Logarithms — Exercise 8.2

Simplify the following :

log a³ - log a²

Answer

Given,

⇒ log a³ - log a²

⇒ 3log a - 2log a

⇒ log a.

Hence, log a³ - log a² = log a

Simplify the following :

log a³ ÷ log a²

Answer

Given,

log a³ ÷ log a²

$\Rightarrow \dfrac{\text{log a}^3}{\text{log a}^2} \\[1em] \Rightarrow \dfrac{3\text{log a}}{2\text{log a}} \\[1em] \Rightarrow \dfrac{3}{2}.$

Hence, log a³ ÷ log a² = $\dfrac{3}{2}.$

Simplify the following :

$\dfrac{\text{log} 4}{\text{log} 2}$

Answer

Given,

$\Rightarrow \dfrac{\text{log 4}}{\text{log 2}} \\[1em] \Rightarrow \dfrac{\text{log 2}^2}{\text{log 2}} \\[1em] \Rightarrow \dfrac{\text{2 log 2}}{\text{log 2}} \\[1em] \Rightarrow 2.$

Hence, $\dfrac{\text{log} 4}{\text{log} 2}$ = 2.

Simplify the following :

$\dfrac{\text{log 8 log 9}}{\text{log 27}}$

Answer

Given,

$\Rightarrow \dfrac{\text{log 8 log 9}}{\text{log 27}} \\[1em] \Rightarrow \dfrac{\text{log 2}^3 \text{log 3}^2}{\text{log 3}^3} \\[1em] \Rightarrow \dfrac{\text{3log 2. 2log 3}}{3\text{log} 3} \\[1em] \Rightarrow \dfrac{\text{6log 2.log 3}}{\text{3log 3}} \\[1em] \Rightarrow 2\text{log 2} = \text{log 2}^2 \\[1em] = \text{log 4}.$

Hence, $\dfrac{\text{log 8 log 9}}{\text{log 27}}$ = log 4.

Simplify the following :

$\dfrac{\text{log 27}}{\text{log } \sqrt{3}}$

Answer

Given,

$\Rightarrow \dfrac{\text{log 27}}{\text{log }\sqrt{3}} \\[1em] \Rightarrow \dfrac{\text{log 3}^3}{\text{log 3}^{\dfrac{1}{2}}} \\[1em] \Rightarrow \dfrac{\text{3log 3}}{\dfrac{1}{2}\text{log 3}} \\[1em] \Rightarrow \dfrac{3}{\dfrac{1}{2}} \\[1em] \Rightarrow 6.$

Hence, $\dfrac{\text{log 27}}{\text{log }\sqrt{3}}$ = 6.

Simplify the following :

$\dfrac{\text{log 9 - log 3}}{\text{ log 27}}$

Answer

Given,

$\Rightarrow \dfrac{\text{log 9 - log 3}}{\text{log 27}} \\[1em] \Rightarrow \dfrac{\text{log 3}^2 - \text{log 3}}{\text{log 3}^3} \\[1em] \Rightarrow \dfrac{\text{2log 3 - log 3}}{\text{3log 3}} \\[1em] \Rightarrow \dfrac{\text{log 3}}{\text{3log 3}} \\[1em] \Rightarrow \dfrac{1}{3}.$

Hence, $\dfrac{\text{log 9 - log 3}}{\text{log 27}} = \dfrac{1}{3}.$

Evaluate the following:

log$(10 ÷ \sqrt[3]{10})$

Answer

Given,

$\Rightarrow \text{log}(10 ÷ \sqrt[3]{10}) \\[1em] \Rightarrow \text{log}\Big(\dfrac{10}{\sqrt[3]{10}}\Big) \\[1em] \Rightarrow \text{log}\Big(\dfrac{10}{10^{\dfrac{1}{3}}}\Big) \\[1em] \Rightarrow \text{log}(10)^{1 - \dfrac{1}{3}} \\[1em] \Rightarrow \text{log}(10)^{\dfrac{2}{3}} \\[1em] \Rightarrow \dfrac{2}{3}\text{log 10} \\[1em] \Rightarrow \dfrac{2}{3}.$

Hence, $\text{log}(10 ÷ \sqrt[3]{10}) = \dfrac{2}{3}$.

Evaluate the following:

2 + $\dfrac{1}{2}$log (10)^-3

Answer

Given,

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

Hence, $2 + \dfrac{1}{2}\text{log} (10)^{-3} = \dfrac{1}{2}$.

Evaluate the following:

2log 5 + log 8 - $\dfrac{1}{2}$log 4

Answer

Given,

$\Rightarrow \text{2log 5 + log 8} - \dfrac{1}{2}\text{log 4} \\[1em] \Rightarrow \text{log 5}^2 + \text{log 8} - \dfrac{1}{2} \times \text{log 2}^2 \\[1em] \Rightarrow \text{log 25 + log 8} - \dfrac{1}{2} \times 2\text{log 2} \\[1em] \Rightarrow \text{log 25 + log 8 - log 2} \\[1em] \Rightarrow \text{log} \dfrac{25 \times 8}{2} \\[1em] \Rightarrow \text{log 100} \\[1em] \Rightarrow \text{log} 10^2 = 2\text{log 10} \Rightarrow 2 \times 1 = 2.$

Hence, $\text{2log 5 + log 8} - \dfrac{1}{2}\text{log 4} = 2.$

Evaluate the following:

2log 10³ + 3log 10^-2 - $\dfrac{1}{3}\text{log 5}^{-3} + \dfrac{1}{2}\text{log 4}$

Answer

Given,

$\Rightarrow 2\text{log} 10^3 + 3\text{log} 10^{-2} - \dfrac{1}{3}\text{log 5}^{-3} + \dfrac{1}{2}\text{log 4} \\[1em] \Rightarrow 2 \times 3\text{log 10} + 3 \times -2\text{log 10} - \dfrac{1}{3} \times (-3) \text{log 5} + \dfrac{1}{2}\text{log 2}^2 \\[1em] \Rightarrow 2 \times 3 \times 1 + 3 \times (-2) \times 1 - (-1)\text{log 5} + \dfrac{1}{2} \times 2 \times \text{log 2} \\[1em] \Rightarrow 6 - 6 + \text{log 5} + \text{log 2} \\[1em] \Rightarrow \text{log 5 × 2} \\[1em] \Rightarrow \text{log 10} \\[1em] 1.$

Hence, $2\text{log} 10^3 + 3\text{log} 10^{-2} - \dfrac{1}{3}\text{log 5}^{-3} + \dfrac{1}{2}\text{log 4}$ = 1.

