Logarithms

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

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

Hence, Option 1 is the correct option.

Given,

⇒ log₄ x = 2.5

⇒ x = 4^2.5

⇒ x = 4^{2 + 0.5}

⇒ x = 4².4^0.5

⇒ x = $16.(4)^{\frac{1}{2}}$

⇒ x = $16.(2^2)^{\frac{1}{2}}$

⇒ x = 16 × 2 = 32.

Hence, Option 2 is the correct option.

Given,

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

Hence, Option 3 is the correct option.

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.

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.

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

⇒ $\frac{x}{10}$.

Hence, Option 2 is the correct option.

(i) Given,

⇒ 5³ = 125

⇒ log₅125 = 3.

Hence, required logarithmic form is log₅ 125 = 3.

(ii) Given,

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

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

(iii) Given,

⇒ 10^-3 = 0.001

⇒ log₁₀ 0.001 = -3.

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

(iv) Given,

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

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

(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.

Given,

⇒ log₁₀ x = -2

⇒ x = 10^-2 = $\frac{1}{100}$ = 0.01

Hence, x = 0.01

Let,

⇒ log₁₀ 100 = x

⇒ 100 = 10^x

⇒ 10² = 10^x

⇒ x = 2.

Hence, required value = 2.

Let,

⇒ log₁₀ 0.1 = x

⇒ (10)^x = 0.1

⇒ 10^x = $\frac{1}{10}$

⇒ 10^x = (10^-1)

⇒ x = -1.

Hence, required value = -1.

Let,

⇒ log₁₀ (0.001) = x

⇒ (10)^x = 0.001

⇒ $10^x = \frac{1}{1000}$

⇒ 10^x = $\frac{1}{10^3}$

⇒ 10^x = 10^-3

⇒ x = -3.

Hence, required value = -3.

Let,

⇒ log₄ 32 = x

⇒ 32 = 4^x

⇒ (2)⁵ = (2²)^x

⇒ (2)⁵ = (2)^2x

⇒ 2x = 5

⇒ x = $\frac{5}{2}$.

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

Let,

⇒ log₂ (0.125) = x

⇒ 0.125 = 2^x

⇒ $\frac{125}{1000} = 2^x$

⇒ $\frac{1}{8} = 2^x$

⇒ $\frac{1}{2^3} = 2^x$

⇒ 2^-3 = 2^x

⇒ x = -3.

Hence, required value = -3.

Let,

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

Hence, required value = -2.

Let,

⇒ log₉ 27 = x

⇒ 27 = 9^x

⇒ 3³ = (3²)^x

⇒ 3³ = 3^2x

⇒ 2x = 3

⇒ x = $\frac{3}{2}$.

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

Let,

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

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

Given,

⇒ log₁₀ x = a

⇒ 10^a = x

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

Given,

⇒ x^y = z

⇒ log_x z = y.

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

Given,

⇒ log₂ 8 = 3

⇒ 2³ = 8

⇒ 8 = 8, which is true.

Given,

⇒ log₈ 2 = $\frac{1}{3}$

⇒ $8^{\frac{1}{3}}$ = 2

⇒ $(2^3)^{\frac{1}{3}}$ = 2

⇒ 2 = 2, which is true.

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

Given,

⇒ log₃ x = 0

⇒ x = 3⁰

⇒ x = 1.

Hence, x = 1.

Given,

⇒ log_x 2 = -1

⇒ (x)^-1 = 2

⇒ $\frac{1}{x} = 2$

⇒ x = $\frac{1}{2}$.

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

Given,

⇒ log₉ 243 = x

⇒ 243 = 9^x

⇒ 3⁵ = (3²)^x

⇒ 3⁵ = 3^2x

⇒ 2x = 5

⇒ x = $\frac{5}{2} = 2\frac{1}{2}$.

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

Given,

⇒ log₅ (x - 7) = 1

⇒ x - 7 = 5¹

⇒ x - 7 = 5

⇒ x = 5 + 7 = 12.

Hence, x = 12.

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 = $\frac{13}{2} = 6\frac{1}{2}$.

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

Given,

⇒ log₇ (2x² - 1) = 2

⇒ 2x² - 1 = 7²

⇒ 2x² - 1 = 49

⇒ 2x² = 49 + 1

⇒ 2x² = 50

⇒ x² = $\frac{50}{2}$

⇒ x² = 25

⇒ x = $\sqrt{25} = \pm 5$.

Hence, x = $\pm 5$.

Let,

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

Hence, log₁₀ 0.01 = -2.

Let,

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

Hence, log₂ (1 ÷ 8) = -3.

Let,

⇒ log₅ 1 = x

⇒ 1 = 5^x

⇒ 5⁰ = 5^x

⇒ x = 0.

Hence, log₅ 1 = 0.

Let,

⇒ log₅ 125 = x

⇒ 125 = 5^x

⇒ 5³ = 5^x

⇒ x = 3.

Hence, log₅ 125 = 3.

Let,

⇒ log₁₆ 8 = x

⇒ 8 = 16^x

⇒ 2³ = (2⁴)^x

⇒ 2³ = 2^4x

⇒ 4x = 3

⇒ x = $\frac{3}{4}$.

Hence, log₁₆ 8 = $\frac{3}{4}$.

Let,

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

Hence, log_0.5 16 = -4.

Given,

⇒ log_a m = n

⇒ m = aⁿ

We need to find the value of:

a^{n - 1}

⇒ aⁿ.a^-1

⇒ m.a^-1

⇒ $\frac{m}{a}$.

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

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

⇒ $\frac{2^m}{2^3}$

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

⇒ $\frac{x}{8}$.

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

(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.

Given,

⇒ 3 + log₅ 5^-2

⇒ 3 + (-2log₅ 5)

⇒ 3 + (-2 × 1)

⇒ 3 - 2

⇒ 1.

Hence, Option 1 is the correct option.

