 1) = 2, the value of x is :

1. 9

2. 8

3. 3

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

Answer

Given,

⇒ log₃ (x + 1) = 2

⇒ x + 1 = 3²

⇒ x + 1 = 9

⇒ x = 9 - 1 = 8.

Hence, Option 2 is the correct option.

If 2 log x = log 250 - 1, the value of x is :

1. 17

2. -5

3. 5

4. 10

Answer

Given,

⇒ 2 log x = log 250 - 1

⇒ 2 log x = log 250 - log 10

⇒ log x² = log $\dfrac{250}{10}$

⇒ x² = 25

⇒ x = $\sqrt{25}$ = 5.

Hence, Option 3 is the correct option.

If log₃ x - log₃ 2 - 1 = 0; the value of x is :

1. 3

2. -3

3. -6

4. 6

Answer

Given,

⇒ log₃ x - log₃ 2 - 1 = 0

⇒ log₃ x = log₃ 2 + 1

⇒ log₃ x = log₃ 2 + log₃ 3

⇒ log₃ x = log₃ (2 × 3)

⇒ log₃ x = log₃ 6

⇒ x = 6.

Hence, Option 4 is the correct option.

The value of 2 log 3 - $\dfrac{1}{3}$ log 64 + log 12 is :

1. log 27

2. 27

3. -27

4. -log 27

Answer

Given,

$\Rightarrow \text{2 log 3} - \dfrac{1}{3} \text{log 64 + log 12} \\[1em] \Rightarrow \text{log 3}^2 - \text{log 64}^{\dfrac{1}{3}} + \text{log 12} \\[1em] \Rightarrow \text{log 3}^2 - (\text{log 4}^3)^{\dfrac{1}{3}} + \text{log 12} \\[1em] \Rightarrow \text{log 9} - (\text{log 4})^{3 \times \dfrac{1}{3}} + \text{log 12} \\[1em] \Rightarrow \text{log 9 - log 4 + log 12} \\[1em] \Rightarrow \text{log 9 + log 12 - log 4} \\[1em] \Rightarrow \text{log } \dfrac{9 \times 12}{4} \\[1em] \Rightarrow \text{log } \dfrac{108}{4} \\[1em] \Rightarrow \text{log } 27.$

Hence, Option 1 is the correct option.

If $\dfrac{3}{2} \text{ log a} + \dfrac{2}{3} \text{ log b } - 1 = 0$, find the value of a⁹.b⁴

Answer

Given,

$\Rightarrow \dfrac{3}{2} \text{ log a} + \dfrac{2}{3} \text{ log b } - 1 = 0 \\[1em] \Rightarrow \text{log a}^{\dfrac{3}{2}} + \text{log b}^{\dfrac{2}{3}} = 1 \\[1em] \Rightarrow \text{log a}^{\dfrac{3}{2}} + \text{log b}^{\dfrac{2}{3}} = \text{log 10} \\[1em] \Rightarrow \text{log } (a^{\dfrac{3}{2}} \times b^{\dfrac{2}{3}}) = \text{log 10} \\[1em] \Rightarrow (a^{\dfrac{3}{2}} \times b^{\dfrac{2}{3}}) = 10$

Cubing and then squaring both sides, we get :

$\Rightarrow [(a^{\dfrac{3}{2}} \times b^{\dfrac{2}{3}})^3]^2 = [(10)^3]^2 \\[1em] \Rightarrow (a^{\dfrac{3}{2}} \times b^{\dfrac{2}{3}})^6 = 10^6 \\[1em] \Rightarrow a^{\dfrac{3}{2} \times 6}.b^{\dfrac{2}{3} \times 6} = 10^6 \\[1em] \Rightarrow a^{\dfrac{18}{2}}.b^{\dfrac{12}{3}} = 10^6 \\[1em] \Rightarrow a^9.b^4 = 10^6.$

Hence, a⁹.b⁴ = 10⁶.

