Indices [Exponents]

Simplifying the expression :

$$\begin{aligned}\Rightarrow \Big(-\frac{2a^2}{b^3}\Big)^4 = \Big(-2 \times \frac{a^2}{b^3}\Big)^4 \\[1em] = (-2)^4 \times \Big(\frac{a^2}{b^3}\Big)^4 = 16 \times \frac{a^{2 \times 4}}{b^{3 \times 4}} \\[1em] = \frac{16a^8}{b^{12}}.\end{aligned}$$

Hence, Option 2 is the correct option.

Simplifying the expression :

$$\begin{aligned}\Rightarrow \sqrt[4]{28} ÷ \sqrt[4]{7} = \frac{\sqrt[4]{28}}{\sqrt[4]{7}} \\[1em] = \Big(\frac{28}{7}\Big)^{\frac{1}{4}} = (4)^{\frac{1}{4}} \\[1em] = (2)^{2 \times \frac{1}{4}} = (2)^{\frac{1}{2}} \\[1em] = \sqrt{2}.\end{aligned}$$

Hence, Option 1 is the correct option.

Simplifying the expression :

$$\begin{aligned}\Rightarrow \Big(\frac{25}{9}\Big)^{-\frac{3}{2}} = \Big[\Big(\frac{5}{3}\Big)^2\Big]^{-\frac{3}{2}} \\[1em] = \Big(\frac{5}{3}\Big)^{-3} = \Big(\frac{3}{5}\Big)^3 \\[1em] = \frac{27}{125}.\end{aligned}$$

Hence, Option 3 is the correct option.

Simplifying the expression :

$$\begin{aligned}\Rightarrow 2 ÷ (243)^{-\frac{1}{5}} = 2 ÷ (3^5)^{-\frac{1}{5}} \\[1em] = 2 ÷ 3^{-1} = 2 ÷ \frac{1}{3} \\[1em] = 2 \times 3 \\[1em] = 6.\end{aligned}$$

Hence, Option 4 is the correct option.

Simplifying the expression :

$$\begin{aligned}\Rightarrow (0.01)^{-\frac{1}{2}} = \Big(\frac{1}{100}\Big)^{-\frac{1}{2}} \\[1em] = \Big(\frac{1}{10^2}\Big)^{-\frac{1}{2}} = (10^{-2})^{-\frac{1}{2}} \\[1em] = 10^1 = 10.\end{aligned}$$

Hence, Option 1 is the correct option.

Simplifying the expression :

$$\begin{aligned}\Rightarrow 3 \times (32)^{\frac{2}{5}} \times 7^0 = 3 \times (2^5)^{\frac{2}{5}} \times 7^0 \\[1em] = 3 \times 2^2 \times 1 \\[1em] = 3 \times 4 \\[1em] = 12.\end{aligned}$$

Hence, Option 2 is the correct option.

Simplifying the expression :

$$\begin{aligned}\Rightarrow 3^3 \times (3^5)^{-\frac{2}{3}} \times (3^2)^{-\frac{1}{3}} \\[1em] = 3^3 \times (3)^{-\frac{10}{3}} \times (3)^{-\frac{2}{3}} \\[1em] = (3)^{3 + \Big(-\frac{10}{3}\Big) + \Big(-\frac{2}{3}\Big)} \\[1em] = (3)^{\frac{9 - 10 - 2}{3}} = (3)^{\frac{-3}{3}} \\[1em] = 3^{-1} = \frac{1}{3}.\end{aligned}$$

Hence, $3^3 \times (243)^{-\frac{2}{3}} \times (9)^{-\frac{1}{3}} = \frac{1}{3}$.