Evaluate the following:

2log 2 + log 5 - $\dfrac{1}{2}\text{log 36} - \text{log}\dfrac{1}{30}$

Answer

Given,

$\Rightarrow \text{2log 2 + log 5} - \dfrac{1}{2}\text{log 36 - log}\dfrac{1}{30} \\[1em] \Rightarrow \text{2log 2 + log 5} - \dfrac{1}{2} \text{log 6}^2 - \text{(log 1 - log 30)} \\[1em] \Rightarrow \text{2log 2 + log 5} - \dfrac{1}{2} \times 2 \times \text{log 6 - log 1 + log 30} \\[1em] \Rightarrow 2\text{log } 2 + \text{log } 5 - \text{log } 6 - 0 + \text{log } (5 \times 6) \\[1em] \Rightarrow \text{log } 2^2 + \text{log 5 - log 6 + log 5 + log 6} \\[1em] \Rightarrow \text{log 4 + log 5 + log 5} \\[1em] \Rightarrow \text{log } (4 \times 5 \times 5) \\[1em] \Rightarrow \text{log 100} = 2.$

Hence, $\text{2log 2 + log 5} - \dfrac{1}{2}\text{log 36 - log}\dfrac{1}{30}$ = 2.

Evaluate the following:

2log 5 + log 3 + 3log 2 - $\dfrac{1}{2}$ log 36 - 2log 10

Answer

Given,

⇒ 2log 5 + log 3 + 3log 2 - $\dfrac{1}{2}$ log 36 - 2log 10

⇒ log 5² + log 3 + log 2³ - $\dfrac{1}{2}$ log 6² - log 10²

⇒ log 25 + log 3 + log 8 - $\dfrac{1}{2} \times 2$ log 6 - log 100

⇒ log 25 + log 3 + log 8 - log 6 - log 100

⇒ log $\dfrac{25 \times 3 \times 8}{6 \times 100}$

⇒ log $\dfrac{600}{600}$

⇒ log 1

⇒ 0.

Hence, 2log 5 + log 3 + 3log 2 - $\dfrac{1}{2}$ log 36 - 2log 10 = 0.

Evaluate the following:

log 2 + 16log $\dfrac{16}{15} + 12\text{log } \dfrac{25}{24} + 7\text{log }\dfrac{81}{80}$

Answer

Given,

$\Rightarrow \text{log 2 + 16log }\dfrac{16}{15} + 12\text{log } \dfrac{25}{24} + 7\text{log }\dfrac{81}{80} \\[1em] \Rightarrow \text{log 2 + 16(log 16 - log 15) + 12(log 25 - log 24) + 7(log 81 - log 80)} \\[1em] \Rightarrow \text{log 2 + 16log 16 - 16log 15 + 12log 25 - 12log 24 + 7log 81 - 7log 80} \\[1em] \Rightarrow \text{log 2 + 16log 2}^4 - \text{16log 3.5 + 12log 5}^2 - \text{12log 2}^3.3 + \text{7log 3}^4 - \text{7log 2}^4.5 \\[1em] \Rightarrow \text{log 2 + 16.4log 2 - 16(log 3 + log 5) + 12.2log 5 - 12(log 2}^3 + \text{log 3) + 7.4log 3} - \text{7(log 2}^4 + \text{log 5)} \\[1em] \Rightarrow \text{log 2 + 64log 2 - 16log 3 - 16log 5 + 24log 5 - 12.3log 2 - 12log 3 + 28log 3 - 28log 2 - 7log 5} \\[1em] \Rightarrow \text{65log 2 - 36log 2 - 28log 2 - 16log 3 - 12log 3 + 28log 3 - 16log 5 + 24log 5 - 7log 5} \\[1em] \Rightarrow \text{log 2 + log 5} \\[1em] \Rightarrow \text{log 2.5} \\[1em] \Rightarrow \text{log 10} = 1.$

Hence, $\text{log 2 + 16log }\dfrac{16}{15} + 12\text{log } \dfrac{25}{24} + 7\text{log }\dfrac{81}{80} = 1$.

Evaluate the following:

2log₁₀5 + log₁₀8 - $\dfrac{1}{2}$log₁₀4

Answer

Given,

⇒ 2log₁₀5 + log₁₀8 - $\dfrac{1}{2}$log₁₀4

⇒ log₁₀5² + log₁₀8 - $\text{log}_{10}4^{\dfrac{1}{2}}$

⇒ log₁₀25 + log₁₀8 - log₁₀2

⇒ log₁₀$\dfrac{25 \times 8}{2}$

⇒ log₁₀100

⇒ log₁₀10²

⇒ 2log₁₀10

⇒ 2.

Hence, 2log₁₀5 + log₁₀8 - $\dfrac{1}{2}$log₁₀4 = 2.

Express the following as a single logarithm:

2log 3 - $\dfrac{1}{2}$log 16 + log 12

Answer

Given,

$\Rightarrow \text{2log 3} - \dfrac{1}{2}\text{log 2}^4 + \text{log 2}^2.3 \\[1em] \Rightarrow \text{2log 3} - \dfrac{1}{2} \times 4 \times \text{log 2} + \text{log 2}^2 + \text{log 3} \\[1em] \Rightarrow \text{2log 3} - 2\text{log 2} + 2\text{log 2} + \text{log 3} \\[1em] \Rightarrow \text{2log 3 + log 3} \\[1em] \Rightarrow \text{3log 3} \\[1em] \Rightarrow \text{log 3}^3 = \text{log 27}.$

Hence, 2log 3 - $\dfrac{1}{2}$log 16 + log 12 = log 27.

Express the following as a single logarithm:

2log₁₀5 - log₁₀2 + 3log₁₀4 + 1

Answer

Given,

⇒ 2log₁₀5 - log₁₀2 + 3log₁₀4 + 1

⇒ log₁₀5² - log₁₀2 + log₁₀4³ + log₁₀10

⇒ log₁₀25 + log₁₀64 + log₁₀10 - log₁₀2

⇒ log₁₀$\dfrac{25 \times 64 \times 10}{2}$

⇒ log₁₀$\dfrac{16000}{2} = \text{log}_{10}8000$.