If x = 1 + log 2 - log 5, y = 2 log 3 and z = log a - log 5; find the value of a, if x + y = 2z.

Answer

Given,

⇒ x + y = 2z

⇒ 1 + log 2 - log 5 + 2 log 3 = 2 (log a - log 5)

⇒ log 10 + log 2 - log 5 + log 3² = 2 log $\dfrac{a}{5}$

⇒ log 10 + log 2 + log 9 - log 5 = log $\Big(\dfrac{a}{5}\Big)^2$

⇒ log $\dfrac{10 \times 2 \times 9}{5}$ = log $\Big(\dfrac{a}{5}\Big)^2$

⇒ 2 × 2 × 9 = $\Big(\dfrac{a}{5}\Big)^2$

⇒ a² = 5² × 2 × 2 × 9

⇒ a² = 900

⇒ a = $\sqrt{900}$ = 30.

Hence, a = 30.

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

(i) x + y - z

(ii) 5^{x + y - z}

Answer

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

⇒ log 0.6 + log 1.25 - (log 3 - 2 log 2)

⇒ log 0.6 + log 1.25 - log 3 + 2 log 2

⇒ log 0.6 + log 1.25 + log 2² - log 3

⇒ log $\dfrac{0.6 \times 1.25 \times 2^2}{3}$

⇒ log $0.2 \times 1.25 \times 4$

⇒ log 1

⇒ 0.

Hence, x + y - z = 0.

(ii) Substituting value of x + y - z from part (i) in 5^{x + y - z}, we get :

⇒ 5⁰

⇒ 1.

Hence, 5^{x + y - z} = 1.

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

Answer

⇒ 3a² - 2b³ = 6 log z

⇒ 3log x - 2log y = 6 log z

⇒ log x³ - log y² = log z⁶

⇒ log $\dfrac{x^3}{y^2}$ = log z⁶

⇒ $\dfrac{x^3}{y^2} = z^6$

⇒ $y^2 = \dfrac{x^3}{z^6}$

⇒ $y = \dfrac{\sqrt{x^3}}{\sqrt{z^6}}$

⇒ y = $x^{\dfrac{3}{2}} ÷ z^3$.

Hence, $y = x^{\dfrac{3}{2}} ÷ z^3$.

If log $\dfrac{a - b}{2} = \dfrac{1}{2}$ (log a + log b), show that :

a² + b² = 6ab

Answer

Given,

$\Rightarrow \text{log } \dfrac{a - b}{2} = \dfrac{1}{2} \text{(log a + log b)} \\[1em] \Rightarrow \text{log } \dfrac{a - b}{2} = \dfrac{1}{2} (\text{log ab}) \\[1em] \Rightarrow \text{log } \dfrac{a - b}{2} = \text{log (ab)} ^{\dfrac{1}{2}} \\[1em] \Rightarrow \dfrac{a - b}{2} = (ab)^{\dfrac{1}{2}}$

Squaring both sides, we get :

$\Rightarrow \Big(\dfrac{a - b}{2}\Big)^2 = (ab)^{\dfrac{1}{2} \times 2} \\[1em] \Rightarrow \dfrac{a^2 + b^2 - 2ab}{4} = ab \\[1em] \Rightarrow a^2 + b^2 - 2ab = 4ab \\[1em] \Rightarrow a^2 + b^2 = 4ab + 2ab \\[1em] \Rightarrow a^2 + b^2 = 6ab.$

Hence, proved that a² + b² = 6ab.

If a² + b² = 23ab, show that :

log $\dfrac{a + b}{5} = \dfrac{1}{2}$ (log a + log b).