Simplifying the expression :

$$\begin{aligned}\Rightarrow 5^{-4} \times (125)^{\frac{5}{3}} ÷ (25)^{-\frac{1}{2}} = 5^{-4} \times (5^3)^{\frac{5}{3}} ÷ (5^2)^{-\frac{1}{2}} \\[1em] = 5^{-4} \times 5^5 ÷ 5^{-1} = 5^{-4} \times 5^5 ÷ \frac{1}{5} \\[1em] = 5^{-4} \times 5^5 \times 5^1 \\[1em] = 5^{-4 + 5 + 1} = 5^2 \\[1em] = 25.\end{aligned}$$

Hence, $5^{-4} \times (125)^{\frac{5}{3}} ÷ (25)^{-\frac{1}{2}} = 25.$

Simplifying the expression :

$$\begin{aligned}\Rightarrow \Big(\frac{27}{125}\Big)^{\frac{2}{3}} \times \Big(\frac{9}{25}\Big)^{-\frac{3}{2}} = \Big[\Big(\frac{3}{5}\Big)^3\Big]^{\frac{2}{3}} \times \Big[\Big(\frac{3}{5}\Big)^2\Big]^{-\frac{3}{2}} \\[1em] = \Big(\frac{3}{5}\Big)^2 \times \Big(\frac{3}{5}\Big)^{-3} = \Big(\frac{3}{5}\Big)^{2 + (-3)} \\[1em] = \Big(\frac{3}{5}\Big)^{-1} = \frac{5}{3} = 1\frac{2}{3}.\end{aligned}$$

Hence, $\Big(\frac{27}{125}\Big)^{\frac{2}{3}} \times \Big(\frac{9}{25}\Big)^{-\frac{3}{2}} = 1\frac{2}{3}$.

Simplifying the expression :

$$\begin{aligned}\Rightarrow 7^0 \times (25)^{-\frac{3}{2}} - 5^{-3} = 1 \times (5^2)^{-\frac{3}{2}} - 5^{-3} \\[1em] = 1 \times 5^{-3} - 5^{-3} \\[1em] = 5^{-3} - 5^{-3} \\[1em] = 0.\end{aligned}$$

Hence, $7^0 \times (25)^{-\frac{3}{2}} - 5^{-3} = 0.$

Simplifying the expression :

$$\begin{aligned}\Rightarrow \Big(\frac{16}{81}\Big)^{-\frac{3}{4}} \times \Big(\frac{49}{9}\Big)^{\frac{3}{2}} ÷ \Big(\frac{343}{216}\Big)^{\frac{2}{3}} = \Big[\Big(\frac{2}{3}\Big)^4\Big]^{-\frac{3}{4}} \times \Big[\Big(\frac{7}{3}\Big)^2\Big]^{\frac{3}{2}} ÷ \Big[\Big(\frac{7}{6}\Big)^3\Big]^{\frac{2}{3}} \\[1em] = \Big(\frac{2}{3}\Big)^{4 \times -\frac{3}{4}} \times \Big(\frac{7}{3}\Big)^{2 \times \frac{3}{2}} ÷ \Big(\frac{7}{6}\Big)^{3 \times \frac{2}{3}} \\[1em] = \Big(\frac{2}{3}\Big)^{-3} \times \Big(\frac{7}{3}\Big)^3 ÷ \Big(\frac{7}{6}\Big)^2 \\[1em] = \Big(\frac{3}{2}\Big)^3 \times \Big(\frac{7}{3}\Big)^3 \times \Big(\frac{6}{7}\Big)^2 \\[1em] = \frac{3^3 \times 7^3 \times 6^2}{2^3 \times 3^3 \times 7^2} \\[1em] = \frac{7 \times 6^2}{2^3} \\[1em] = \frac{252}{8} \\[1em] = 31.5\end{aligned}$$

Hence, $\Big(\frac{16}{81}\Big)^{-\frac{3}{4}} \times \Big(\frac{49}{9}\Big)^{\frac{3}{2}} ÷ \Big(\frac{343}{216}\Big)^{\frac{2}{3}} = 31.5$

Simplifying the expression :

$$\begin{aligned}\Rightarrow \Big(\frac{8x^3}{125y^3}\Big)^{\frac{2}{3}} =\Big[\frac{(2x)^3}{(5y)^3}\Big]^{\frac{2}{3}} \\[1em] = \Big[\Big(\frac{2x}{5y}\Big)^3\Big]^{\frac{2}{3}} = \Big(\frac{2x}{5y}\Big)^2 \\[1em] = \frac{4x^2}{25y^2}.\end{aligned}$$

Hence, $(8x^3 ÷ 125y^3)^{\frac{2}{3}} = \frac{4x^2}{25y^2}.$

Simplifying the expression :

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

Hence, $(a + b)^{-1}(a^{-1} + b^{-1}) = \frac{1}{ab}$.

