Logarithms

(i) 5² = 25

⇒ log₅ 25 = 2.

(ii) a⁵ = 64

⇒ log_a 64 = 5.

(iii) 7^x = 100

⇒ log₇ 100 = x.

(iv) 9⁰ = 1

⇒ log₉ 1 = 0.

(v) 6¹ = 6

⇒ log₆ 6 = 1.

(vi) 3^-2 = $\frac{1}{9}$

⇒ log₃ $\frac{1}{9}$ = -2.

(vii) 10^-2 = 0.01

⇒ log₁₀ 0.01 = -2.

(viii) $(81)^{\frac{3}{4}}$ = 27

⇒ log₈₁ 27 = $\frac{3}{4}$

(i) log₂32 = 5

⇒ 2⁵ = 32

(ii) log₃81 = 4

⇒ 3⁴ = 81

(iii) log₃$\frac{1}{3}$ = -1

⇒ 3^-1 = $\frac{1}{3}$

(iv) log₈4 = $\frac{2}{3}$

⇒ $(8)^{\frac{2}{3}}$ = 4

(v) log₈ 32 = $\frac{5}{3}$

⇒ $(8)^{\frac{5}{3}} = 32$

(vi) log₁₀ (0.001) = -3

⇒ 10^-3 = 0.001

(vii) log₂ 0.25 = -2

⇒ (2)^-2 = 0.25

(viii) log_a $\Big(\frac{1}{a}\Big)$ = -1

⇒ a^-1 = $\frac{1}{a}$.

(i) log₂16 = x

⇒ 2^x = 16

⇒ 2^x = 2⁴

∴ x = 4.

Hence, log₂16 = 4.

(ii) log₅125 = x

⇒ 5^x = 125

⇒ 5^x = 5³

∴ x = 3.

Hence, log₅125 = 3.

(iii) log₄8 = x

⇒ 4^x = 8

⇒ (2²)^x = 2³

⇒ 2^2x = 2³

⇒ 2x = 3

⇒ x = $\frac{3}{2}.$

Hence, log₄8 = $\frac{3}{2}$.

(iv) log₉27 = x

⇒ 9^x = 27

⇒ (3²)^x = 3³

⇒ 3^2x = 3³

⇒ 2x = 3

⇒ x = $\frac{3}{2}.$

Hence, log₉27 = $\frac{3}{2}$.

(v) log₁₀ (0.01) = x

⇒ 10^x = 0.01

⇒ 10^x = 10^-2

∴ x = -2.

Hence, log₁₀ (0.01) = -2.

(vi) log₇$\frac{1}{7}$ = x

⇒ 7^x = $\frac{1}{7}$

⇒ 7^x = 7^-1

∴ x = -1.

Hence, log₇$\frac{1}{7}$ = -1.

(vii) log_0.5 256 = x

⇒ (0.5)^x = 256

⇒ $\Big(\frac{5}{10}\Big)^x$ = (2)⁸

⇒ $\Big(\frac{1}{2}\Big)^x$ = (2)⁸

⇒ (2)^-x = (2)⁸

∴ -x = 8 ⇒ x = -8.

Hence, log_0.5256 = -8.

(viii) log₂ 0.25 = x

⇒ 2^x = 0.25

⇒ 2^x = $\frac{25}{100}$

⇒ 2^x = $\frac{1}{4} = \frac{1}{2^2}$

⇒ 2^x = 2^-2

⇒ x = -2.

Hence, log₂ 0.25 = -2.

Given,

⇒ log₃x = 2

⇒ x = 3² = 9.

Hence, x = 9.

Given,

⇒ log_x25 = 2

⇒ 25 = x²

⇒ 5² = x²

⇒ x = 5.

Hence, x = 5.

Given,

⇒ log₁₀x = -2

⇒ x = 10^-2

⇒ x = $\frac{1}{10^2} = \frac{1}{100}$ = 0.01.

Hence, x = 0.01.

Given,

⇒ log₄x = $\frac{1}{2}$

⇒ x = $4^{\frac{1}{2}} = \sqrt{4}$ = 2.

Hence, x = 2.

Given,

⇒ log_x11 = 1

⇒ x¹ = 11

⇒ x = 11.

Hence, x = 11.

Given,

⇒ log_x$\frac{1}{4}$ = -1

$$\begin{aligned}\Rightarrow x^{-1} = \frac{1}{4} \\[1em] \Rightarrow \frac{1}{x} = \frac{1}{4} \\[1em] \Rightarrow x = 4.\end{aligned}$$

Hence, x = 4.

Given,

⇒ log₈₁x = $\frac{3}{2}$

$$\begin{aligned}\Rightarrow x = (81)^{\frac{3}{2}} \\[1em] \Rightarrow x = (9)^2{\frac{3}{2}} \\[1em] \Rightarrow x = 9^3 = 729.\end{aligned}$$

Hence, x = 729.

Given,

⇒ log₉x = 2.5

⇒ x = 9^2.5

⇒ x = $(3^2)^{\frac{5}{2}} = 3^{2 \times \frac{5}{2}}$

⇒ x = 3⁵ = 243.

Hence, x = 243.