Given,

$$\begin{aligned}\Rightarrow \text{log}_{5} \space 75 - \text{log}_{5} \space 3 \\[1em] \Rightarrow \text{log}_{5} \space {\frac{75}{3}} \\[1em] \Rightarrow \text{log}_{5} \space 25 \\[1em] \Rightarrow \text{log}_{5} \space 5^2 \\[1em] \Rightarrow 2 \times \text{log}_{5} \space 5 \\[1em] \Rightarrow 2 \times 1 \\[1em] \Rightarrow 2.\end{aligned}$$

Hence, Option 2 is the correct option.

Given,

$$\begin{aligned}\Rightarrow \text{log}_{5} \space 125 ÷ \text{log}_{5} \space \sqrt{5} \\[1em] \Rightarrow \text{log}_{5} \space 5^3 ÷ \text{log}_{5} \space (5)^{\frac{1}{2}} \\[1em] \Rightarrow 3 \times \text{log}_{5} \space 5 ÷ \frac{1}{2} \times \text{log}_{5} \space 5 \\[1em] \Rightarrow 3 \times 1 ÷ \frac{1}{2} \times 1 \\[1em] \Rightarrow 3 ÷ \frac{1}{2} \\[1em] \Rightarrow 3 \times 2 \\[1em] \Rightarrow 6.\end{aligned}$$

Hence, Option 3 is the correct option.

Given,

$$\begin{aligned}\Rightarrow (\sqrt{x})^{\text{4log}_{x} \space a} \\[1em] \Rightarrow (x^{\frac{1}{2}})^{\text{4log}_{x} \space a} \\[1em] \Rightarrow (x)^{\frac{1}{2} \times \text{4log}_{x} \space a} \\[1em] \Rightarrow (x)^{\text{2log}_{x} \space a} \\[1em] \Rightarrow (x)^{\text{log}_{x} \space a^2} \\[1em] \Rightarrow a^2.\end{aligned}$$

Hence, Option 4 is the correct option.

Given,

$$\begin{aligned}\Rightarrow \frac{\text{log }125}{\text{log } \frac{1}{5}} = \text{log x} \\[1em] \Rightarrow \frac{\text{log }5^3}{\text{log } 5^{-1}} = \text{log x} \\[1em] \Rightarrow \frac{3\text{log }5}{-1\text{log } 5} = \text{log x} \\[1em] \Rightarrow -3 = \text{log x} \\[1em] \Rightarrow x = 10^{-3} \\[1em] \Rightarrow x = \frac{1}{10^3} \\[1em] \Rightarrow x = \frac{1}{1000} \\[1em] \Rightarrow x = 0.001\end{aligned}$$

Hence, Option 1 is the correct option.

Given,

⇒ log (x - 5) + log (x + 5) = 2 log 12

⇒ log (x - 5)(x + 5) = log 12²

⇒ log (x² + 5x - 5x - 25) = log 144

⇒ log (x² - 25) = log 144

⇒ x² - 25 = 144

⇒ x² = 144 + 25

⇒ x² = 169

⇒ x = $\sqrt{169}$

⇒ x = $\pm 13$.

So, positive value of x = 13.

Hence, Option 2 is the correct option.

Simplifying the expression,

⇒ log 36

⇒ log (2² × 3²)

⇒ log 2² + log 3²

⇒ 2 log 2 + 2 log 3.

Hence, log 36 = 2 log 2 + 2 log 3.

Simplifying the expression,

⇒ log 144

⇒ log (2⁴ × 3²)

⇒ log 2⁴ + log 3²

⇒ 4 log 2 + 2 log 3.

Hence, log 144 = 4 log 2 + 2 log 3.

Simplifying the expression,

⇒ log 4.5

⇒ log $\frac{45}{10}$

⇒ log $\frac{9}{2}$

⇒ log 9 - log 2

⇒ log 3² - log 2

⇒ 2 log 3 - log 2.

Hence, log 4.5 = 2 log 3 - log 2.

Simplifying the expression,

$$\begin{aligned}\Rightarrow \text{log } \frac{26}{51} - \text{log } \frac{91}{119} \\[1em] \Rightarrow \text{log } \Big(\frac{26}{51} ÷ \frac{91}{119}\Big) \\[1em] \Rightarrow \text{log } \Big(\frac{26}{51} \times \frac{119}{91}\Big) \\[1em] \Rightarrow \text{log } \frac{3094}{4641} \\[1em] \Rightarrow \text{log } \frac{2}{3} \\[1em] \Rightarrow \text{log } 2 - \text{log 3}.\end{aligned}$$

Hence, $\text{log } \frac{26}{51} - \text{log } \frac{91}{119} = \text{log 2 - log 3}$.

Simplifying the expression,

$$\begin{aligned}\Rightarrow \text{log } \frac{75}{16} - \text{2 log } \frac{5}{9} + \text{log } \frac{32}{243} \\[1em]\end{aligned}$$

⇒ (log 75 - log 16) - 2 (log 5 - log 9) + (log 32 - log 243)

⇒ log 75 - log 16 - 2 log 5 + 2 log 9 + log 32 - log 243

⇒ log 75 - log 16 - log 5² + log 9² + log 32 - log 243

⇒ log 75 - log 16 - log 25 + log 81 + log 32 - log 243

⇒ log 75 + log 81 + log 32 - log 16 - log 25 - log 243

⇒ log 75 + log 81 + log 32 - (log 16 + log 25 + log 243)

⇒ log (75 × 81 × 32) - log (16 × 25 × 243)

⇒ log $\frac{75 \times 81 \times 32}{16 \times 25 \times 243}$

⇒ log 2.

Hence, $\text{log } \frac{75}{16} - \text{2 log } \frac{5}{9} + \text{log } \frac{32}{243}$ = log 2.