Answer

Given,

⇒ a² + b² = 23ab

⇒ a² + b² + 2ab = 23ab + 2ab

⇒ (a + b)² = 25ab

⇒ $\dfrac{(a + b)^2}{25}$ = ab

⇒ $\Big(\dfrac{a + b}{5}\Big)^2$ = ab

Taking log on both sides, we get :

⇒ log $\Big(\dfrac{a + b}{5}\Big)^2$ = log ab

⇒ 2 log $\dfrac{a + b}{5}$ = log a + log b

⇒ log $\dfrac{a + b}{5}$ = $\dfrac{1}{2}$ (log a + log b).

Hence, proved that log $\dfrac{a + b}{5}$ = $\dfrac{1}{2}$ (log a + log b).

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

Answer

Given,

⇒ 2 log (x - 4) = 2m - n

Substituting value of m and n in above equation, we get :

⇒ 2 log (x - 4) = 2 log 20 - log 25

⇒ log (x - 4)² = log 20² - log 25

⇒ log (x - 4)² = log 400 - log 25

⇒ log (x - 4)² = log $\dfrac{400}{25}$

⇒ log (x - 4)² = log 16

⇒ (x - 4)² = 16

⇒ x² + 16 - 8x = 16

⇒ x² - 8x + 16 - 16 = 0

⇒ x² - 8x = 0

⇒ x(x - 8) = 0

⇒ x = 0 or x - 8 = 0

⇒ x = 0 or x = 8.

x cannot be zero as then (x - 4) will be negative.

Hence, x = 8.

Solve for x and y; if x > 0 and y > 0 :

log xy = log $\dfrac{x}{y}$ + 2 log 2 = 2.

Answer

Given,

⇒ log xy = log $\dfrac{x}{y}$ + 2 log 2 = 2 ........(1)

Solving L.H.S. of the equation :

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

From equation (1), we get :

⇒ log xy = 2

⇒ log x(2) = 2

⇒ log 2x = 2

⇒ log 2x = 2log 10

⇒ log 2x = log 10²

⇒ 2x = 10²

⇒ 2x = 100

⇒ x = $\dfrac{100}{2}$ = 50.

Hence, x = 50 and y = 2.

Find x, if :

log_x 625 = -4

Answer

Given,

⇒ log_x 625 = -4

⇒ x^-4 = 625

⇒ $\dfrac{1}{x^4}$ = 625

⇒ $x^4 = \dfrac{1}{625}$

⇒ $x^4 = \Big(\dfrac{1}{5}\Big)^4$

⇒ x = $\dfrac{1}{5}$ = 0.2

Hence, x = 0.2

Find x, if :

log_x (5x - 6) = 2

Answer

Given,

⇒ log_x (5x - 6) = 2

⇒ (5x - 6) = x²

⇒ x² - 5x + 6 = 0

⇒ x² - 3x - 2x + 6 = 0

⇒ x(x - 3) - 2(x - 3) = 0

⇒ (x - 2)(x - 3) = 0

⇒ x - 2 = 0 or x - 3 = 0

⇒ x = 2 or x = 3.

Hence, x = 2 or 3.

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

Answer

Given,

⇒ 2 log (x + 1) = 2p - q

Substituting value of p and q in above equation, we get :

⇒ 2 log (x + 1) = 2 log 20 - log 25

⇒ log (x + 1)² = log 20² - log 25

⇒ log (x + 1)² = log 400 - log 25

⇒ log (x + 1)² = log $\dfrac{400}{25}$

⇒ log (x + 1)² = log 16

⇒ (x + 1)² = 16

⇒ x² + 1 + 2x = 16

⇒ x² + 2x + 1 - 16 = 0

⇒ x² + 2x - 15 = 0

⇒ x² + 5x - 3x - 15 = 0

⇒ x(x + 5) - 3(x + 5) = 0

⇒ (x - 3)(x + 5) = 0

⇒ x - 3 = 0 or x + 5 = 0

⇒ x = 3 or x = -5.

But x cannot be negative, then x + 1 will be negative which is not possible.

Hence, x = 3.

If log₂ (x + y) = log₃ (x - y) = $\dfrac{\text{log 25}}{\text{log 0.2}}$, find the values of x and y.