Hence, 2log₁₀5 - log₁₀2 + 3log₁₀4 + 1 = log₁₀8000.

Express the following as a single logarithm:

$\dfrac{1}{2}$log 36 + 2log 8 - log 1.5

Answer

Given,

$\Rightarrow \dfrac{1}{2}\text{log 36 + 2log 8 - log 1.5} \\[1em] \Rightarrow \dfrac{1}{2}\text{log 6}^2 + \text{log 8}^2 - \text{log} \dfrac{15}{10} \\[1em] \Rightarrow \dfrac{1}{2} \times 2 \times \text{log 6 + log 64 - (log 15 - log 10)} \\[1em] \Rightarrow \text{log 6 + log 64 - log 15 + log 10} \\[1em] \Rightarrow \text{log} \dfrac{6 \times 64 \times 10}{15} \\[1em] \Rightarrow \text{log 256}.$

Hence, $\dfrac{1}{2}\text{log 36 + 2log 8 - log 1.5}$ = log 256.

Express the following as a single logarithm:

$\dfrac{1}{2}$log 25 - 2log 3 + 1

Answer

Given,

$\Rightarrow \dfrac{1}{2}\text{log 25 - 2log 3 + 1} \\[1em] \Rightarrow \dfrac{1}{2}\text{log 5}^2 - \text{2log 3 + log 10} \\[1em] \Rightarrow \dfrac{1}{2} \times 2\text{log 5} - \text{log 3}^2 + \text{log 10} \\[1em] \Rightarrow \text{log 5 - log 9 + log 10} \\[1em] \Rightarrow \text{log }\dfrac{5 \times 10}{9} \\[1em] \Rightarrow \text{log }\dfrac{50}{9}.$

Hence, $\dfrac{1}{2}\text{log 25 - 2log 3 + 1} = \text{log }\dfrac{50}{9}.$

Express the following as a single logarithm:

$\dfrac{1}{2}$log 9 + 2log 3 - log 6 + log 2 - 2

Answer

Given,

$\Rightarrow \dfrac{1}{2}\text{log 9 + 2log 3 - log 6 + log 2 - 2} \\[1em] \Rightarrow \dfrac{1}{2}\text{log 3}^2 + \text{log 3}^2 - \text{log 6 + log 2 - 2log 10} \\[1em] \Rightarrow \dfrac{1}{2} \times 2 \times \text{log 3 + log 9 - log 6 + log 2 - log 10}^2 \\[1em] \Rightarrow \text{log 3 + log 9 + log 2 - log 6 - log 100} \\[1em] \Rightarrow \text{log }\dfrac{3 \times 9 \times 2}{6 \times 100} \\[1em] \Rightarrow \text{log } \dfrac{9}{100}.$

Hence, $\dfrac{1}{2}\text{log 9 + 2log 3 - log 6 + log 2 - 2} = \text{log }\dfrac{9}{100}$.

Prove the following:

log₁₀4 ÷ log₁₀2 = log₃9

Answer

Given,

log₁₀4 ÷ log₁₀2 = log₃9

Simplifying L.H.S. we get,

⇒ log₁₀4 ÷ log₁₀2 = log₁₀2² ÷ log₁₀2

= $\dfrac{2\text{log}_{10}2}{\text{log}_{10}2}$

= 2.

Simplifying R.H.S. we get,

⇒ log₃9 = log₃3²

= 2log₃3

= 2.

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

Hence, proved that log₁₀4 ÷ log₁₀2 = log₃9.

Prove the following:

log₁₀25+ log₁₀4 = log₅25

Answer

Given,

log₁₀25+ log₁₀4 = log₅25

Simplifying L.H.S. we get,

⇒ log₁₀25+ log₁₀4 = log₁₀(25 × 4)

= log₁₀100

= log₁₀10²

= 2log₁₀10 = 2.

Simplifying R.H.S. we get,

⇒ log₅25 = log₅5²

= 2log₅5

= 2(1) = 2.

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

Hence, proved that log₁₀25+ log₁₀4 = log₅25.

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

Answer

Given,

x = (100)^a = (10²)^a = 10^2a,

y = (10000)^b = (10⁴)^b = 10^4b,

z = (10)^c.

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

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

If a = log₁₀x, find the following in terms of a:

(i) x

(ii) log₁₀$\sqrt[5]{x^2}$

(iii) log₁₀ 3x

Answer

(i) Given,

a = log₁₀x

⇒ x = 10^a.

(ii) Given,

$\Rightarrow \text{log}_{10}(x^2)^{\dfrac{1}{5}} \\[1em] \Rightarrow \dfrac{1}{5} \times 2\text{log}_{10}x \\[1em] \Rightarrow \dfrac{1}{5} \times 2\text{log}_{10}10^a \\[1em] \Rightarrow \dfrac{2a}{5}\text{log}_{10}10 \\[1em] \Rightarrow \dfrac{2a}{5} \times 1 \\[1em] \Rightarrow \dfrac{2a}{5}.$

Hence, log₁₀$\sqrt[5]{x^2} = \dfrac{2a}{5}$.

(iii) Given,

a = log₁₀ x

Now,

⇒ log₁₀ 3x

⇒ log₁₀ (3 × x)

⇒ log₁₀ 3 + log₁₀ x

⇒ log₁₀ 3 + a.

Hence, log₁₀ 3x = log₁₀ 3 + a.

If a = log$\dfrac{2}{3}$, b = log$\dfrac{3}{5}$ and c = 2log$\sqrt{\dfrac{5}{2}}$, find the value of

(i) a + b + c

(ii) 5^{a + b + c}

Answer

(i) Given,

$\Rightarrow a + b + c = \text{log }\dfrac{2}{3} + \text{log }\dfrac{3}{5} + 2\text{log }\sqrt{\dfrac{5}{2}} \\[1em] = \text{log 2 - log 3 + log 3 - log 5} + 2 \times \text{log}\Big(\dfrac{5}{2}\Big)^{\dfrac{1}{2}} \\[1em] = \text{log } 2 - \text{log } 5 + 2 \times \dfrac{1}{2} \times \text{(log 5 - log 2)} \\[1em] = \text{log 2 - log 5 + log 5 - log 2} \\[1em] = 0.$

Hence, a + b + c = 0.