Given,

⇒ log₄x = -1.5

⇒ x = 4^-1.5

⇒ x = $4^{-\frac{3}{2}} = (2^2)^{-\frac{3}{2}}$

⇒ x = 2^-3 = $\frac{1}{2^3} = \frac{1}{8}$.

Hence, x = $\frac{1}{8}$.

Given,

⇒ log_√5x = 2

x = $(\sqrt{5})^2$ = 5.

Hence, x = 5.

Given,

⇒ log_x 0.001 = -3

⇒ x^-3 = 0.001

⇒ x^-3 = $\frac{1}{1000} = \frac{1}{10^3}$

⇒ x^-3 = 10^-3

⇒ x = 10.

Hence, x = 10.

Given,

⇒ log_√3(x + 1) = 2

⇒ (x + 1) = $(\sqrt{3})^2$

⇒ (x + 1) = 3

⇒ x = 2.

Hence, x = 2.

Given,

⇒ log₄(2x + 3) = $\frac{3}{2}$

⇒ 2x + 3 = $4^{\frac{3}{2}}$

⇒ 2x + 3 = $(2^2)^{\frac{3}{2}}$

⇒ 2x + 3 = 2³

⇒ 2x + 3 = 8

⇒ 2x = 5

⇒ x = $\frac{5}{2}$.

Hence, x = $\frac{5}{2}$.

Given,

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

Hence, x = 2.

Given,

⇒ $\text{log}_2(x^2 - 1)$ = 3

⇒ x² - 1 = 2³

⇒ x² - 1 = 8

⇒ x² = 9

⇒ x = $\sqrt{9} = \pm3$.

Hence, x = ±3.

Given,

⇒ log x = -1

⇒ x = 10^-1

⇒ x = $\frac{1}{10}.$

Hence, x = $\frac{1}{10}$.

Given,

⇒ log(2x - 3) = 1

⇒ 2x - 3 = 10¹

⇒ 2x - 3 = 10

⇒ 2x = 13

⇒ x = $\frac{13}{2} = 6\frac{1}{2}.$

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

Given,

⇒ log x = -2

⇒ x = 10^-2 = $\frac{1}{10^2} = \frac{1}{100}$.

⇒ log x = 0

⇒ x = 10⁰ = 1.

⇒ log x = $\frac{1}{3}$

⇒ x = $10^{\frac{1}{3}} = \sqrt[3]{10}.$

Hence, x = $\frac{1}{100}, 1, \sqrt[3]{10}.$

Given,

⇒ log₁₀a = b

∴ a = 10^b.

Simplifying 10^{2b - 3} we get,

⇒ 10^{2b - 3} = 10^2b.10^-3

= $\frac{10^{2b}}{10^3} = \frac{(10^b)^2}{10^3}$

= $\frac{a^2}{10^3} = \frac{a^2}{1000}$.

Hence, 10^{2b - 3} = $\frac{a^2}{1000}.$

(i) Given,

⇒ log₁₀x = a

∴ x = 10^a.

Simplifying 10^{2a - 3} we get,

⇒ 10^{2a - 3} = 10^2a.10^-3

= $\frac{10^{2a}}{10^3} = \frac{(10^a)^2}{10^3}$

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

Hence, 10^{2a - 3} = $\frac{x^2}{1000}.$

(ii) Given,

⇒ log₁₀y = b

∴ y = 10^b.

Simplifying 10^{3b - 1} we get,

⇒ 10^{3b - 1} = 10^3b.10^-1

= $\frac{10^{3b}}{10^1} = \frac{(10^b)^3}{10}$

= $\frac{y^3}{10}$.

Hence, 10^{3b - 1} = $\frac{y^3}{10}.$

(iii) Given,

log₁₀z = c

⇒ z = 10^c.

log₁₀P = 2a + $\frac{b}{2}$ - 3c

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

Hence, P = $\frac{x^2\sqrt{y}}{z^3}.$

Given,

log₁₀x = a

⇒ x = 10^a.

log₁₀y = b

⇒ y = 10^b

xy = 10^a.10^b = 10^{a + b}.

Hence, xy = 10^{a + b}.

Given,

log₁₀a = m

⇒ a = 10^m

log₁₀b = n

⇒ b = 10ⁿ

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

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

(i) Given,

log₁₀x = 2a

⇒ x = 10^2a

⇒ x = (10^a)²

⇒ 10^a = $\sqrt{x}$.

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

(ii) Given,

log₁₀y = $\frac{b}{2}$

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

Squaring both sides we get,

⇒ y² = 10^b

Simplifying 10^{2b + 1} we get,

10^{2b + 1} = 10^2b.10

= (10^b)².10

= (y²)².10

= 10y⁴.

Hence, 10^{2b + 1} = 10y⁴.

(iii) Given,

log₁₀P = 3a - 2b

⇒ P = 10^{3a - 2b}

⇒ P = 10^3a.10^-2b

= (10^a)³.(10^b)^-2

$$\begin{aligned}= (\sqrt{x})^3 \times (y^2)^{-2} \\[1em] = (x^3)^{\frac{1}{2}} \times \frac{1}{(y^2)^2} \\[1em] = \frac{x^{\frac{3}{2}}}{y^4}\end{aligned}$$

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

Given,

log₂y = x and log₃z = x

⇒ y = 2^x and z = 3^x.