Given,

⇒ 2 log x - log y = 1

⇒ log x² - log y = 1

⇒ log $\frac{x^2}{y}$ = 1

⇒ $\frac{x^2}{y}$ = 10¹

⇒ x² = 10y.

Hence, required equation is x² = 10y.

Given,

⇒ 2 log x + 3 log y = log a

⇒ log x² + log y³ = log a

⇒ log (x².y³) = log a

⇒ x².y³ = a

Hence, required equation is x²y³ = a.

Given,

⇒ a log x - b log y = 2 log 3

⇒ log x^a - log y^b = log 3²

⇒ log $\Big(\frac{x^a}{y^b}\Big)$ = log 3²

⇒ log $\Big(\frac{x^a}{y^b}\Big)$ = log 9

⇒ $\Big(\frac{x^a}{y^b}\Big) = 9$

⇒ x^a = 9y^b.

Hence, required equation is x^a = 9y^b.

Evaluating the expression,

⇒ log 5 + log 8 - 2 log 2

⇒ log 5 + log 8 - log 2²

⇒ log $\frac{5 \times 8}{2^2}$

⇒ log $\frac{40}{4}$

⇒ log 10

⇒ 1.

Hence, log 5 + log 8 - 2 log 2 = 1.

Evaluating the expression,

⇒ log₁₀ 8 + log₁₀ 25 + 2 log₁₀ 3 - log₁₀ 18

⇒ log₁₀ 8 + log₁₀ 25 + log₁₀ 3² - log₁₀ 18

⇒ log₁₀ $\frac{8 \times 25 \times 3^2}{18}$

⇒ log₁₀ $\frac{1800}{18}$

⇒ log₁₀ 100

⇒ log₁₀ 10²

⇒ 2 × log₁₀ 10

⇒ 2 × 1

⇒ 2.

Hence, log₁₀ 8 + log₁₀ 25 + 2 log₁₀ 3 - log₁₀ 18 = 2.

Evaluating the expression,

$$\begin{aligned}\Rightarrow \text{log 4} + \frac{1}{3}\text{ log 125} - \frac{1}{5}\text{ log 32} \\[1em] \Rightarrow \text{log 4} + \frac{1}{3}\text{ log 5}^3 - \frac{1}{5}\text{ log 2}^5 \\[1em] \Rightarrow \text{log 4 + log (5)}^{\Big(3 \times \frac{1}{3}\Big)} - \text{log (2)}^{\Big(5 \times \frac{1}{5}\Big)} \\[1em] \Rightarrow \text{log 4 + log 5 - log 2} \\[1em] \Rightarrow \text{log } \frac{4 \times 5}{2} \\[1em] \Rightarrow \text{log } \frac{20}{2} \\[1em] \Rightarrow \text{log } 10 \\[1em] \Rightarrow 1.\end{aligned}$$

Hence, $\text{log 4} + \frac{1}{3}\text{ log 125} - \frac{1}{5}\text{ log 32}$ = 1.

To prove:

$\text{2 log }\frac{15}{18} - \text{log } \frac{25}{162} + \text{log } \frac{4}{9} = \text{log 2}$

Solving L.H.S. of the equation, we get :

$$\begin{aligned}\Rightarrow \text{2 log }\frac{15}{18} - \text{log } \frac{25}{162} + \text{log } \frac{4}{9} \\[1em] \Rightarrow \text{log } \Big(\frac{15}{18}\Big)^2 + \text{log } \frac{4}{9} - \text{log } \frac{25}{162} \\[1em] \Rightarrow \text{log } \frac{225}{324} + \text{log } \frac{4}{9} - \text{log } \frac{25}{162} \\[1em] \Rightarrow \text{log } \frac{\frac{225}{324} \times \frac{4}{9}}{\frac{25}{162}} \\[1em] \Rightarrow \text{log } \frac{\frac{900}{2916}}{\frac{25}{162}} \\[1em] \Rightarrow \text{log } \frac{900 \times 162}{25 \times 2916} \\[1em] \Rightarrow \text{log } \frac{145800}{72900} \\[1em] \Rightarrow \text{log } 2.\end{aligned}$$

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

Hence, proved that $\text{2 log }\frac{15}{18} - \text{log } \frac{25}{162} + \text{log } \frac{4}{9} = \text{log 2}$.

Given,

⇒ x - log 48 + 3 log 2 = $\frac{1}{3}$ log 125 - log 3

⇒ x - log 48 + log 2³ = log $(125)^{\frac{1}{3}}$ - log 3

⇒ x - log 48 + log 8 = log $(5^3)^{\frac{1}{3}}$ - log 3

⇒ x - log 48 + log 8 = log 5 - log 3

⇒ x = log 5 + log 48 - log 3 - log 8

⇒ x = (log 5 + log 48) - (log 3 + log 8)

⇒ x = log (5 × 48) - log (3 × 8)

⇒ x = log 240 - log 24

⇒ x = log $\frac{240}{24}$

⇒ x = log 10

⇒ x = 1.

Hence, x = 1.

Given,

⇒ log₁₀ 2 + 1

⇒ log₁₀ 2 + log₁₀ 10

⇒ log₁₀ (2 × 10)

⇒ log₁₀ 20.

Hence, log₁₀ 2 + 1 = log₁₀ 20.

Given,

⇒ log₁₀ (x - 10) = 1

⇒ x - 10 = 10¹

⇒ x - 10 = 10

⇒ x = 10 + 10 = 20.

Hence, x = 20.

Given,

⇒ log (x² - 21) = 2

⇒ x² - 21 = 10²

⇒ x² - 21 = 100

⇒ x² = 100 + 21

⇒ x² = 121

⇒ x = $\sqrt{121}$

⇒ x = $\pm 11$.

Hence, x = $\pm 11$.