Simplifying the expression :

$$\begin{aligned}\Rightarrow \frac{5^{n + 3} - 6 \times 5^{n + 1}}{9 \times 5^n - 5^n \times 2^2} = \frac{5^n.5^3 - 6 \times 5^n \times 5^1}{5^n(9 - 2^2)} \\[1em] = \frac{5^n(5^3 - 6 \times 5)}{5^n(9 - 4)} = \frac{125 - 30}{5} = \frac{95}{5} \\[1em] = 19.\end{aligned}$$

Hence, $\frac{5^{n + 3} - 6 \times 5^{n + 1}}{9 \times 5^n - 5^n \times 2^2} = 19.$

Simplifying the expression :

$$\begin{aligned}\Rightarrow (3x^2)^{-3} \times (x^9)^{\frac{2}{3}} = \Big(\frac{1}{3x^2}\Big)^3 \times x^{9 \times \frac{2}{3}} \\[1em] = \frac{1}{27x^6} \times x^6 \\[1em] = \frac{1}{27}.\end{aligned}$$

Hence, $(3x^2)^{-3} \times (x^9)^{\frac{2}{3}} = \frac{1}{27}$.

Simplifying the expression :

$$\begin{aligned}\Rightarrow \sqrt{\frac{1}{4}} + (0.01)^{-\frac{1}{2}} - (27)^{\frac{2}{3}} = \frac{1}{2} + \Big(\frac{1}{100}\Big)^{-\frac{1}{2}} - (3^3)^{\frac{2}{3}} \\[1em] = \frac{1}{2} + \Big(\frac{1}{10^2}\Big)^{-\frac{1}{2}} - 3^2 = \frac{1}{2} + (10^{-2})^{-\frac{1}{2}} - 3^{3 \times \frac{2}{3}} \\[1em] = \frac{1}{2} + 10^{-2 \times -\frac{1}{2}} - 3^2 \\[1em] = \frac{1}{2} + 10 - 9 \\[1em] = \frac{1}{2} + 1 \\[1em] = \frac{1 + 2}{2} \\[1em] = \frac{3}{2} \\[1em] = 1\frac{1}{2}.\end{aligned}$$

Hence, $\sqrt{\frac{1}{4}} + (0.01)^{-\frac{1}{2}} - (27)^{\frac{2}{3}} = 1\frac{1}{2}.$

Simplifying the expression :

$$\begin{aligned}\Rightarrow \Big(\frac{27}{8}\Big)^{\frac{2}{3}} - \Big(\frac{1}{4}\Big)^{-2} + 5^0 = \Big[\Big(\frac{3}{2}\Big)^3\Big]^{\frac{2}{3}} - \Big(\frac{1}{2^2}\Big)^{-2} + 5^0\\[1em] = \Big(\frac{3}{2}\Big)^{3 \times \frac{2}{3}} - (2^2)^2 + 1 \\[1em] = \Big(\frac{3}{2}\Big)^2 - 2^4 + 1 \\[1em] = \frac{9}{4} - 16 + 1 \\[1em] = \frac{9 - 64 + 4}{4} \\[1em] = -\frac{51}{4}.\end{aligned}$$

Hence, $\Big(\frac{27}{8}\Big)^{\frac{2}{3}} - \Big(\frac{1}{4}\Big)^{-2} + 5^0 = -\frac{51}{4}$.

Simplifying the expression :

$$\begin{aligned}\Rightarrow \Big(\frac{3^{-4}}{2^{-8}}\Big)^{\frac{1}{4}} = \frac{3^{-4\times \frac{1}{4}}}{2^{-8 \times \frac{1}{4}}} \\[1em] = \frac{3^{-1}}{2^{-2}} = \frac{\frac{1}{3}}{\frac{1}{2^2}} \\[1em] = \frac{2^2}{3} = \frac{4}{3}.\end{aligned}$$

Hence, $\Big(\frac{3^{-4}}{2^{-8}}\Big)^{\frac{1}{4}} = \frac{4}{3}$.