Answer

Given,

log₂ (x + y) = log₃ (x - y) = $\dfrac{\text{log 25}}{\text{log 0.2}}$

Solving, equation :

$\Rightarrow \text{log}_2 \space (x + y) = \dfrac{\text{log 25}}{\text{log 0.2}} \\[1em] \Rightarrow \text{log}_2 \space (x + y) = \text{log}_{0.2} \space 25 \\[1em] \Rightarrow \text{log}_2 \space (x + y) = \text{log}_{\dfrac{2}{10}} \space 25 \\[1em] \Rightarrow \text{log}_2 \space (x + y) = \text{log}_{\dfrac{1}{5}} \space 5^2 \\[1em] \Rightarrow \text{log}_2 \space (x + y) = \text{log}_{5^{-1}} \space 5^2 \\[1em] \Rightarrow \text{log}_2 \space (x + y) = \dfrac{2}{-1}\text{log}_{5} \space 5 \\[1em] \Rightarrow \text{log}_2 \space (x + y) = -2 \times 1 \\[1em] \Rightarrow \text{log}_2 \space (x + y) = -2 \\[1em] \Rightarrow x + y = 2^{-2} \\[1em] \Rightarrow x + y = \dfrac{1}{2^2} \\[1em] \Rightarrow x + y = \dfrac{1}{4}\text{ ......(1)}$

Solving, equation :

$\Rightarrow \text{log}_3 \space (x - y) = \dfrac{\text{log 25}}{\text{log 0.2}} \\[1em] \Rightarrow \text{log}_3 \space (x - y) = \text{log}_{0.2} \space 25 \\[1em] \Rightarrow \text{log}_3 \space (x - y) = \text{log}_{\dfrac{2}{10}} \space 25 \\[1em] \Rightarrow \text{log}_3 \space (x - y) = \text{log}_{\dfrac{1}{5}} \space 5^2 \\[1em] \Rightarrow \text{log}_3 \space (x - y) = \text{log}_{5^{-1}} \space 5^2 \\[1em] \Rightarrow \text{log}_3 \space (x - y) = \dfrac{2}{-1}\text{log}_{5}5 \\[1em] \Rightarrow \text{log}_3 \space (x - y) = -2 \times 1 \\[1em] \Rightarrow \text{log}_3 \space (x - y) = -2 \\[1em] \Rightarrow x - y = 3^{-2} \\[1em] \Rightarrow x - y = \dfrac{1}{3^2} \\[1em] \Rightarrow x - y = \dfrac{1}{9}\text{ ......(2)}$

Adding equation (1) and (2), we get :

$\Rightarrow (x + y) + (x - y) = \dfrac{1}{4} + \dfrac{1}{9} \\[1em] \Rightarrow x + x + y - y = \dfrac{9 + 4}{36} \\[1em] \Rightarrow 2x = \dfrac{13}{36} \\[1em] \Rightarrow x = \dfrac{13}{36 \times 2} = \dfrac{13}{72}.$

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

$\Rightarrow x + y = \dfrac{1}{4} \\[1em] \Rightarrow \dfrac{13}{72} + y = \dfrac{1}{4} \\[1em] \Rightarrow y = \dfrac{1}{4} - \dfrac{13}{72} \\[1em] \Rightarrow y = \dfrac{18 - 13}{72} \\[1em] \Rightarrow y = \dfrac{5}{72}.$

Hence, x = $\dfrac{13}{72}\text{ and y} = \dfrac{5}{72}$.

Given :

$\dfrac{\text{log x}}{\text{log y}} = \dfrac{3}{2}$ and log (xy) = 5; find the values of x and y.