Given,

⇒ log (x - 2) + log (x + 2) = log 5

⇒ log (x - 2)(x + 2) = log 5

⇒ log (x² + 2x - 2x - 4) = log 5

⇒ log (x² - 4) = log 5

⇒ x² - 4 = 5

⇒ x² = 5 + 4

⇒ x² = 9

⇒ x = $\sqrt{9} = \pm 3$.

Since, x cannot be negative as that will make (x - 2) and (x + 2) negative.

Hence, x = 3.

Given,

⇒ log (x + 5) + log (x - 5) = 4 log 2 + 2 log 3

⇒ log (x + 5)(x - 5) = log 2⁴ + log 3²

⇒ log (x² - 5x + 5x - 25) = log (2⁴ × 3²)

⇒ log (x² - 25) = log (16 × 9)

⇒ log (x² - 25) = log 144

⇒ x² - 25 = 144

⇒ x² = 144 + 25

⇒ x² = 169

⇒ x = $\sqrt{169} = \pm 13$.

Since, x cannot be negative as that will make (x + 5) and (x - 5) negative.

Hence, x = 13.

Given,

$$\begin{aligned}\Rightarrow \frac{\text{log 81}}{\text{log 27}} = x \\[1em] \Rightarrow \frac{\text{log 3}^4}{\text{log 3}^3} = x \\[1em] \Rightarrow \frac{\text{4 log 3}}{\text{3 log 3}} = x \\[1em] \Rightarrow x = \frac{4}{3} = 1\frac{1}{3}.\end{aligned}$$

Hence, x = $1\frac{1}{3}$.

Given,

$$\begin{aligned}\Rightarrow \frac{\text{log 128}}{\text{log 32}} = x \\[1em] \Rightarrow \frac{\text{log 2}^7}{\text{log 2}^5} = x \\[1em] \Rightarrow \frac{\text{7 log 2}}{\text{5 log 2}} = x \\[1em] \Rightarrow x = \frac{7}{5} = 1.4\end{aligned}$$

Hence, x = 1.4

Given,

$$\begin{aligned}\Rightarrow \frac{\text{log 64}}{\text{log 8}} = \text{log x} \\[1em] \Rightarrow \frac{\text{log 2}^6}{\text{log 2}^3} = \text{log x} \\[1em] \Rightarrow \frac{\text{6 log 2}}{\text{3 log 2}} = \text{log x} \\[1em] \Rightarrow \text{log x} = 2 \\[1em] \Rightarrow x = 10^2 = 100.\end{aligned}$$

Hence, x = 100.

Given,

$$\begin{aligned}\Rightarrow \frac{\text{log 225}}{\text{log 15}} = \text{log x} \\[1em] \Rightarrow \frac{\text{log 15}^2}{\text{log 15}} = \text{log x} \\[1em] \Rightarrow \frac{\text{2 log 15}}{\text{log 15}} = \text{log x} \\[1em] \Rightarrow \text{log x} = 2 \\[1em] \Rightarrow x = 10^2 = 100.\end{aligned}$$

Hence, x = 100.

Given,

⇒ log x = m + n

⇒ x = 10^{m + n} ........(1)

Given,

⇒ log y = m - n

⇒ y = 10^{m - n} ........(2)

Substituting value of x and y from equation (1) and equation (2) in log $\frac{10x}{y^2}$, we get :

$$\begin{aligned}\Rightarrow \text{log} \frac{10x}{y^2} = \text{log } \frac{10 \times 10^{m + n}}{(10^{m - n})^2} \\[1em] = \text{log } \frac{10^{m + n + 1}}{10^{2(m - n)}} \\[1em] = \text{log } 10^{m + n + 1 - 2(m - n)} \\[1em] = \text{log } 10^{m + n + 1 - 2m + 2n} \\[1em] = \text{log } 10^{3n - m + 1} \\[1em] = (3n - m + 1) \text{ log} 10 \\[1em] = (3n - m + 1) \times 1 \\[1em] = 3n - m + 1 \\[1em] = 1 - m + 3n\end{aligned}$$

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

Given,

log 1 × log 1000 = 0

Solving L.H.S. of the above equation, we get :

⇒ log 1 × log 1000

⇒ 0 × log 1000

⇒ 0.

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

Hence, the statement "log 1 × log 1000 = 0" is true.

Given,

$\frac{\text{log x}}{\text{log y}}$ = log x - log y

Solving R.H.S. of the above equation, we get :

⇒ log x - log y

⇒ log $\frac{x}{y}$.

Since, L.H.S. ≠ R.H.S.

Hence, the statement $\frac{\text{log x}}{\text{log y}} = \text{log x} - \text{log y}$ is false.

Given,

$\frac{\text{log 25}}{\text{log 5}}$ = log x

Solving the equation, we get :

$$\begin{aligned}\Rightarrow \frac{\text{log 25}}{\text{log 5}} = \text{log x} \\[1em] \Rightarrow \frac{\text{log 5}^2}{\text{log 5}} = \text{log x} \\[1em] \Rightarrow \frac{\text{2 log 5}}{\text{log 5}} = \text{log x} \\[1em] \Rightarrow \text{log x} = 2 \\[1em] \Rightarrow x = 10^2 = 100.\end{aligned}$$

Since, x is not equal to 2.

Hence, the statement $\frac{\text{log 25}}{\text{log 5}}$ = log x, then x = 2 is false.

Given,

log x × log y = log x + log y

Solving the R.H.S. of the above equation, we get :

⇒ log x + log y

⇒ log xy

Since, L.H.S. ≠ R.H.S.

Hence, the statement log x × log y = log x + log y is false.

Given,

log₁₀2 = a and log₁₀3 = b

Simplifying the expression :

⇒ log 12

⇒ log (2² × 3)

⇒ log 2² + log 3

⇒ 2 log 2 + log 3

⇒ 2a + b.

Hence, log 12 = 2a + b.