Simplifying the expression :

$$\begin{aligned}\Rightarrow \Big(\frac{27^{-3}}{9^{-3}}\Big)^{\frac{1}{5}} = \Big[\frac{(3^3)^{-3}}{(3^2)^{-3}}\Big]^{\frac{1}{5}}\\[1em] = \Big(\frac{3^{3 \times -3}}{3^{2 \times -3}}\Big)^{\frac{1}{5}} = \Big(\frac{3^{-9}}{3^{-6}}\Big)^{\frac{1}{5}} \\[1em] = (3^{-9 - (-6)})^{\frac{1}{5}} = (3^{-9 + 6})^{\frac{1}{5}} \\[1em] = (3^{-3})^{\frac{1}{5}} = (3)^{-\frac{3}{5}} \\[1em] = \Big(\frac{1}{3^3}\Big)^{\frac{1}{5}} = \frac{1}{3^{\frac{3}{5}}}.\end{aligned}$$

Hence, $\Big(\frac{27^{-3}}{9^{-3}}\Big)^{\frac{1}{5}} = \frac{1}{3^{\frac{3}{5}}}$.

Simplifying the expression :

$$\begin{aligned}\Rightarrow (32)^{-\frac{2}{5}} ÷ (125)^{-\frac{2}{3}} = (2^5)^{-\frac{2}{5}} ÷ (5^3)^{-\frac{2}{3}} \\[1em] = (2)^{5 \times -\frac{2}{5}} ÷ (5)^{3 \times -\frac{2}{3}} \\[1em] = (2)^{-2} ÷ 5^{-2} \\[1em] = \frac{1}{2^2} ÷ \frac{1}{5^2} \\[1em] = \frac{1}{2^2} \times 5^2 \\[1em] = \frac{1}{4} \times 25 \\[1em] = \frac{25}{4} = 6\frac{1}{4}.\end{aligned}$$

Hence, $(32)^{-\frac{2}{5}} ÷ (125)^{-\frac{2}{3}} = 6\frac{1}{4}$.

Simplifying the expression :

$$\begin{aligned}\Rightarrow [1 - {1 - (1 - n)^{-1}}^{-1}]^{-1} = \Big[1 - \text{\textbraceleft}1 - \frac{1}{1 - n}\text{\textbraceright}^{-1}\Big]^{-1} \\[1em] = \Big[1 - \text{\textbraceleft}\frac{1 - n - 1}{1 - n}\text{\textbraceright}^{-1}\Big]^{-1} \\[1em] = \Big[1 - \text{\textbraceleft}\frac{-n}{1 - n}\text{\textbraceright}^{-1}\Big]^{-1} \\[1em] = \Big[1 - \text{\textbraceleft}-\frac{1 - n}{n}\text{\textbraceright}\Big]^{-1} \\[1em] = \Big[1 + \frac{1 - n}{n}\Big]^{-1} \\[1em] = \Big[\frac{n + 1 - n}{n}\Big]^{-1} \\[1em] = \Big[\frac{1}{n}\Big]^{-1} \\[1em] = n.\end{aligned}$$

Hence, $[1 - {1 - (1 - n)^{-1}}^{-1}]^{-1} = n$.

Factorizing 2160, we get :

⇒ 2160 = 2⁴ × 3³ × 5¹

⇒ 2^a.3^b.5^c = 2⁴ × 3³ × 5¹

⇒ a = 4, b = 3 and c = 1.

Substituting values of a, b and c in 3^a × 2^-b × 5^-c, we get :

$$\begin{aligned}\Rightarrow 3^4 \times 2^{-3} \times 5^{-1} = 81 \times \frac{1}{2^3} \times \frac{1}{5} \\[1em] = 81 \times \frac{1}{8} \times \frac{1}{5} \\[1em] = \frac{81}{40} = 2\frac{1}{40}.\end{aligned}$$

Hence, a = 4, b = 3 and c = 1 and 3^a × 2^-b × 5^-c = $2\frac{1}{40}$.