(72)^x = (2³.3²)^x

= (2)^3x.(3)^2x

= (2^x)³.(3^x)²

= (y)³.(z)²

Hence, (72)^x = y³.z².

Given,

log₂x = a and log₅y = a

⇒ x = 2^a and y = 5^a

100^{2a - 1} = 100^2a.100^-1

= (100^a)².100^-1

= [(2².5²)^a]².100^-1

= [(2^2a).(5^2a)]².100^-1

= [(2^a)².(5^a)²]².(100)^-1

= [x².y²]².100^-1

= $\frac{x^4y^4}{100}$.

Hence, 100^{2a - 1} = $\frac{x^4y^4}{100}$.

Given,

⇒ log a³ - log a²

⇒ 3log a - 2log a

⇒ log a.

Hence, log a³ - log a² = log a

Given,

log a³ ÷ log a²

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

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

Given,

$$\begin{aligned}\Rightarrow \frac{\text{log 4}}{\text{log 2}} \\[1em] \Rightarrow \frac{\text{log 2}^2}{\text{log 2}} \\[1em] \Rightarrow \frac{\text{2 log 2}}{\text{log 2}} \\[1em] \Rightarrow 2.\end{aligned}$$

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

Given,

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

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

Given,

$$\begin{aligned}\Rightarrow \frac{\text{log 27}}{\text{log }\sqrt{3}} \\[1em] \Rightarrow \frac{\text{log 3}^3}{\text{log 3}^{\frac{1}{2}}} \\[1em] \Rightarrow \frac{\text{3log 3}}{\frac{1}{2}\text{log 3}} \\[1em] \Rightarrow \frac{3}{\frac{1}{2}} \\[1em] \Rightarrow 6.\end{aligned}$$

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

Given,

$$\begin{aligned}\Rightarrow \frac{\text{log 9 - log 3}}{\text{log 27}} \\[1em] \Rightarrow \frac{\text{log 3}^2 - \text{log 3}}{\text{log 3}^3} \\[1em] \Rightarrow \frac{\text{2log 3 - log 3}}{\text{3log 3}} \\[1em] \Rightarrow \frac{\text{log 3}}{\text{3log 3}} \\[1em] \Rightarrow \frac{1}{3}.\end{aligned}$$

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

Given,

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

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

Given,

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

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

Given,

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

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

Given,

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

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

Given,

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

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

Given,

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

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

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

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

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

⇒ log $\frac{600}{600}$

⇒ log 1

⇒ 0.

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

Given,

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

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

Given,

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

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

⇒ log₁₀25 + log₁₀8 - log₁₀2

⇒ log₁₀$\frac{25 \times 8}{2}$

⇒ log₁₀100

⇒ log₁₀10²

⇒ 2log₁₀10

⇒ 2.

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

Given,

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

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

Given,

⇒ 2log₁₀5 - log₁₀2 + 3log₁₀4 + 1

⇒ log₁₀5² - log₁₀2 + log₁₀4³ + log₁₀10

⇒ log₁₀25 + log₁₀64 + log₁₀10 - log₁₀2

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

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

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

Given,

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

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

Given,

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

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

Given,

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

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

Given,

log₁₀4 ÷ log₁₀2 = log₃9

Simplifying L.H.S. we get,

⇒ log₁₀4 ÷ log₁₀2 = log₁₀2² ÷ log₁₀2

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

= 2.

Simplifying R.H.S. we get,

⇒ log₃9 = log₃3²

= 2log₃3

= 2.

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

Hence, proved that log₁₀4 ÷ log₁₀2 = log₃9.

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.

Given,

x = (100)^a = (10²)^a = 10^2a,

y = (10000)^b = (10⁴)^b = 10^4b,

z = (10)^c.

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

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

(i) Given,

a = log₁₀x

⇒ x = 10^a.

(ii) Given,

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

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

(iii) Given,

a = log₁₀ x

Now,

⇒ log₁₀ 3x

⇒ log₁₀ (3 × x)

⇒ log₁₀ 3 + log₁₀ x

⇒ log₁₀ 3 + a.

Hence, log₁₀ 3x = log₁₀ 3 + a.

(i) Given,

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

Hence, a + b + c = 0.

(ii) Given,

⇒ 5^{a + b + c} = 5⁰ = 1.

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

Given,

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

Hence, x + y - z = 0.

(ii) Given,

3^{x + y - z} = 3⁰ = 1.

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

(i) Given,

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

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

Hence, x - y - z = 1.

(ii) Given,

7^{x - y - z} = 7¹ = 7.

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

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 $\frac{π4r^3}{3}$

⇒ V = $\frac{4}{3}πr^3$.

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

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

⇒ log 100 - log 3 = log 100 - log n

⇒ log n = log 3

⇒ n = 3.

Hence, n = 3.

Given,

log₁₀y + 2log₁₀x = 2

⇒ log₁₀y + log₁₀x² = 2

⇒ log₁₀yx² = 2log₁₀10

⇒ log₁₀yx² = log₁₀10²

⇒ log₁₀yx² = log₁₀100

⇒ yx² = 100

⇒ y = $\frac{100}{x^2}.$

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

Given,

⇒ log₁₀2 + 1

⇒ log₁₀2 + log₁₀10

⇒ log₁₀2 × 10

⇒ log₁₀20

Hence, log₁₀2 + 1 = log₁₀20.