(ii) Given,

⇒ 5^{a + b + c} = 5⁰ = 1.

Hence, 5^{a + b + c} = 1.

If x = log $\dfrac{3}{5}, \text{ y = log }\dfrac{5}{4}\text{ and z = 2log }\dfrac{\sqrt{3}}{2}$, find the values of

(i) x + y - z

(ii) 3^{x + y - z}

Answer

Given,

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

Hence, x + y - z = 0.

(ii) Given,

3^{x + y - z} = 3⁰ = 1.

Hence, 3^{x + y - z} = 1.

If x = log₁₀12, y = log₄2 × log₁₀9 and z = log₁₀0.4, find the values of

(i) x - y - z

(ii) 7^{x - y - z}

Answer

(i) Given,

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

$= \text{log}_{10} \space 3.4 - \text{log}_{4} \space (4)^{\dfrac{1}{2}} \times \text{log}_{10} \space 3^2 - \text{log}_{10} \space \dfrac{4}{10} \\[1em] = \text{log}_{10} \space 3 + \text{log}_{10} \space 4 - \dfrac{1}{2}\text{log}_{4}4 \space \times 2\text{log}_{10} \space 3 - (\text{log}_{10} \space 4 - \text{log}_{10} \space 10) \\[1em] = \text{log}_{10} \space 3 + \text{log}_{10} \space 4 - 1 \times \text{log}_{10} \space 3 - \text{log}_{10} \space 4 + \text{log}_{10} \space 10 \\[1em] = \text{log}_{10} \space 3 - \text{log}_{10} \space 3 + 1 \\[1em] = 1.$

Hence, x - y - z = 1.

(ii) Given,

7^{x - y - z} = 7¹ = 7.

Hence, 7^{x - y - z} = 1.

If log V + log 3 = log π + log 4 + 3log r, find V in terms of other quantities.

Answer

Given,

log V + log 3 = log π + log 4 + 3log r

⇒ log V = log π + log 4 + 3log r - log 3

⇒ log V = log π + log 4 + log r³ - log 3

⇒ log V = log $\dfrac{π4r^3}{3}$

⇒ V = $\dfrac{4}{3}πr^3$.

Given 3(log 5 - log 3) - (log 5 - 2log 6) = 2 - log n, find n.

Answer

Given,

3(log 5 - log 3) - (log 5 - 2log 6) = 2 - log n

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

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

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

⇒ log 25 - log 27 + log 36 = 2log 10 - log n

⇒ log$\dfrac{25 \times 36}{27}$ = log 10² - log n

⇒ log$\dfrac{100}{3}$ = log 100 - log n

⇒ log 100 - log 3 = log 100 - log n

⇒ log n = log 3

⇒ n = 3.

Hence, n = 3.

Given that log₁₀y + 2log₁₀x = 2, express y in terms of x.

Answer

Given,

log₁₀y + 2log₁₀x = 2

⇒ log₁₀y + log₁₀x² = 2

⇒ log₁₀yx² = 2log₁₀10

⇒ log₁₀yx² = log₁₀10²

⇒ log₁₀yx² = log₁₀100

⇒ yx² = 100

⇒ y = $\dfrac{100}{x^2}.$

Hence, y = $\dfrac{100}{x^2}$.

Express log₁₀2 + 1 in the form of log₁₀x.

Answer

Given,

⇒ log₁₀2 + 1

⇒ log₁₀2 + log₁₀10

⇒ log₁₀2 × 10

⇒ log₁₀20

Hence, log₁₀2 + 1 = log₁₀20.

If a² = log₁₀x, b³ = log₁₀y and $\dfrac{a^2}{2} - \dfrac{b^3}{3}$ = log₁₀z, express z in terms of x and y.

Answer

Given,

a² = log₁₀x

b³ = log₁₀y

Substituting above values in $\dfrac{a^2}{2} - \dfrac{b^3}{3} = \text{log}_{10}z$ we get,

$\Rightarrow \dfrac{\text{log}_{10}x}{2} - \dfrac{\text{log}_{10}y}{3} = \text{log}_{10}z \\[1em] \Rightarrow \dfrac{3\text{log}_{10}x - 2\text{log}_{10}y}{6} = \text{log}_{10}z \\[1em] \Rightarrow \dfrac{\text{log}_{10}x^3 - \text{log}_{10}y^2}{6} = \text{log}_{10}z \\[1em] \Rightarrow \dfrac{1}{6}\text{log}_{10}\dfrac{x^3}{y^2} = \text{log}_{10}z \\[1em] \Rightarrow \text{log}_{10}\Big(\dfrac{x^3}{y^2}\Big)^{\dfrac{1}{6}} = \text{log}_{10}z \\[1em] \Rightarrow \text{log}_{10}\Big(\dfrac{x^\dfrac{1}{2}}{y^\dfrac{1}{3}}\Big) = \text{log}_{10}z \\[1em] \Rightarrow \text{log}_{10}\dfrac{\sqrt{x}}{\sqrt[3]y} = \text{log}_{10}z \\[1em] \Rightarrow z = \dfrac{\sqrt{x}}{\sqrt[3]y}.$

Hence, $z = \dfrac{\sqrt{x}}{\sqrt[3]y}.$

Given that log m = x + y and log n = x - y, express the value of log m²n in terms of x and y.

Answer

Given,

log m = x + y .......(i)

log n = x - y ........(ii)

Multiplying (i) by 2 we get,

2log m = log m² = 2x + 2y ......(iii)

Adding (ii) and (iii) we get,

⇒ log m² + log n = 2x + 2y + (x - y)

⇒ log m²n = 3x + y.

Hence, log m²n = 3x + y.

Given that log x = m + n and log y = m - n, express the value of log$\Big(\dfrac{10x}{y^2}\Big)$ in terms of m and n.

Answer

Given,

$\Rightarrow \text{log }\Big(\dfrac{10x}{y^2}\Big) = \text{log }10x - \text{log } y^2$

= log 10 + log x - 2log y

= 1 + m + n - 2(m - n)

= 1 + m + n - 2m + 2n

= 1 - m + 3n.