Given,

log₁₀2 = a and log₁₀3 = b

Simplifying the expression :

⇒ log 2.25

⇒ log $\frac{225}{100}$

⇒ log $\frac{9}{4}$

⇒ log 9 - log 4

⇒ log 3² - log 2²

⇒ 2 log 3 - 2 log 2

⇒ 2b - 2a.

Hence, log 2.25 = 2b - 2a.

Given,

log₁₀ 2 = a and log₁₀ 3 = b

Simplifying the expression :

$$\begin{aligned}\Rightarrow \text{log } 2\frac{1}{4} \\[1em] \Rightarrow \text{log } \frac{9}{4} \\[1em] \Rightarrow \text{log 9 - log 4} \\[1em] \Rightarrow \text{log 3}^2 - \text{log 2}^2 \\[1em] \Rightarrow \text{2 log 3 - 2 log 2} \\[1em] \Rightarrow \text{2b - 2a}.\end{aligned}$$

Hence, $\text{log } 2\frac{1}{4}$ = 2b - 2a.

Given,

log₁₀ 2 = a and log₁₀ 3 = b

Simplifying the expression :

$$\begin{aligned}\Rightarrow \text{log } 5.4 \\[1em] \Rightarrow \text{log } \frac{54}{10} \\[1em] \Rightarrow \text{log 54 - log 10} \\[1em] \Rightarrow \text{log }(2 \times 3^3) - 1 \\[1em] \Rightarrow \text{log 2} + \text{log }3^3 - 1 \\[1em] \Rightarrow \text{log 2 + 3 log 3} - 1 \\[1em] \Rightarrow a + 3b - 1.\end{aligned}$$

Hence, log 5.4 = a + 3b - 1.

Given,

log₁₀ 2 = a and log₁₀ 3 = b

Simplifying the expression :

⇒ log 60

⇒ log (2 × 3 × 10)

⇒ log 2 + log 3 + log 10

⇒ a + b + 1.

Hence, log 60 = a + b + 1.

Given,

log₁₀ 2 = a and log₁₀ 3 = b

Simplifying the expression :

$$\begin{aligned}\Rightarrow \text{log } 3\frac{1}{8} \\[1em] \Rightarrow \text{log } \frac{25}{8} \\[1em] \Rightarrow \text{log 25 - log 8} \\[1em] \Rightarrow \text{log 5}^2 - \text{log 2}^3 \\[1em] \Rightarrow \text{2 log 5 - 3 log 2} \\[1em] \Rightarrow \text{2 log } \frac{10}{2} - \text{3 log 2} \\[1em] \Rightarrow \text{2 (log 10 - log 2)} - \text{3 log 2} \\[1em] \Rightarrow \text{2 log 10 - 2 log 2 - 3 log 2} \\[1em] \Rightarrow 2 \times 1 - \text{5 log 2} \\[1em] \Rightarrow 2 - 5a.\end{aligned}$$

Hence, $\text{log } 3\frac{1}{8}$ = 2 - 5a.

Simplifying the expression,

⇒ log 12

⇒ log (2² × 3)

⇒ log 2² + log 3

⇒ 2 log 2 + log 3

⇒ 2 × 0.3010 + 0.4771

⇒ 0.6020 + 0.4771

⇒ 1.0791

Hence, log 12 = 1.0791

Simplifying the expression,

⇒ log 1.2

⇒ log $\frac{12}{10}$

⇒ log 12 - log 10

⇒ log (2² × 3) - log 10

⇒ log 2² + log 3 - log 10

⇒ 2 log 2 + log 3 - 1

⇒ 2 × 0.3010 + 0.4771 - 1

⇒ 0.6020 + 0.4771 - 1

⇒ 1.0791 - 1

⇒ 0.0791

Hence, log 1.2 = 0.0791

Simplifying the expression,

⇒ log 3.6

⇒ log $\frac{36}{10}$

⇒ log 36 - log 10

⇒ log (12 × 3) - log 10

⇒ log 12 + log 3 - log 10

⇒ log (2² × 3) + log 3 - log 10

⇒ log 2² + log 3 + log 3 - log 10

⇒ 2 log 2 + 2 log 3 - 1

⇒ 2 × 0.3010 + 2 × 0.4771 - 1

⇒ 0.6020 + 0.9542 - 1

⇒ 1.5562 - 1

⇒ 0.5562

Hence, log 3.6 = 0.5562

Simplifying the expression,

⇒ log 15

⇒ log (3 × 5)

⇒ log 3 + log 5

⇒ log 3 + log $\frac{10}{2}$

⇒ log 3 + log 10 - log 2

⇒ 0.4771 + 1 - 0.3010

⇒ 1.1761

Hence, log 15 = 1.1761

Simplifying the expression,

⇒ log 25

⇒ log 5²

⇒ 2 log 5

⇒ 2 log $\frac{10}{2}$

⇒ 2(log 10 - log 2)

⇒ 2(1 - 0.3010)

⇒ 2 × 0.699

⇒ 1.3980

Hence, log 25 = 1.3980

Simplifying the expression,

$$\begin{aligned}\Rightarrow \frac{2}{3} \text{ log 8} \\[1em] \Rightarrow \frac{2}{3} \text{ log 2}^3 \\[1em] \Rightarrow \frac{2}{3} \times 3 \text{ log 2} \\[1em] \Rightarrow 2 \text{log 2} \\[1em] \Rightarrow 2 \times 0.3010 \\[1em] \Rightarrow 0.6020\end{aligned}$$

Hence, $\frac{2}{3}$ log 8 = 0.6020

(i) Given,

⇒ 2 log₁₀ x + 1 = log₁₀ 250

⇒ log₁₀ x² + log₁₀ 10 = log₁₀ 250

⇒ log₁₀ (x² × 10) = log₁₀ 250

⇒ log₁₀ (10x²) = log₁₀ 250

⇒ 10x² = 250

⇒ x² = $\frac{250}{10}$

⇒ x² = 25

⇒ x = $\sqrt{25}$ = 5.