Given,

a² = log₁₀x

b³ = log₁₀y

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

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

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

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,

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

= log 10 + log x - 2log y

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

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

= 1 - m + 3n.

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

Given,

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

⇒ 3log x = 2log y

⇒ log x³ = log y²

⇒ x³ = y²

Squaring both sides we get,

⇒ x⁶ = y⁴

⇒ $\frac{x^6}{y^4} = 1$.

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

Given,

⇒ log x + log 5 = 2log 3

⇒ log x = log 3² - log 5

⇒ log x = log 9 - log 5

⇒ log x = log $\frac{9}{5}$

⇒ x = $\frac{9}{5}$.

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

Given,

⇒ log₃x - log₃2 = 1

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

⇒ $\frac{x}{2} = 3$

⇒ x = 6.

Hence, x = 6.

Given,

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

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

Given,

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

Hence, x = 1000.

(i) Given,

2log₁₀x + 1 = log₁₀250

⇒ log₁₀x² + log₁₀10 = log₁₀250

⇒ log₁₀x² = log₁₀250 - log₁₀10

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

⇒ log₁₀x² = log₁₀25

⇒ x² = 25

⇒ x = 5.

Hence, x = 5.

(ii) Given,

log₁₀2x

⇒ log₁₀2(5) = log₁₀10 = 1.

Hence, log₁₀2x = 1.

Given,

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

Considering,

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

Considering,

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

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

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}.

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}.

Given,

⇒ log(2x + 3) = log 7

⇒ 2x + 3 = 7

⇒ 2x = 4

⇒ x = 2.

Hence, x = 2.

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.

Given,

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

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

Hence, x = 4.5

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.

Given,

⇒ log(4y - 3) = log(2y + 1) - log 3

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

⇒ $\frac{2y + 1}{3} = 4y - 3$

⇒ 2y + 1 = 3(4y - 3)

⇒ 2y + 1 = 12y - 9

⇒ 12y - 2y = 1 + 9

⇒ 10y = 10

⇒ y = 1.

Hence, y = 1.

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.

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.

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)

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

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

⇒ $\frac{x + 1}{3} = 27(x - 1)$

⇒ x + 1 = 81(x - 1)

⇒ x + 1 = 81x - 81

⇒ 81x - x = 1 + 81

⇒ 80x = 82

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

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

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

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

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

Hence, x = 10.

Given,

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

Hence, proved that x² + y² = 6xy.

Given,

x² + y² = 23xy

Above equation can be written as,

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

Taking log on both sides we get,

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

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

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 = $\text{log}_{10}\space\frac{400}{25}$

⇒ 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.

Given,

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

Simplifying L.H.S. we get,

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

= log₄₂ (2 × 3 × 7)

= log₄₂ 42

= 1.

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

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

Given,

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

Simplifying L.H.S. we get,

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

Given,

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

Simplifying L.H.S. we get,

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

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

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

Given,

log_ba . log_cb . log_dc = log_da

Simplifying L.H.S. we get,

$$\begin{aligned}\Rightarrow \text{log}_\text{b}\text{ a}\space.\space\text{log}_\text{c}\text{ b}\space.\space\text{log}_\text{d}\text{ c} \\[1em] \Rightarrow \frac{\text{log a}}{\text{log b}} \times \frac{\text{log b}}{\text{log c}} \times \frac{\text{log c}}{\text{log d}} \\[1em] \Rightarrow \frac{\text{log a}}{\text{log d}} \\[1em] \Rightarrow \text{log}\_\text{d}\text{ a}.\end{aligned}$$

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

Given,

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

Solving log_abc x we get,

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

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

Given,

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

Hence, x = 3.

Given,

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

Hence, x = 2.

Given,

$$\begin{aligned}\Rightarrow x = \text{log}_{\sqrt{3}} \space 27 \\[1em] = \frac{\text{log}_{10} \space 27}{\text{log}_{10} \space \sqrt{3}} \\[1em] = \frac{\text{log}_{10} \space 3^3}{\text{log}_{10} \space 3^{\frac{1}{2}}} \\[1em] = \frac{3\text{log}_{10} \space 3}{\frac{1}{2}\text{log}_{10} \space 3} \\[1em] = \frac{3}{\frac{1}{2}} \\[1em] = 3 \times 2 = 6.\end{aligned}$$

Hence, Option 3 is the correct option.

Given,

log₅ (0.04) = x

⇒ 5^x = 0.04

⇒ 5^x = $\frac{4}{100} = \frac{1}{25}$

⇒ 5^x = $\frac{1}{5^2}$

⇒ 5^x = 5^-2

⇒ x = -2.

Hence, Option 4 is the correct option.