Hence, log$\Big(\dfrac{10x}{y^2}\Big)$ = 1 - m + 3n.

If $\dfrac{\text{log x}}{2} = \dfrac{\text{log y}}{3}$, find the value of $\dfrac{y^4}{x^6}$.

Answer

Given,

$\dfrac{\text{log x}}{2} = \dfrac{\text{log y}}{3}$

⇒ 3log x = 2log y

⇒ log x³ = log y²

⇒ x³ = y²

Squaring both sides we get,

⇒ x⁶ = y⁴

⇒ $\dfrac{x^6}{y^4} = 1$.

Hence, $\dfrac{x^6}{y^4} = 1$.

Solve for x:

log x + log 5 = 2log 3

Answer

Given,

⇒ log x + log 5 = 2log 3

⇒ log x = log 3² - log 5

⇒ log x = log 9 - log 5

⇒ log x = log $\dfrac{9}{5}$

⇒ x = $\dfrac{9}{5}$.

Hence, $x = \dfrac{9}{5}$.

Solve for x:

log₃x - log₃2 = 1

Answer

Given,

⇒ log₃x - log₃2 = 1

⇒ $\text{log}_3 \dfrac{x}{2} = \text{log}_3 3$

⇒ $\dfrac{x}{2} = 3$

⇒ x = 6.

Hence, x = 6.

Solve for x:

x = $\dfrac{\text{log 125}}{\text{log 25}}$

Answer

Given,

$\Rightarrow x = \dfrac{\text{log 125}}{\text{log 25}} \\[1em] \Rightarrow x = \dfrac{\text{log 5}^3}{\text{log 5}^2} \\[1em] \Rightarrow x = \dfrac{\text{3log 5}}{\text{2log 5}} \\[1em] \Rightarrow x = \dfrac{3}{2}.$

Hence, x = $\dfrac{3}{2}$.

Solve for x:

$\dfrac{\text{log 8}}{\text{log 2}} \times \dfrac{\text{log 3}}{\text{log}\sqrt{3}} = 2\text{log} x$

Answer

Given,

$\Rightarrow \dfrac{\text{log 8}}{\text{log 2}} \times \dfrac{\text{log 3}}{log\sqrt{3}} = 2log x \\[1em] \Rightarrow \dfrac{\text{log 2}^3}{\text{log 2}} \times \dfrac{\text{log 3}}{\text{log} 3^{\dfrac{1}{2}}} = 2log x \\[1em] \Rightarrow \dfrac{\text{3log 2}}{\text{log 2}} \times \dfrac{\text{log 3}}{\dfrac{1}{2}\text{log 3}} = \text{2 log x} \\[1em] \Rightarrow 3 \times 2 = 2\text{log } x \\[1em] \Rightarrow \text{log } x = \dfrac{6}{2} \\[1em] \Rightarrow \text{log } x = 3 \\[1em] \Rightarrow x = 10^3 = 1000.$

Hence, x = 1000.

Given 2log₁₀x + 1 = log₁₀250, find

(i) x

(ii) log₁₀2x

Answer

(i) Given,

2log₁₀x + 1 = log₁₀250

⇒ log₁₀x² + log₁₀10 = log₁₀250

⇒ log₁₀x² = log₁₀250 - log₁₀10

⇒ log₁₀x² = $\text{log}_{10}\dfrac{250}{10}$

⇒ log₁₀x² = log₁₀25

⇒ x² = 25

⇒ x = 5.

Hence, x = 5.

(ii) Given,

log₁₀2x

⇒ log₁₀2(5) = log₁₀10 = 1.

Hence, log₁₀2x = 1.

If $\dfrac{\text{log x}}{\text{log 5}} = \dfrac{\text{log y}^2}{\text{log 2}} = \dfrac{\text{log 9}}{\text{log}\dfrac{1}{3}}$, find x and y.

Answer

Given,

$\dfrac{\text{log x}}{\text{log 5}} = \dfrac{\text{log y}^2}{\text{log 2}} = \dfrac{\text{log 9}}{\text{log}\dfrac{1}{3}}$

Considering,

$\Rightarrow \dfrac{\text{log x}}{\text{log 5}} = \dfrac{\text{log 9}}{\text{log}\dfrac{1}{3}} \\[1em] \Rightarrow \text{log x} = \dfrac{\text{log 9 × log 5}}{\text{log}\dfrac{1}{3}} \\[1em] \Rightarrow \text{log x} = \dfrac{\text{log 9 × log 5}}{\text{log 1 - log 3}} \\[1em] \Rightarrow \text{log x} = \dfrac{\text{log 3}^2 × \text{log } 5}{\text{0 - log 3}} \\[1em] \Rightarrow \text{log x} = \dfrac{\text{2log 3 × log 5}}{\text{-log 3}} \\[1em] \Rightarrow \text{log x} = \text{-2log 5} \\[1em] \Rightarrow \text{log x} = \text{log 5}^{-2} \\[1em] \Rightarrow x = 5^{-2} = \dfrac{1}{25}.$

Considering,

$\Rightarrow \dfrac{\text{log y}^2}{\text{log 2}} = \dfrac{\text{log 9}}{\text{log}\dfrac{1}{3}} \\[1em] \Rightarrow \text{log y}^2 = \dfrac{\text{log 9} \times \text{log 2}}{\text{log 1 - log 3}} \\[1em] \Rightarrow \text{log y}^2 = \dfrac{\text{log 3}^2 \times \text{log 2}}{\text{0 - log 3}} \\[1em] \Rightarrow \text{log y}^2 = \dfrac{\text{2log 3} \times \text{log 2}}{\text{-log 3}} \\[1em] \Rightarrow \text{log y}^2 = \text{-2log 2} \\[1em] \Rightarrow \text{log y}^2 = \text{log 2}^{-2} \\[1em] \Rightarrow y^2 = 2^{-2} \\[1em] \Rightarrow y = \sqrt{\dfrac{1}{2^2}} = \dfrac{1}{2}.$

Hence, $x = \dfrac{1}{25} \text{and y} = \dfrac{1}{2}.$

Prove the following:

3^{log 4} = 4^{log 3}

Answer

Given,

3^{log 4} = 4^{log 3}

Taking log on both sides we get,

⇒ log 3^{log 4} = log 4^{log 3}

⇒ log 4.log 3 = log 3.log 4

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

Hence, proved that 3^{log 4} = 4^{log 3}.