Factorizing 1960, we get :

⇒ 1960 = 2³.5¹.7²

⇒ 2^a.5^b.7^c = 2³.5¹.7²

⇒ a = 3, b = 1 and c = 2.

Substituting values of a, b and c in 2^-a.7^b.5^-c, we get :

$$\begin{aligned}\Rightarrow 2^{-3} \times 7^1 \times 5^{-2} = \frac{1}{2^3} \times 7 \times \frac{1}{5^2} \\[1em] = \frac{1}{8} \times 7 \times \frac{1}{25} \\[1em] = \frac{7}{200}.\end{aligned}$$

Hence, 2^-a.7^b.5^-c = $\frac{7}{200}$.

Simplifying the expression :

$$\begin{aligned}\Rightarrow \frac{8^{3a} \times 2^5 \times 2^{2a}}{4 \times 2^{11a} \times 2^{-2a}} = \frac{(2^3)^{3a} \times 2^5 \times 2^{2a}}{2^2 \times 2^{11a} \times 2^{-2a}} \\[1em] = \frac{2^{9a} \times 2^5 \times 2^{2a}}{2^2 \times 2^{11a} \times 2^{-2a}} \\[1em] = \frac{2^{9a + 5 + 2a}}{2^{2 + 11a + (-2a)}} \\[1em] = \frac{2^{11a + 5}}{2^{9a + 2}} \\[1em] = 2^{(11a + 5) - (9a + 2)} \\[1em] = 2^{11a - 9a + 5 - 2} \\[1em] = 2^{2a + 3}.\end{aligned}$$

Hence, $\frac{8^{3a} \times 2^5 \times 2^{2a}}{4 \times 2^{11a} \times 2^{-2a}} = 2^{2a + 3}$.

Simplifying the expression

$$\begin{aligned}\Rightarrow \frac{3 \times 27^{n + 1} + 9 \times 3^{3n - 1}}{8 \times 3^{3n} - 5 \times 27^n} = \frac{3 \times (3^3)^{n + 1} + (3^2) \times 3^{3n - 1}}{8 \times 3^{3n} - 5 \times (3^3)^n}\\[1em] = \frac{3 \times 3^{3(n + 1)} + 3^{2 + 3n - 1}}{8 \times 3^{3n} - 5 \times 3^{3n}} \\[1em] = \frac{3^{1 + 3(n + 1)} + 3^{3n + 1}}{3^{3n}(8 - 5)} \\[1em] = \frac{3^{1 + 3n + 3} + 3^{3n + 1}}{3 \times 3^{3n}} \\[1em] = \frac{3^{3n + 1}.3^3 + 3^{3n + 1}}{3^{3n + 1}} \\[1em] = \frac{3^{3n + 1}(3^3 + 1)}{3^{3n + 1}} \\[1em] = 3^3 + 1 \\[1em] = 27 + 1 \\[1em] = 28.\end{aligned}$$

Hence, $\frac{3 \times 27^{n + 1} + 9 \times 3^{3n - 1}}{8 \times 3^{3n} - 5 \times 27^n} = 28$.

Solving L.H.S. of the above equation :

$$\begin{aligned}\Rightarrow \Big(\frac{a^m}{a^{-n}}\Big)^{m - n} \times \Big(\frac{a^n}{a^{-l}}\Big)^{n - l} \times (\frac{a^l}{a^{-m}}\Big)^{l - m} \\[1em] \Rightarrow (a^{m - (-n)})^{m - n} \times (a^{n - (-l)})^{n - l} \times (a^{l - (-m)})^{l - m} \\[1em] \Rightarrow (a^{(m + n)})^{m - n} \times (a^{(n + l)})^{n - l} \times (a^{(l + m)})^{l - m} \\[1em] \Rightarrow a^{(m + n)(m - n)} \times a^{(n + l)(n - l)} \times a^{(l + m)(l - m)} \\[1em] \Rightarrow a^{m^2 - n^2} \times a^{n^2 - l^2} \times a^{l^2 - m^2} \\[1em] \Rightarrow a^{m^2 - n^2 + n^2 - l^2 +l^2 - m^2} \\[1em] \Rightarrow a^{m^2 - m^2 - n^2 + n^2 - l^2 + l^2} \\[1em] \Rightarrow a^0 \\[1em] \Rightarrow 1.\end{aligned}$$

Since, L.H.S. = R.H.S. = 1.