Given,

log_0.5 64 = x

$$\begin{aligned}\Rightarrow (0.5)^x = 64 \\[1em] \Rightarrow \Big(\frac{5}{10}\Big)^x = 64 \\[1em] \Rightarrow \Big(\frac{1}{2}\Big)^x = (2)^6 \\[1em] \Rightarrow (2^{-1})^x = (2)^6 \\[1em] \Rightarrow 2^{-x} = (2)^6 \\[1em] \Rightarrow -x = 6 \\[1em] \Rightarrow x = -6.\end{aligned}$$

Hence, Option 2 is the correct option.

Given,

$$\begin{aligned}\Rightarrow \text{log}_{\sqrt[3]{5}} \space x = -3 \\[1em] \Rightarrow x = (\sqrt[3]{5})^{-3} \\[1em] \Rightarrow x = [(5)^{\frac{1}{3}}]^{-3} \\[1em] \Rightarrow x = (5)^{\frac{1}{3} \times (-3)} \\[1em] \Rightarrow x = (5)^{-1} = \frac{1}{5}.\end{aligned}$$

Hence, Option 1 is the correct option.

Given,

log (3x + 1) = 2

⇒ 10² = 3x + 1

⇒ 100 = 3x + 1

⇒ 3x = 100 - 1

⇒ 3x = 99

⇒ x = 33.

Hence, Option 3 is the correct option.

Given,

2 + log₁₀ (0.01)

⇒ 2 + log₁₀ (10^-2)

⇒ 2 + (-2)log₁₀ 10

⇒ 2 + (-2)1

⇒ 2 + (-2)

⇒ 0

Hence, Option 4 is the correct option.

Given,

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

Hence, Option 1 is the correct option.

According to statement 1; log₅ 150 = log₅ 25 + log₅ 125

Solving R.H.S.,

⇒ log₅ 25 + log₅ 125

⇒ log₅ (25 x 125)

⇒ log₅ 3125.

As, log₅ 150 ≠ log₅ 3125

∴ Statement 1 is false.

According to statement 2; log_a (b + c) = log_a b + log_a c

Solving R.H.S.

⇒ log_a b + log_a c

⇒ log_a (b x c)

⇒ log_a bc.

As, log_a (b + c) ≠ log_a bc

∴ Statement 2 is false.

∴ Both statements are false.

Hence, option 2 is the correct option.

log_a (bc) = log_a b + log_a c

This is a fundamental property of logarithms, known as the product rule. It states that the logarithm of a product is the sum of the logarithms of the individual factors.

∴ Reason (R) is true.

Given, log₂ 16

⇒ log₂ (4 x 4)

⇒ log₂ 4 + log₂ 4

⇒ log₂ 2² + log₂ 2²

⇒ 2log₂ 2 + 2log₂ 2

⇒ 2 + 2

⇒ 4.

∴ Assertion (A) is true.

∴ Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason (or explanation) for Assertion (A).

Hence, option 3 is the correct option.

Given,

⇒ log_a $\Big(\frac{1}{b}\Big)$

⇒ log_a b^-1

⇒ (-1) x log_a b

⇒ -log_a b

∴ Reason (R) is true.

Given, log₃ $\Big(\frac{1}{9}\Big)$

⇒ log₃ $\Big(\frac{1}{3^2}\Big)$

⇒ log₃ 3^-2

⇒ -2log₃ 3

⇒ -2 x 1

⇒ -2.

∴ Assertion (A) is true.

∴ Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason (or explanation) for Assertion (A).

Hence, option 3 is the correct option.

Given,

log _a^m bⁿ

By changing the base, we get :

$$\begin{aligned}\Rightarrow \frac{\text{ log } b^n}{\text{ log } a^m}\\[1em] \Rightarrow \frac{n\text{ log } b}{m\text{ log } a}\\[1em] \Rightarrow \frac{n}{m} \text{ log } _a b\end{aligned}$$

∴ Reason (R) is true.

Given,

$$\begin{aligned}\Rightarrow \text{log }_{\sqrt{2} } 2^5\\[1em] \Rightarrow \frac{\text{log } 2^5}{\text{log }\sqrt{2}} \\[1em] \Rightarrow \frac{\text{log } 2^5}{\text{log }2^{\frac{1}{2}}} \\[1em] \Rightarrow \frac{5 \times \text{ log } 2}{\frac{1}{2} \times \text{ log } 2} \\[1em] \Rightarrow \frac{10}{1} \times 1\\[1em] \Rightarrow 10.\end{aligned}$$

∴ Assertion (A) is true.

∴ Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason (or explanation) for Assertion (A).

Hence, option 3 is the correct option.

According to reason,

If log _a b = log _a c, then b = c.

If the logarithms of two numbers are equal, and they share the same base, then the numbers themselves must be equal.

∴ Reason (R) is true.

According to assertion,

$$\begin{aligned}\Rightarrow \text{log x} = \frac{\text{ log } 8}{\text{ log } 0.25}\\[1em] \Rightarrow \text{log x} = \frac{\text{ log } 8}{\text{ log } \Big(\frac{25}{100}\Big)}\\[1em] \Rightarrow \text{log x} = \frac{\text{ log } 2^3}{\text{ log } \Big(\frac{1}{4}\Big)}\\[1em] \Rightarrow \text{log x} = \frac{\text{ log } 2^3}{\text{ log } \Big(\frac{1}{2^2}\Big)}\\[1em] \Rightarrow \text{log x} = \frac{\text{ log } 2^3}{\text{ log } 2^{-2}}\\[1em] \Rightarrow \text{log x} = \frac{3\text{ log } 2}{-2\text{ log } 2}\\[1em] \Rightarrow \text{log x} = \frac{-3}{2} \\[1em] \Rightarrow x = (10)^{-\frac{3}{2}}.\end{aligned}$$

∴ Assertion (A) is false.