Prove the following:

27^{log 2} = 8^{log 3}

Answer

Given,

27^{log 2} = 8^{log 3}

Taking log on both sides we get,

⇒ log 27^{log 2} = log 8^{log 3}

⇒ log 2.log 27 = log 3.log 8

⇒ log 2.log 3³ = log 3.log 2³

⇒ 3.log 2.log 3 = 3log 3.log 2

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

Hence, proved that 27^{log 2} = 8^{log 3}.

Solve the following equation:

log(2x + 3) = log 7

Answer

Given,

⇒ log(2x + 3) = log 7

⇒ 2x + 3 = 7

⇒ 2x = 4

⇒ x = 2.

Hence, x = 2.

Solve the following equation:

log(x + 1) + log(x - 1) = log 24

Answer

Given,

⇒ log(x + 1) + log(x - 1) = log 24

⇒ log((x + 1)(x - 1)) = log 24

⇒ log(x² - 1) = log 24

⇒ x² - 1 = 24

⇒ x² = 25

⇒ x = 5.

Hence, x = 5.

Solve the following equation:

log(10x + 5) - log(x - 4) = 2

Answer

Given,

log(10x + 5) - log(x - 4) = 2

$\Rightarrow \text{log}(10x + 5) - \text{log}(x - 4) = 2\text{log } 10 \\[1em] \Rightarrow \text{log}\dfrac{10x + 5}{x - 4} = \text{log } 10^2 \\[1em] \Rightarrow \dfrac{10x + 5}{x - 4} = 100 \\[1em] \Rightarrow 10x + 5 = 100(x - 4) \\[1em] \Rightarrow 10x + 5 = 100x - 400 \\[1em] \Rightarrow 100x - 10x = 405 \\[1em] \Rightarrow 90x = 405 \\[1em] \Rightarrow x = \dfrac{405}{90} \\[1em] \Rightarrow x = 4.5$

Hence, x = 4.5

Solve the following equation:

log₁₀5 + log₁₀(5x + 1) = log₁₀(x + 5) + 1

Answer

Given,

⇒ log₁₀5 + log₁₀(5x + 1) = log₁₀(x + 5) + 1

⇒ log₁₀5(5x + 1) = log₁₀(x + 5) + log₁₀10

⇒ log₁₀(25x + 5) = log₁₀10(x + 5)

⇒ log₁₀(25x + 5) = log₁₀(10x + 50)

⇒ 25x + 5 = 10x + 50

⇒ 25x - 10x = 50 - 5

⇒ 15x = 45

⇒ x = 3.

Hence, x = 3.

Solve the following equation:

log(4y - 3) = log(2y + 1) - log 3

Answer

Given,

⇒ log(4y - 3) = log(2y + 1) - log 3

⇒ $\text{log}(4y - 3) = \text{log}\dfrac{2y + 1}{3}$

⇒ $\dfrac{2y + 1}{3} = 4y - 3$

⇒ 2y + 1 = 3(4y - 3)

⇒ 2y + 1 = 12y - 9

⇒ 12y - 2y = 1 + 9

⇒ 10y = 10

⇒ y = 1.

Hence, y = 1.

Solve the following equation:

log₁₀(x + 2) + log₁₀(x - 2) = log₁₀3 + 3log₁₀4

Answer

Given,

⇒ log₁₀ (x + 2) + log₁₀ (x - 2) = log₁₀ 3 + 3log₁₀ 4

⇒ log₁₀ (x + 2)(x - 2) = log₁₀ 3 + log₁₀ 4³

⇒ log₁₀ (x² - 4) = log₁₀ 3 + log₁₀ 64

⇒ log₁₀ (x² - 4) = log₁₀ (3 × 64)

⇒ log₁₀ (x² - 4) = log₁₀ 192

⇒ x² - 4 = 192

⇒ x² = 196

⇒ x = $\sqrt{196}$ = 14.

Hence, x = 14.

Solve the following equation:

log(3x + 2) + log(3x - 2) = 5log 2

Answer

Given,

⇒ log(3x + 2) + log(3x - 2) = 5log 2

⇒ log(3x + 2)(3x - 2) = log 2⁵

⇒ log(9x² - 4) = log 32

⇒ 9x² - 4 = 32

⇒ 9x² = 36

⇒ x² = 4

⇒ x = 2.

Hence, x = 2.

Solve for x: log₃(x + 1) - 1 = 3 + log₃(x - 1)

Answer

Given,

log₃(x + 1) - 1 = 3 + log₃(x - 1)

Since log₃3 = 1, the above equation can be written as

⇒ log₃(x + 1) - log₃3 = 3log₃3 + log₃(x - 1)

⇒ log₃(x + 1) - log₃3 = log₃3³ + log₃(x - 1)

⇒ $\text{log}_3\dfrac{(x + 1)}{3} = \text{log}_3\space[3^3 \times (x - 1)]$

⇒ $\text{log}_3\dfrac{(x + 1)}{3} = \text{log}_3\space[27(x - 1)]$

⇒ $\dfrac{x + 1}{3} = 27(x - 1)$

⇒ x + 1 = 81(x - 1)

⇒ x + 1 = 81x - 81

⇒ 81x - x = 1 + 81

⇒ 80x = 82

⇒ x = $\dfrac{82}{80} = \dfrac{41}{40} = 1\dfrac{1}{40}.$

Hence, x = $1\dfrac{1}{40}$.

Solve for x: 5^{log x} + 3^{log x} = 3^{log x + 1} - 5^{log x - 1}.