Hence, x = 5.

(ii) Substituting value of x in log₁₀ 2x, we get :

⇒ log₁₀ 2x

⇒ log₁₀ 2(5)

⇒ log₁₀ 10

⇒ 1.

Hence, log₁₀ 2x = 1.

Given,

$$\begin{aligned}\Rightarrow \text{3 log x} + \frac{1}{2} \text{log y} = 2 \\[1em] \Rightarrow \text{log } x^3 + \text{log } y^{\frac{1}{2}} = 2 \\[1em] \Rightarrow \text{log } x^3\sqrt{y} = 2 \\[1em] \Rightarrow x^3\sqrt{y} = 10^2\end{aligned}$$

Squaring both sides, we get :

$$\begin{aligned}\Rightarrow (x^3\sqrt{y})^2 = (10^2)^2 \\[1em] \Rightarrow x^6y = 10^4 \\[1em] \Rightarrow y = \frac{10^4}{x^6} = 10000x^{-6}.\end{aligned}$$

Hence, y = 10000x^-6.

Given,

⇒ log₂ (log₃ x) = 4

⇒ log₃ x = 2⁴

⇒ log₃ x = 16

⇒ x = 3¹⁶.

Hence, Option 1 is the correct option.

Given,

⇒ log (5x - 4) - log (x + 1) = log 4

⇒ log $\frac{5x - 4}{x + 1}$ = log 4

⇒ $\frac{5x - 4}{x + 1} = 4$

⇒ 5x - 4 = 4(x + 1)

⇒ 5x - 4 = 4x + 4

⇒ 5x - 4x = 4 + 4

⇒ x = 8.

Hence, Option 2 is the correct option.

Given,

$$\begin{aligned}\Rightarrow \text{log } x - \text{log } (2x - 1) = 1 \\[1em] \Rightarrow \text{log } \frac{x}{2x - 1} = \text{log } 10 \\[1em] \Rightarrow \frac{x}{2x - 1} = 10 \\[1em] \Rightarrow x = 10(2x - 1) \\[1em] \Rightarrow x = 20x - 10 \\[1em] \Rightarrow 20x - x = 10 \\[1em] \Rightarrow 19x = 10 \\[1em] \Rightarrow x = \frac{10}{19}.\end{aligned}$$

Hence, Option 3 is the correct option.

Solving,

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

Hence, Option 4 is the correct option.

Given,

⇒ log v + log 3 = log π + log 4 + 3 log r

⇒ log 3v = log π + log 4 + log r³

⇒ log 3v = log 4πr³

⇒ 3v = 4πr³

⇒ v = $\frac{4}{3}πr^3$

Hence, Option 1 is the correct option.

Given,

⇒ log y + 2 log x = 2

⇒ log y + log x² = 2

⇒ log yx² = 2

⇒ yx² = 10²

⇒ yx² = 100

⇒ y = 100 ÷ x².

Hence, Option 2 is the correct option.

Given,

⇒ log₁₀ 8 = 0.90

⇒ log₁₀ 2³ = 0.90

⇒ 3log₁₀ 2 = 0.90

⇒ log₁₀ 2 = $\frac{0.90}{3}$

⇒ log₁₀ 2 = 0.30 .........(1)

(i) Given,

⇒ log₁₀ 4

⇒ log₁₀ 2²

⇒ 2 log₁₀ 2

Substituting value of log₁₀ 2 from equation (1) in above equation, we get :

⇒ 2 × 0.30

⇒ 0.60

Hence, log₁₀ 4 = 0.60

(ii) Given,

$$\begin{aligned}\Rightarrow \text{log } \sqrt{32} \\[1em] \Rightarrow \text{log } 4\sqrt{2} \\[1em] \Rightarrow \text{log } 4 + \text{log } \sqrt{2} \\[1em] \Rightarrow \text{log } 2^2 + \text{log } 2^{\frac{1}{2}} \\[1em] \Rightarrow 2 \times \text{log 2} + \frac{1}{2} \times \text{log 2}\end{aligned}$$

Substituting value of log 2 from equation (1), in above equation, we get :

$$\begin{aligned}\Rightarrow 2 \times 0.30 + \frac{1}{2} \times 0.30 \\[1em] \Rightarrow 0.60 + 0.15 \\[1em] \Rightarrow 0.75\end{aligned}$$

Hence, log $\sqrt{32}$ = 0.75

(iii) Given,

$$\begin{aligned}\Rightarrow \text{log } 0.125 \\[1em] \Rightarrow \text{log } \frac{125}{1000} \\[1em] \Rightarrow \text{log } \frac{1}{8} \\[1em] \Rightarrow \text{log } \frac{1}{2^3} \\[1em] \Rightarrow \text{log } 2^{-3} \\[1em] \Rightarrow -3 \times \text{log 2} \\[1em] \Rightarrow -3 \times 0.30 \\[1em] \Rightarrow -0.90\end{aligned}$$

Hence, log 0.125 = -0.90

Given,

⇒ log 27 = 1.431

⇒ log 3³ = 1.431

⇒ 3 log 3 = 1.431

⇒ log 3 = $\frac{1.431}{3}$ = 0.477 .......(1)

(i) Solving,

⇒ log 9

⇒ log 3²

⇒ 2 log 3

Substituting value of log 3 from equation (1) in above equation, we get :

⇒ 2 × 0.477

⇒ 0.954

Hence, log 9 = 0.954

(ii) Solving,

⇒ log 300

⇒ log 3 × 100

⇒ log 3 + log 100

⇒ 0.477 + log 10²

⇒ 0.477 + 2 log 10

⇒ 0.477 + 2 × 1

⇒ 0.477 + 2

⇒ 2.477

Hence, log 300 = 2.477

Given,

⇒ log₁₀ a = b

⇒ 10^b = a

Cubing both sides, we get :

⇒ (10^b)³ = a³

⇒ 10^3b = a³

Dividing both sides by 10², we get :

$$\begin{aligned}\Rightarrow \frac{10^{3b}}{10^2} = \frac{a^3}{10^2} \\[1em] \Rightarrow 10^{3b - 2} = \frac{a^3}{100}.\end{aligned}$$

Hence, 10^{3b - 2} = $\frac{a^3}{100}.$

Given,

⇒ log₅ x = y

⇒ x = 5^y ......(1)

Solving, equation 5^{2y + 3}, we get :

⇒ 5^2y.5³

⇒ (5^y)² × 125

Substituting value of 5^y from equation (1) in above equation, we get :

⇒ 125x².