∴ Assertion (A) is false, Reason (R) is true.

Hence, option 2 is the correct option.

Given,

$$\begin{aligned}\Rightarrow \text{log}_a \space {\sqrt[3]{x^7y^8 ÷ \sqrt[4]{z}}} \\[1em] \Rightarrow \text{log}_a \space {(x^7y^8 ÷ \sqrt[4]{z})^{\frac{1}{3}}} \\[1em] \Rightarrow \frac{1}{3}\text{log}_a \space {(x^7y^8 ÷ \sqrt[4]{z})} \\[1em] \Rightarrow \frac{1}{3}[\text{log}_a \space x^7y^8 - \text{log}_a \space \sqrt[4]{z}] \\[1em] \Rightarrow \frac{1}{3}[\text{log}_a \space x^7 + \text{log}_a \space y^8 - \text{log}_a \space z^{\frac{1}{4}}] \\[1em] \Rightarrow \frac{1}{3}[7\text{log}_a \space x + 8\text{log}_a \space y - \frac{1}{4}\text{log}_a \space z] \\[1em] \Rightarrow \frac{7}{3}\text{log}_a \space x + \frac{8}{3}\text{log}_a \space y - \frac{1}{12}\text{log}_a \space z.\end{aligned}$$

Hence, $\text{log}_a \space {\sqrt[3]{x^7y^8 ÷ \sqrt[4]{z}}} = \frac{7}{3}\text{log}_a \space x + \frac{8}{3}\text{log}_a \space y - \frac{1}{12}\text{log}_a \space z.$

Given,

$$\begin{aligned}\Rightarrow \text{log}_{\sqrt{3}} \space 3\sqrt{3} - \text{log}_5 \space (0.04) \\[1em] \Rightarrow \text{log}_{\sqrt{3}} \space (\sqrt{3})^3 - \text{log}_5 \space \Big(\frac{4}{100}\Big) \\[1em] \Rightarrow 3\text{log}_{\sqrt{3}}\sqrt{3} - \text{log}_5\Big(\frac{1}{25}\Big) \\[1em] \Rightarrow 3(1) - \text{log}_5(5)^{-2} \\[1em] \Rightarrow 3 - (-2)\text{log}_55 \\[1em] \Rightarrow 3 + 2(1) \\[1em] \Rightarrow 5.\end{aligned}$$

Hence, $\text{log}_{\sqrt{3}} \space 3\sqrt{3} - \text{log}_5 \space (0.04)$ = 5.

Given,

$(\text{log} \space x)^2 - (\text{log} \space y)^2 = \text{log} \space \frac{x}{y}.\text{log} \space xy$

Simplifying L.H.S. of above equation we get,

$$\begin{aligned}\Rightarrow (\text{log} \space x)^2 - (\text{log} \space y)^2 \\[1em] \Rightarrow (\text{log} \space x - \text{log} \space y)(\text{log} \space x + \text{log} \space y) \\[1em] \Rightarrow \text{log} \space\frac{x}{y}.\text{log} \space xy\end{aligned}$$

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

Hence, proved that $(\text{log} \space x)^2 - (\text{log} \space y)^2 = \text{log} \space \frac{x}{y}.\text{log} \space xy$

Given,

$2\text{log} \space \frac{11}{13} + \text{log} \space \frac{130}{77} - \text{log} \space \frac{55}{91} = \text{log} \space 2.$

Simplifying L.H.S. of the above equation we get,

$$\begin{aligned}\Rightarrow 2\text{log} \space \frac{11}{13} + \text{log} \space \frac{130}{77} - \text{log} \space \frac{55}{91} \\[1em] \Rightarrow 2(\text{log} \space 11 - \text{log} \space 13) + (\text{log} \space 130 - \text{log} \space 77) - (\text{log} \space 55 - \text{log} \space 91) \\[1em] \Rightarrow 2(\text{log} \space 11 - \text{log} \space 13) + (\text{log} \space 13.10 - \text{log} \space 11.7) - (\text{log} \space 11.5 - \text{log} \space 13.7) \\[1em] \Rightarrow 2(\text{log} \space 11 - \text{log} \space 13) + (\text{log} \space 13 + \text{log} \space 10 - (\text{log} \space 11 + \text{log} \space 7) - (\text{log} \space 11 + \text{log} \space 5 - (\text{log} \space 13 + \text{log} \space 7))) \\[1em] \Rightarrow 2\text{log} \space 11 - 2\text{log} \space 13 + \text{log} \space 13 + \text{log} \space 10 - \text{log} \space 11 - \cancel{\text{log} \space 7} - \text{log} \space 11 - \text{log} \space 5 + \text{log} \space 13 + \cancel{\text{log} \space 7} \\[1em] \Rightarrow \cancel{2\text{log} \space 11} - \cancel{2\text{log} \space 11} + \cancel{2\text{log} \space 13} - \cancel{2\text{log} \space 13} + \text{log} \space 10 - \text{log} \space 5 \\[1em] \Rightarrow \text{log} \space 10 - \text{log} \space 5 \\[1em] \Rightarrow \text{log} \space 2.5 - \text{log} \space 5 \\[1em] \Rightarrow \text{log} \space 2 + \cancel{\text{log} \space 5} - \cancel{\text{log} \space 5} \\[1em] \Rightarrow \text{log} \space 2.\end{aligned}$$