Answer

Given,

5^{log x} + 3^{log x} = 3^{log x + 1} - 5^{log x - 1}

⇒ 5^{log x} + 3^{log x} = 3^{log x}.3¹ - 5^{log x}.5^-1

⇒ 5^{log x} + 5^{log x}.5^-1 = 3^{log x}.3¹ - 3^{log x}

⇒ 5^{log x}(1 + 5^-1) = 3^{log x}(3 - 1)

⇒ 5^{log x}$\Big(1 + \dfrac{1}{5}\Big)$ = 2.3^{log x}

⇒ 5^{log x}.$\dfrac{6}{5}$ = 2.3^{log x}

$\Rightarrow \dfrac{5^{\text{log } x}}{3^{\text{log } x}} = \dfrac{5 \times 2}{6} \\[1em] \Rightarrow \Big(\dfrac{5}{3}\Big)^{\text{log } x} = \dfrac{5}{3} \\[1em] \Rightarrow \text{log } x = 1 \\[1em] \Rightarrow \text{log } x = \text{log } 10 \\[1em] \Rightarrow x = 10.$

Hence, x = 10.

If log$\dfrac{x - y}{2} = \dfrac{1}{2}$(log x + log y), prove that x² + y² = 6xy.

Answer

Given,

$\Rightarrow \text{log}\dfrac{x - y}{2} = \dfrac{1}{2}(\text{log x + log y}) \\[1em] \Rightarrow 2\text{log}\dfrac{x - y}{2} = \text{log xy} \\[1em] \Rightarrow \Big(\dfrac{x - y}{2}\Big)^2 = xy \\[1em] \Rightarrow (x - y)^2 = 4xy \\[1em] \Rightarrow x^2 + y^2 - 2xy = 4xy \\[1em] \Rightarrow x^2 + y^2 = 6xy.$

Hence, proved that x² + y² = 6xy.

If x² + y² = 23xy, prove that log$\dfrac{x + y}{5} = \dfrac{1}{2}$(log x + log y).

Answer

Given,

x² + y² = 23xy

Above equation can be written as,

$\Rightarrow x^2 + y^2 = 25xy - 2xy \\[1em] \Rightarrow x^2 + y^2 + 2xy = 25xy \\[1em] \Rightarrow \dfrac{x^2 + y^2 + 2xy}{25} = xy \\[1em] \Rightarrow \Big(\dfrac{x + y}{5}\Big)^2 = xy$

Taking log on both sides we get,

$\Rightarrow \text{log }\Big(\dfrac{x + y}{5}\Big)^2 = \text{log }xy \\[1em] \Rightarrow 2\text{log }\dfrac{x + y}{5} = \text{log x + log y} \\[1em] \Rightarrow \text{log }\dfrac{x + y}{5} = \dfrac{1}{2}(\text{log x + log y}) \\[1em]$

Hence, proved that $\text{log}\dfrac{x + y}{5} = \dfrac{1}{2}(\text{log x + log y})$.

If p = log₁₀ 20 and q = log₁₀ 25, find the value of x if

2log₁₀ (x + 1) = 2p - q

Answer

Given,

2log₁₀(x + 1) = 2p - q

⇒ 2log₁₀(x + 1) = 2log₁₀20 - log₁₀25

⇒ log₁₀(x + 1)² = log₁₀20² - log₁₀25

⇒ log₁₀(x + 1)² = log₁₀400 - log₁₀25

⇒ log₁₀(x + 1)² = $\text{log}_{10}\space\dfrac{400}{25}$

⇒ log₁₀(x + 1)² = log₁₀16

⇒ (x + 1)² = 16

⇒ x² + 1 + 2x = 16

x² + 2x - 15 = 0

x² + 5x - 3x - 15 = 0

x(x + 5) - 3(x + 5) = 0

(x - 3)(x + 5) = 0

x = 3 or -5.

But x ≠ -5 as then (x + 1) will be negative.

Hence, x = 3.

Show that:

$\dfrac{1}{\text{log}_2\space42} + \dfrac{1}{\text{log}_3\space42} + \dfrac{1}{\text{log}_7\space42} = 1$

Answer

Given,

$\dfrac{1}{\text{log}_2\space42} + \dfrac{1}{\text{log}_3\space42} + \dfrac{1}{\text{log}_7\space42} = 1$

Simplifying L.H.S. we get,

$\dfrac{1}{\text{log}_2\space42} + \dfrac{1}{\text{log}_3\space42} + \dfrac{1}{\text{log}_7\space42}$ = log₄₂ 2 + log₄₂ 3 + log₄₂ 7

= log₄₂ (2 × 3 × 7)

= log₄₂ 42

= 1.

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

Hence, proved that $\dfrac{1}{\text{log}_2\space42} + \dfrac{1}{\text{log}_3\space42} + \dfrac{1}{\text{log}_7\space42} = 1$

Show that:

$\dfrac{1}{\text{log}_8\space36} + \dfrac{1}{\text{log}_9\space36} + \dfrac{1}{\text{log}_{18}\space36} = 2$

Answer

Given,

$\dfrac{1}{\text{log}_8\space36} + \dfrac{1}{\text{log}_9\space36} + \dfrac{1}{\text{log}_{18}\space36} = 2$

Simplifying L.H.S. we get,

$\dfrac{1}{\text{log}_8\space36} + \dfrac{1}{\text{log}_9\space36} + \dfrac{1}{\text{log}_{18}\space36}$ = log₃₆ 8 + log₃₆ 9 + log₃₆ 18

= log₃₆ (8 × 9 × 18)

= log₃₆ (36)²

= 2log₃₆ 36

= 2.

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

Hence, proved that $\dfrac{1}{\text{log}_8\space36} + \dfrac{1}{\text{log}_9\space36} + \dfrac{1}{\text{log}_{18}\space36} = 2$.

Prove the following identities:

$\dfrac{1}{\text{log}_a\text{ abc}} + \dfrac{1}{\text{log}_b\text{ abc}} + \dfrac{1}{\text{log}_c\text{ abc}} = 1$

Answer

Given,

$\dfrac{1}{\text{log}_a\text{ abc}} + \dfrac{1}{\text{log}_b\text{ abc}} + \dfrac{1}{\text{log}_c\text{ abc}} = 1$

Simplifying L.H.S. we get,

$\Rightarrow \dfrac{1}{\text{log}_a\text{ abc}} + \dfrac{1}{\text{log}_b\text{ abc}} + \dfrac{1}{\text{log}\_c\text{ abc}} \\[1em] \Rightarrow \text{log}_\text{abc}\text{ a} + \text{log}_\text{abc}\text{ b} + \text{log}_\text{abc}\text{ c} \\[1em] \Rightarrow \text{log}_\text{abc}\space(\text{a }\times \text{b } \times \text{c}) \\[1em] \Rightarrow \text{log}_\text{abc}\text{ abc} \\[1em] \Rightarrow 1.$

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

Hence, proved that $\dfrac{1}{\text{log}_a\text{ abc}} + \dfrac{1}{\text{log}_b\text{ abc}} + \dfrac{1}{\text{log}_c\text{ abc}} = 1$.