Hence, 5^{2y + 3} = 125x².

Given,

1st equation :

⇒ log₃ m = x

⇒ m = 3^x ......(1)

2nd equation :

⇒ log₃ n = y

⇒ n = 3^y ......(2)

(i) Given,

3^{2x - 3}

⇒ 3^2x.3^-3

⇒ (3^x)².3^-3

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

⇒ m².3^-3

⇒ $\frac{m^2}{3^3}$

⇒ $\frac{m^2}{27}$.

Hence, 3^{2x - 3} = $\frac{m^2}{27}$.

(ii) Simplifying the expression,

⇒ 3^{1 - 2y + 3x}

⇒ 3¹.3^-2y.3^3x

⇒ 3.(3^y)^-2.(3^x)³

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

⇒ 3.n^-2.m³

⇒ $\frac{3m^3}{n^2}$.

Hence, 3^{1 - 2y + 3x} = $\frac{3m^3}{n^2}$.

(iii) Given,

⇒ 2log₃ A = 5x - 3y

⇒ log₃ A² = 5x - 3y

⇒ A² = 3^{5x - 3y}

⇒ A² = 3^5x.3^-3y

⇒ A² = (3^x)⁵.(3^y)^-3

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

⇒ A² = m⁵.n^-3

⇒ A² = $\frac{m^5}{n^3}$

⇒ A = $\sqrt{\frac{m^5}{n^3}}$.

Hence, A = $\sqrt{\frac{m^5}{n^3}}$.

Simplifying the expression,

⇒ log (a)³ - log a

⇒ 3log a - log a

⇒ 2log a

⇒ log a²

⇒ 2 log a.

Hence, log (a)³ - log a = 2 log a.

Simplifying the expression,

⇒ log (a)³ ÷ log a

⇒ 3log a ÷ log a

⇒ $\frac{\text{3 log a}}{\text{log a}}$

⇒ 3.

Hence, log (a)³ ÷ log a = 3.

Given,

⇒ log (a + b) = log a + log b

⇒ log (a + b) = log ab

⇒ a + b = ab

⇒ ab - a = b

⇒ a(b - 1) = b

⇒ a = $\frac{b}{b - 1}$.

Hence, a = $\frac{b}{b - 1}$.

To prove:

(log a)² - (log b)² = log $\Big(\frac{a}{b}\Big)$. log (ab)

Solving L.H.S. of the above equation, we get :

⇒ (log a)² - (log b)²

⇒ (log a + log b)(log a - log b)

⇒ log ab. log $\Big(\frac{a}{b}\Big)$

Hence, proved that (log a)² - (log b)² = log $\Big(\frac{a}{b}\Big)$. log (ab)

Given,

⇒ a log b + b log a - 1 = 0

⇒ log b^a + log a^b = 1

⇒ log b^a + log a^b = log 10

⇒ log b^a.a^b = log 10

⇒ b^a.a^b = 10.

Hence, proved that b^a.a^b = 10.

Given,

⇒ log (a + 1) = log (4a - 3) - log 3

⇒ log (a + 1) = log $\frac{4a - 3}{3}$

⇒ (a + 1) = $\frac{4a - 3}{3}$

⇒ 4a - 3 = 3(a + 1)

⇒ 4a - 3 = 3a + 3

⇒ 4a - 3a = 3 + 3

⇒ a = 6.

Hence, a = 6.

Given,

⇒ 2 log y - log x - 3 = 0

⇒ 2 log y - log x = 3

⇒ log y² - log x = 3

⇒ log $\frac{y^2}{x}$ = 3

⇒ $\frac{y^2}{x} = 10^3$

⇒ y² = x.10³

⇒ x = $\frac{y^2}{10^3} = \frac{y^2}{1000}$.

Hence, X = $\frac{y^2}{1000}$.

To prove:

log₁₀ 125 = 3(1 - log₁₀ 2)

Solving R.H.S. of the equation, we get :

⇒ 3(1 - log₁₀ 2)

⇒ 3(log₁₀ 10 - log₁₀ 2)

⇒ 3(log₁₀ $\frac{10}{2}$)

⇒ 3 log₁₀ 5

⇒ log₁₀ 5³

⇒ log₁₀ 125

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

Hence, proved that log₁₀ 125 = 3(1 - log₁₀ 2).

Simplifying the expression :

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

Hence, Option 1 is the correct option.

Given,

⇒ log₃ (x + 1) = 2

⇒ x + 1 = 3²

⇒ x + 1 = 9

⇒ x = 9 - 1 = 8.

Hence, Option 2 is the correct option.

Given,

⇒ 2 log x = log 250 - 1

⇒ 2 log x = log 250 - log 10

⇒ log x² = log $\frac{250}{10}$

⇒ x² = 25

⇒ x = $\sqrt{25}$ = 5.

Hence, Option 3 is the correct option.

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.

Given,

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

Hence, Option 1 is the correct option.

Given,

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

Cubing and then squaring both sides, we get :

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

Hence, a⁹.b⁴ = 10⁶.

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 $\frac{a}{5}$

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

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

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

⇒ a² = 5² × 2 × 2 × 9

⇒ a² = 900

⇒ a = $\sqrt{900}$ = 30.

Hence, a = 30.