Hence, proved that $\Big(\frac{a^m}{a^{-n}}\Big)^{m - n} \times \Big(\frac{a^n}{a^{-l}}\Big)^{n - l} \times (\frac{a^l}{a^{-m}}\Big)^{l - m}$ = 1.

Substituting values of a, b and c in L.H.S. of equation a^{m - n}.b^{n - l}.c^{l - m} = 1, we get :

⇒ a^{m - n}.b^{n - l}.c^{l - m} = (x^{m + n}.y^l)^{m - n}.(x^{n + l}.y^m)^{n - l}.(x^{l + m}.yⁿ)^{l- m}

= (x^{(m + n)(m - n)}.y^{l(m - n)}).(x^{(n + l)(n - l)}.y^{m(n - l)}).(x^{(l + m)(l - m)}.y^{n(l - m)})

= (x^{m2 - n2}).(x^{n2 - l2}).(x^{l2 - m2}).(y^{lm - ln}).(y^{mn - ml}).(y^{nl - nm})

= x^{m2 - n2 + n2 - l2 + l2 - m2}.y^{lm - ln + mn - ml + nl - nm}

= x⁰.y⁰

= 1.1

= 1.

Since, L.H.S. = R.H.S. = 1.

Hence, proved that a^{m - n}.b^{n - l}.c^{l- m} = 1.

Simplifying the expression :

$$\begin{aligned}\Rightarrow \Big(\frac{x^a}{x^b}\Big)^{a^2 + ab + b^2} \times \Big(\frac{x^b}{x^c}\Big)^{b^2 + bc + c^2} \times \Big(\frac{x^c}{x^a}\Big)^{c^2 + ca + a^2} \\[1em] = (x^{a - b})^{a^2 + ab + b^2} \times (x^{b - c})^{b^2 + bc + c^2} \times (x^{c - a})^{c^2 + ca + a^2} \\[1em] = x^{(a - b)(a^2 + ab + b^2)} \times x^{(b - c)(b^2 + bc + c^2)} \times x^{(c - a)(c^2 + ca + a^2)} \\[1em] = x^{a^3 - b^3} \times x^{b^3 - c^3} \times x^{c^3 - a^3} \\[1em] = x^{a^3 - b^3 + b^3 - c^3 + c^3 - a^3} \\[1em] = x^0 \\[1em] = 1.\end{aligned}$$

Hence, $\Big(\frac{x^a}{x^b}\Big)^{a^2 + ab + b^2} \times \Big(\frac{x^b}{x^c}\Big)^{b^2 + bc + c^2} \times \Big(\frac{x^c}{x^a}\Big)^{c^2 + ca + a^2}$ = 1.

Simplifying the expression :

$$\begin{aligned}\Rightarrow \Big(\frac{x^a}{x^{-b}}\Big)^{a^2 - ab + b^2} \times \Big(\frac{x^b}{x^{-c}}\Big)^{b^2 - bc + c^2} \times \Big(\frac{x^c}{x^{-a}}\Big)^{c^2 - ca + a^2} \\[1em] = (x^{a - (-b)})^{a^2 - ab + b^2} \times (x^{b - (-c)})^{b^2 - bc + c^2} \times (x^{c - (-a)})^{c^2 - ca + a^2} \\[1em] = x^{(a + b)(a^2 - ab + b^2)} \times x^{(b + c)(b^2 - bc + c^2)} \times x^{(c + a)(c^2 - ca + a^2)} \\[1em] = x^{a^3 + b^3} \times x^{b^3 + c^3} \times x^{c^3 + a^3} \\[1em] = x^{a^3 + b^3 + b^3 + c^3 + c^3 + a^3} \\[1em] = x^{2a^3 + 2b^3 + 2c^3} \\[1em] = x^{2(a^3 + b^3 + c^3)}.\end{aligned}$$

Hence, $\Big(\frac{x^a}{x^{-b}}\Big)^{a^2 - ab + b^2} \times \Big(\frac{x^b}{x^{-c}}\Big)^{b^2 - bc + c^2} \times \Big(\frac{x^c}{x^{-a}}\Big)^{c^2 - ca + a^2} = x^{2(a^3 + b^3 + c^3)}.$

Simplifying the expression :

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

Hence, Option 1 is the correct option.