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

Hence, proved that $2\text{log} \space \frac{11}{13} + \text{log} \space \frac{130}{77} - \text{log} \space \frac{55}{91} = \text{log} \space 2.$

Given,

log (m + n) = log m + log n

⇒ log (m + n) = log mn

⇒ m + n = mn

⇒ m = mn - n

⇒ m = n(m - 1)

⇒ n = $\frac{m}{m - 1}$.

Hence, proved that n = $\frac{m}{m - 1}$.

Given,

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

Squaring both sides we get,

$$\begin{aligned}\Rightarrow (x + y)^2 = 4(xy) \\[1em] \Rightarrow (x^2 + y^2 + 2xy) = 4xy \\[1em] \Rightarrow x^2 + y^2 + 2xy - 4xy = 0 \\[1em] \Rightarrow x^2 + y^2 - 2xy = 0 \\[1em] \Rightarrow (x - y)^2 = 0 \\[1em] \Rightarrow x - y = 0 \\[1em] \Rightarrow x = y.\end{aligned}$$

Hence, proved that x = y.

Given,

a² + b² = 27ab

⇒ a² + b² = 2ab + 25ab

⇒ a² + b² - 2ab = 25ab

$\Rightarrow ab = \frac{a^2 + b^2 - 2ab}{25} = \Big(\frac{a - b}{5}\Big)^2$

Taking log on both sides:

$$\begin{aligned}\Rightarrow \text{log} \space ab = \text{log} \space \Big(\frac{a - b}{5}\Big)^2 \\[1em] \Rightarrow \text{log} \space a + \text{log} \space b = 2 \text{log} \space \Big(\frac{a - b}{5}\Big) \\[1em] \Rightarrow \text{log} \space \Big(\frac{a - b}{5}\Big) = \frac{1}{2}(\text{log} \space a + \text{log} \space b).\end{aligned}$$

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

Given,

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

Hence, x = 7.

Given,

$$\begin{aligned}\Rightarrow \text{log}_x \space \frac{1}{4\sqrt{2}} = -5 \\[1em] \Rightarrow \text{log}_x \space \frac{1}{2^2.2^{\frac{1}{2}}} = -5 \\[1em] \Rightarrow -\frac{1}{5}\text{log}_x \space \frac{1}{2^{2 + \frac{1}{2}}} = 1 \\[1em] \Rightarrow -\frac{1}{5}\text{log}_x \space \frac{1}{2^{\frac{5}{2}}} = 1 \\[1em] \Rightarrow -\frac{1}{5}\text{log}_x \space 2^{-\frac{5}{2}} = 1 \\[1em] \Rightarrow -\frac{1}{5} \times -\frac{5}{2} \text{log}_x \space 2 = 1 \\[1em] \Rightarrow \frac{1}{2}\text{log}_x \space 2 = 1 \\[1em] \Rightarrow \text{log}_x \space 2 = 2 \\[1em] \Rightarrow x^2 = 2 \\[1em] \Rightarrow x = \sqrt{2}.\end{aligned}$$

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

Given,

$$\begin{aligned}\Rightarrow \text{log}_x \space \frac{1}{243} = 10 \\[1em] \Rightarrow \text{log}_x \space 1 - \text{log}_x \space 243 = 10 \\[1em] \Rightarrow 0 - \text{log}_x \space (3)^5 = 10 \\[1em] \Rightarrow -5\text{log}_x \space 3 = 10 \\[1em] \Rightarrow -\text{log}_x \space 3 = \frac{10}{5} \\[1em] \Rightarrow \text{log}_x \space 3 = -2 \\[1em] \Rightarrow x^{-2} = 3 \\[1em] \Rightarrow \frac{1}{x^2} = 3 \\[1em] \Rightarrow \frac{1}{x} = \sqrt{3} \\[1em] \Rightarrow x = \frac{1}{\sqrt{3}}.\end{aligned}$$

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

Given,

⇒ log₄ 32 = x - 4

⇒ 4^{x - 4} = 32

⇒ (2²)^{x - 4} = 2⁵

⇒ 2^{2x - 8} = 2⁵

⇒ 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 = 50

⇒ x² = 25

⇒ x = 5, -5.

Hence, x = 5, -5.

Given,

⇒ log (x² - 21) = 2

⇒ x² - 21 = 10²

⇒ x² = 100 + 21

⇒ x² = 121

⇒ x = 11, -11.

Hence, x = 11, -11.