(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 $\frac{0.6 \times 1.25 \times 2^2}{3}$

⇒ 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.

⇒ 3a² - 2b³ = 6 log z

⇒ 3log x - 2log y = 6 log z

⇒ log x³ - log y² = log z⁶

⇒ log $\frac{x^3}{y^2}$ = log z⁶

⇒ $\frac{x^3}{y^2} = z^6$

⇒ $y^2 = \frac{x^3}{z^6}$

⇒ $y = \frac{\sqrt{x^3}}{\sqrt{z^6}}$

⇒ y = $x^{\frac{3}{2}} ÷ z^3$.

Hence, $y = x^{\frac{3}{2}} ÷ z^3$.

Given,

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

Squaring both sides, we get :

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

Hence, proved that a² + b² = 6ab.

Given,

⇒ a² + b² = 23ab

⇒ a² + b² + 2ab = 23ab + 2ab

⇒ (a + b)² = 25ab

⇒ $\frac{(a + b)^2}{25}$ = ab

⇒ $\Big(\frac{a + b}{5}\Big)^2$ = ab

Taking log on both sides, we get :

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

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

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

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

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 $\frac{400}{25}$

⇒ 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.

Given,

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

Solving L.H.S. of the equation :

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

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 = $\frac{100}{2}$ = 50.

Hence, x = 50 and y = 2.

Given,

⇒ log_x 625 = -4

⇒ x^-4 = 625

⇒ $\frac{1}{x^4}$ = 625

⇒ $x^4 = \frac{1}{625}$

⇒ $x^4 = \Big(\frac{1}{5}\Big)^4$

⇒ x = $\frac{1}{5}$ = 0.2

Hence, x = 0.2

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.

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 $\frac{400}{25}$

⇒ 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.

Given,

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

Solving, equation :

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

Solving, equation :

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

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

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

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

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

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

Solving, equation :

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

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

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

Squaring both sides, we get :

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

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

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

Hence, x = 1000 and y = 100.

(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,

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

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 = $\frac{10^{3a}}{10^{2b}}$

We know that: (shown above)

y⁴ = 10^2b

and

$\sqrt{x}$ = 10^a

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

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

Given,

⇒ log₅ (x + 1) - 1 = 1 + log₅ (x - 1)

⇒ log₅ (x + 1) - log₅ (x - 1) = 1 + 1

⇒ log₅ $\frac{x + 1}{x - 1}$ = 2

⇒ $\frac{x + 1}{x - 1} = 5^2$

⇒ x + 1 = 25(x - 1)

⇒ x + 1 = 25x - 25

⇒ 25x - x = 1 + 25

⇒ 24x = 26

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

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

Given,

⇒ log₃ 81

⇒ log₃ (3)⁴

⇒ 4.log₃ 3

⇒ 4 x 1

⇒ 4.

Hence, option 1 is the correct option.

Let, log₁₆ 2 = x

⇒ 16^x = 2

⇒ (2⁴)^x = 2

⇒ 2^4x = 2¹

⇒ 4x = 1

⇒ x = $\frac{1}{4}$.

Hence, option 3 is the correct option.

Given, log(3x - 2) = 2

⇒ log₁₀(3x - 2) = 2

⇒ 3x - 2 = 10²

⇒ 3x - 2 = 100

⇒ 3x = 100 + 2

⇒ 3x = 102

⇒ x = $\frac{102}{3}$

⇒ x = 34.

Hence, option 1 is the correct option.

Given,

⇒ 2 + $\frac{1}{2}$ log 9 - 2 log 5

⇒ 2log 10 + $\text{log 9}^\frac{1}{2}$ - 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 $\frac{300}{25}$

⇒ log 12.

Hence, option 3 is the correct option.

Given,

⇒ 3 + log 10^-2

⇒ 3 + (-2) log 10

⇒ 3 + (-2) × 1

⇒ 3 - 2

⇒ 1.

Hence, option 2 is the correct option.

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.

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.

Given,

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

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

Hence, option 3 is the correct option.

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.

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

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 = $\sqrt{2}$.

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 = $\pm 11$.

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 $\frac{10\sqrt{y}}{x^2z^3}$, we get :

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

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

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 $\frac{25 \times 36 \times x}{27}$ = 2

⇒ $\frac{100x}{3} = 10^2$

⇒ 100x = 3 × 10²

⇒ 100x = 300

⇒ x = $\frac{300}{100}$ = 3.

Hence, x = 3.

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 $\frac{x^2y^3}{z^4}$, we get :

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

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

Given,

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

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

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

⇒ log_x $\frac{25}{625}$ = 2

⇒ x² = $\frac{25}{625}$

⇒ $x^2 = \frac{1}{25}$

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

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

Given,

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

Hence, x = 7.

Given,

a² = log x and b³ = log y

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

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

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

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 - $\frac{1}{2}$ log ₂ 2 × log₁₀ 9 - log₁₀ 0.4

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

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

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

= log₁₀12 - log₁₀3 - log₁₀0.4

= log₁₀12 - (log₁₀3 + log₁₀0.4)

= log₁₀ $\frac{12}{3 \times 0.4}$

= log₁₀ $\frac{12}{1.2}$

= 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.

Given,

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

Hence, x = 15.

Simplifying the expression,

$$\begin{aligned}\Rightarrow \text{log}_b a \times \text{log}_c b \times \text{log}_a c \\[1em] \Rightarrow \frac{\text{log a}}{\text{log b}} \times \frac{\text{log b}}{\text{log c}} \times \frac{\text{log c}}{\text{log a}} \\[1em] \Rightarrow \frac{\text{log a.log b.log c}}{\text{log b.log c.log a}} \\[1em] \Rightarrow 1.\end{aligned}$$

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

Simplifying the expression,

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

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

Simplifying the expression,

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

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

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

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

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

Given,

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

Hence, x = 81.

Evaluating,

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

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