Answer

Solving, equation :

$\Rightarrow \dfrac{\text{log x}}{\text{log y}} = \dfrac{3}{2} \\[1em] \Rightarrow \text{2 log x = 3 log y} \\[1em] \Rightarrow \text{log } x^2 = \text{log } y^3 \\[1em] \Rightarrow x^2 = y^3 \\[1em] \Rightarrow x = \sqrt{y^3}\text{ ........(1)}$

Substituting value of x from equation (1) in log (xy) = 5, we get :

$\Rightarrow \text{log } (\sqrt{y^3} \times y) = 5 \\[1em] \Rightarrow \text{log } (y^{\dfrac{3}{2}}\times y) = 5 \\[1em] \Rightarrow \text{log } y^{\dfrac{3}{2} + 1} = 5 \\[1em] \Rightarrow y^{\dfrac{5}{2}} = 10^5 \\[1em] \Rightarrow (y^{\dfrac{1}{2}})^5 =10^5 \\[1em] \Rightarrow (\sqrt{y})^5 = 10^5 \\[1em] \Rightarrow \sqrt{y} = 10$

Squaring both sides, we get :

$\Rightarrow y = 10^2$ = 100.

Substituting value of y in equation (1), we get :

$\Rightarrow x = \sqrt{(10^2)^3} \\[1em] = \sqrt{10^6} \\[1em] = 10^3 \\[1em] = 1000.$

Hence, x = 1000 and y = 100.

Given log₁₀ x = 2a and log₁₀ y = $\dfrac{b}{2}$.

(i) Write 10^a in terms of x.

(ii) Write 10^{2b + 1} in terms of y.

(iii) If log₁₀P = 3a - 2b, express P in terms of x and y.

Answer

(i) Given,

⇒ log₁₀ x = 2a

⇒ x = 10^2a

⇒ x = (10^a)²

Taking square root on both sides, we get :

$\Rightarrow \sqrt{x} = 10^a$

Hence, 10^a = $\sqrt{x}$.

(ii) Given,

⇒ 10^{2b + 1}

⇒ 10^2b.10¹ .........(1)

Given,

$\Rightarrow \text{log}_{10}y = \dfrac{b}{2} \\[1em] \Rightarrow y = 10^{\dfrac{b}{2}} \\[1em] \Rightarrow y^4 = (10^{\dfrac{b}{2}})^{4} \\[1em] \Rightarrow y^4 = 10^{2b} \text{ ......(2)}$

Substituting value of 10^2b from equation (2) in equation (1), we get :

⇒ y⁴.10¹

⇒ 10y⁴.

Hence, 10^{2b + 1} = 10y⁴.

(iii) Given,

⇒ log₁₀ P = 3a - 2b

⇒ P = 10^{3a - 2b}

⇒ P = 10^3a.10^-2b

⇒ P = $\dfrac{10^{3a}}{10^{2b}}$

We know that: (shown above)

y⁴ = 10^2b

and

$\sqrt{x}$ = 10^a

∴ P = $\dfrac{10^{3a}}{10^{2b}}$ = $\dfrac{(10^{a})^3}{y^4} = \dfrac{(\sqrt{x})^3}{y^4} = \dfrac{x^{\dfrac{3}{2}}}{y^4}$.

Hence, P = $\dfrac{x^{\dfrac{3}{2}}}{y^4}$

Solve :

log₅ (x + 1) - 1 = 1 + log₅ (x - 1).

Answer

Given,

⇒ log₅ (x + 1) - 1 = 1 + log₅ (x - 1)

⇒ log₅ (x + 1) - log₅ (x - 1) = 1 + 1

⇒ log₅ $\dfrac{x + 1}{x - 1}$ = 2

⇒ $\dfrac{x + 1}{x - 1} = 5^2$

⇒ x + 1 = 25(x - 1)

⇒ x + 1 = 25x - 25

⇒ 25x - x = 1 + 25

⇒ 24x = 26

⇒ x = $\dfrac{26}{24} = \dfrac{13}{12} = 1\dfrac{1}{12}$.

Hence, x = $1\dfrac{1}{12}$.