Prove the following identities:

log_b a . log_c b . log_d c = log_d a

Answer

Given,

log_ba . log_cb . log_dc = log_da

Simplifying L.H.S. we get,

$\Rightarrow \text{log}_\text{b}\text{ a}\space.\space\text{log}_\text{c}\text{ b}\space.\space\text{log}_\text{d}\text{ c} \\[1em] \Rightarrow \dfrac{\text{log a}}{\text{log b}} \times \dfrac{\text{log b}}{\text{log c}} \times \dfrac{\text{log c}}{\text{log d}} \\[1em] \Rightarrow \dfrac{\text{log a}}{\text{log d}} \\[1em] \Rightarrow \text{log}\_\text{d}\text{ a}.$

Hence, proved that log_b a . log_c b . log_d c = log_d a.

Given that log_a x = $\dfrac{1}{α}$, log_b x = $\dfrac{1}{β}$, log_c x = $\dfrac{1}{γ}$, find log_abc x.

Answer

Given,

$\Rightarrow \dfrac{1}{α} = \text{log}_a\space x = \dfrac{\text{log x}}{\text{log a}} \\[1em] \Rightarrow \text{log a} = α\text{log x} \\[1em] \\[1em] \Rightarrow \dfrac{1}{β} = \text{log}_b\space x = \dfrac{\text{log x}}{\text{log b}} \\[1em] \Rightarrow \text{log b} = β\text{log x} \\[1em] \\[1em] \Rightarrow \dfrac{1}{γ} = \text{log}_c\space x = \dfrac{\text{log x}}{\text{log c}} \\[1em] \Rightarrow \text{log c} = γ\text{log x} \\[1em] \\[1em]$

Solving log_abc x we get,

$\Rightarrow \text{log}_\text{abc}\space x = \dfrac{\text{log x}}{\text{log abc}} \\[1em] = \dfrac{\text{log x}}{\text{log a + log b + log c}} \\[1em] = \dfrac{\text{log x}}{\text{αlog x + βlog x + γlog x}} \\[1em] = \dfrac{\text{log x}}{\text{log x(α + β + γ)}} \\[1em] = \dfrac{1}{(\text{α} + \text{β} + \text{γ})}.$

Hence, log_abc x = $\dfrac{1}{(\text{α} + \text{β} + \text{γ})}$.

Solve for x:

log₃ x + log₉ x + log₈₁ x = $\dfrac{7}{4}$

Answer

Given,

$\Rightarrow \text{log}_3\space x + \text{log}_9\space x + \text{log}_{81} \space x = \dfrac{7}{4} \\[1em] \Rightarrow \dfrac{1}{\text{log}_x\space 3} + \dfrac{1}{\text{log}_x\space 9} + \dfrac{1}{\text{log}_x\space 81} = \dfrac{7}{4} \\[1em] \Rightarrow \dfrac{1}{\text{log}_x\space 3} + \dfrac{1}{\text{log}_x\space 3^2} + \dfrac{1}{\text{log}_x\space 3^4} = \dfrac{7}{4} \\[1em] \Rightarrow \dfrac{1}{\text{log}_x\space 3} + \dfrac{1}{2\text{log}_x\space 3} + \dfrac{1}{4\text{log}_x\space 3} = \dfrac{7}{4} \\[1em] \Rightarrow \dfrac{1}{\text{log}_x\space 3}\Big[1 + \dfrac{1}{2} + \dfrac{1}{4}\Big] = \dfrac{7}{4} \\[1em] \Rightarrow \dfrac{1}{\text{log}_x\space 3} \times \dfrac{7}{4} = \dfrac{7}{4} \\[1em] \Rightarrow \dfrac{1}{\text{log}_x\space 3} = \dfrac{7}{4} \times \dfrac{4}{7} \\[1em] \Rightarrow \dfrac{1}{\text{log}_x\space 3} = 1 \\[1em] \Rightarrow \text{log}_3\space x = \text{log}_3\space 3 \\[1em] \Rightarrow x = 3.$

Hence, x = 3.

Solve for x:

log₂ x + log₈ x + log₃₂ x = $\dfrac{23}{15}$

Answer

Given,

$\Rightarrow \text{log}_2 \space x + \text{log}_8 \space x + \text{log}_{32} \space x = \dfrac{23}{15} \\[1em] \Rightarrow \dfrac{1}{\text{log}_x \space 2} + \dfrac{1}{\text{log}_x \space 8} + \dfrac{1}{\text{log}_x \space 32} = \dfrac{23}{15} \\[1em] \Rightarrow \dfrac{1}{\text{log}_x \space 2} + \dfrac{1}{\text{log}_x \space 2^3} + \dfrac{1}{\text{log}_x \space 2^5} = \dfrac{23}{15} \\[1em] \Rightarrow \dfrac{1}{\text{log}_x \space 2} + \dfrac{1}{3\text{log}_x \space 2} + \dfrac{1}{5\text{log}_x \space 2} = \dfrac{23}{15} \\[1em] \Rightarrow \dfrac{1}{\text{log}_x \space 2}\Big[1 + \dfrac{1}{3} + \dfrac{1}{5}\Big] = \dfrac{23}{15} \\[1em] \Rightarrow \text{log}_2 \space x\Big[\dfrac{15 + 5 + 3}{15}\Big] = \dfrac{23}{15} \\[1em] \Rightarrow \text{log}_2 \space x \times \dfrac{23}{15} = \dfrac{23}{15} \\[1em] \Rightarrow \text{log}_2 \space x = \dfrac{23}{15} \times \dfrac{15}{23} \\[1em] \Rightarrow \text{log}_2 \space x = 1 \\[1em] \Rightarrow \text{log}_2 \space x = \text{log}_2 \space 2 \\[1em] \Rightarrow x = 2.$

Hence, x = 2.