Given,

⇒ log₆ (x - 2)(x + 3) = 1

⇒ (x - 2)(x + 3) = 6¹

⇒ (x² + 3x - 2x - 6) = 6

⇒ x² + x - 6 = 6

⇒ x² + x - 12 = 0

⇒ x² + 4x - 3x - 12 = 0

⇒ x(x + 4) - 3(x + 4) = 0

⇒ (x - 3)(x + 4) = 0

⇒ x - 3 = 0 or x + 4 = 0

⇒ x = 3 or x = -4.

Hence, x = 3, -4.

Given,

⇒ log₆ (x - 2) + log₆ (x + 3) = 1

⇒ log₆ (x - 2)(x + 3) = 1

⇒ (x - 2)(x + 3) = 6¹

⇒ (x² + 3x - 2x - 6) = 6

⇒ x² + x - 6 = 6

⇒ x² + x - 12 = 0

⇒ x² + 4x - 3x - 12 = 0

⇒ x(x + 4) - 3(x + 4) = 0

⇒ (x - 3)(x + 4) = 0

⇒ x - 3 = 0 or x + 4 = 0

⇒ x = 3 or x = -4.

In this case x ≠ -4 as (x + 3) and (x - 2) will be negative and log of only positive numbers are defined.

Hence, x = 3.

Given,

⇒ log (x + 1) + log (x - 1) = log 11 + 2log 3

⇒ log (x + 1)(x - 1) = log 11 + log 3²

⇒ log (x² - 1) = log (11 x 9)

⇒ log (x² - 1) = log 99

⇒ (x² - 1) = 99

⇒ x² = 99 + 1

⇒ x² = 100

⇒ x = 10, -10

In this case x ≠ -10 as (x + 1) and (x - 1) will be negative and log of only positive numbers are defined.

Hence, x = 10.

Given,

$\frac{\text{log} \space x}{3} = \frac{\text{log} \space y}{2}$

⇒ 2 log x = 3 log y

⇒ 2 log x - 3 log y = 0 .......(i)

Given,

log (xy) = 5

⇒ log x + log y = 5 ......(ii)

Multiplying (ii) by 2 we get,

⇒ 2 log x + 2 log y = 10 ......(iii)

Subtracting (i) from (iii) we get,

⇒ 2 log x + 2 log y - (2 log x - 3 log y) = 10 - 0

⇒ 2 log x - 2 log x + 2 log y + 3 log y = 10

⇒ 5 log y = 10

⇒ log y = 2

⇒ y = 10² = 100.

Substituting value of y in (ii) we get,

⇒ log x + log 100 = 5

⇒ log x + 2 = 5

⇒ log x = 3

⇒ x = 10³ = 1000.

Hence, x = 1000 and y = 100.

a = 1 + log_x yz = log_x x + log_x yz = log_x xyz.

$\therefore \frac{1}{a} = \text{log}_{xyz} \space x$

b = 1 + log_y zx = log_y y + log_y zx = log_y xyz.

$\therefore \frac{1}{b} = \text{log}_{xyz} \space y$

c = 1 + log_z xy = log_z z + log_z xy = log_z xyz.

$\therefore \frac{1}{c} = \text{log}_{xyz} \space z$

$$\begin{aligned}\Rightarrow \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \text{log}_{xyz} \space x + \text{log}_{xyz} \space y + \text{log}_{xyz} \space z \\[1em] \Rightarrow \frac{bc + ac + ab}{abc} = \frac{\text{log x}}{\text{log xyz}} + \frac{\text{log y}}{\text{log xyz}} + \frac{\text{log z}}{\text{log xyz}} \\[1em] \Rightarrow \frac{bc + ac + ab}{abc} = \frac{\text{log x + log y + log z}}{\text{log xyz}} \\[1em] \Rightarrow \frac{bc + ac + ab}{abc} = \frac{\text{log xyz}}{\text{log xyz}} \\[1em] \Rightarrow \frac{bc + ac + ab}{abc} = 1 \\[1em] \Rightarrow ab + bc + ca = abc.\end{aligned}$$

Hence, proved that ab + bc + ca = abc.

Given,

$$\begin{aligned}\Rightarrow \frac{1}{\text{ log }_a x} + \frac{1}{\text{ log }_b x} = \frac{2}{\text{ log }_c x}\\[1em] \Rightarrow \frac{1}{\frac{\text{log }x}{\text{log }a}} + \frac{1}{\frac{\text{log } x}{\text{log } b}} = \frac{2}{\frac{\text{log }x}{\text{log }c}} \\[1em] \Rightarrow \frac{\text{ log } a}{\text{ log } x} + \frac{\text{ log } b}{\text{ log } x} = \frac{2\text{ log } c}{\text{ log } x}\\[1em] \Rightarrow \frac{\text{ log } a + \text{ log } b}{\text{ log } x} = \frac{2\text{ log } c}{\text{ log } x}\\[1em] \Rightarrow \frac{\text{ log } (a \times b)}{\text{ log } x} = \frac{2\text{ log } c}{\text{ log } x}\\[1em] \Rightarrow \frac{\text{ log } ab}{\text{ log } x} = \frac{2\text{ log } c}{\text{ log } x}\\[1em] \Rightarrow \text{ log } ab = 2\text{ log } c\\[1em] \Rightarrow \text{ log } ab = \text{ log } c^2\\[1em] \Rightarrow ab = c^2.\end{aligned}$$

Hence, proved that c² = ab.