Solving, the given expression :

$$\begin{aligned}\Rightarrow 4^{2x} = \frac{1}{32} \\[1em] \Rightarrow (2^2)^{2x} = \frac{1}{2^5} \\[1em] \Rightarrow 2^{4x} = 2^{-5} \\[1em] \Rightarrow 4x = -5 \\[1em] \Rightarrow x = -\frac{5}{4} = -1.25\end{aligned}$$

Hence, Option 2 is the correct option.

Simplifying the expression :

$$\begin{aligned}\Rightarrow \frac{3x}{y^{-1}} + \frac{2y}{x^{-1}} = \frac{3x}{\frac{1}{y}} + \frac{2y}{\frac{1}{x}} \\[1em] = 3xy + 2xy \\[1em] = 5xy.\end{aligned}$$

Hence, Option 3 is the correct option.

Simplifying the expression :

$$\begin{aligned}\Rightarrow \Big(8^{-\frac{4}{3}} ÷ 2^{-2}\Big) = \Big[(2^3)^{-\frac{4}{3}} ÷ \frac{1}{2^2}\Big] \\[1em] = \Big[2^{-4} ÷ \frac{1}{2^2}\Big] \\[1em] = \frac{1}{2^4} ÷ \frac{1}{2^2} \\[1em] = \frac{1}{2^4} \times 2^2 \\[1em] = \frac{1}{2^2} \\[1em] = \frac{1}{4}.\end{aligned}$$

Hence, Option 1 is the correct option.

Given,

⇒ 8^{2x + 5} = 1

⇒ 8^{2x + 5} = 8⁰

⇒ 2x + 5 = 0

⇒ 2x = -5

⇒ x = $-\frac{5}{2}$.

Hence, Option 1 is the correct option.

Given,

⇒ 3^{x + 1} = 9^{x - 2}

⇒ 3^{x + 1} = (3²)^{x - 2}

⇒ 3^{x + 1} = 3^{2x - 4}

⇒ x + 1 = 2x - 4

⇒ 2x - x = 1 + 4

⇒ x = 5.

Hence, Option 2 is the correct option.

Given,

$$\begin{aligned}\Rightarrow \sqrt{\frac{a}{b}} = \Big(\frac{b}{a}\Big)^{1 - 2x} \\[1em] \Rightarrow \Big(\frac{a}{b}\Big)^{\frac{1}{2}} = \Big(\frac{b}{a}\Big)^{1 - 2x} \\[1em] \Rightarrow \Big(\frac{a}{b}\Big)^{\frac{1}{2}} = \Big(\frac{a}{b}\Big)^{-(1 - 2x)} \\[1em] \Rightarrow \Big(\frac{a}{b}\Big)^{\frac{1}{2}} = \Big(\frac{a}{b}\Big)^{(2x - 1)} \\[1em] \Rightarrow \frac{1}{2} = 2x - 1 \\[1em] \Rightarrow 2x = 1 + \frac{1}{2} \\[1em] \Rightarrow 2x = \frac{2 + 1}{2} \\[1em] \Rightarrow x = \frac{3}{2 \times 2} \\[1em] \Rightarrow x = \frac{3}{4}.\end{aligned}$$

Hence, Option 1 is the correct option.

Given,

⇒ 2^{2x + 1} = 8

⇒ 2^{2x + 1} = 2³

⇒ 2x + 1 = 3

⇒ 2x = 3 - 1

⇒ 2x = 2

⇒ x = $\frac{2}{2}$ = 1.

Hence, x = 1.

Given,

⇒ 2^{5x - 1} = 4 × 2^{3x + 1}

⇒ 2^{5x - 1} = 2² × 2^{3x + 1}

⇒ 2^{5x - 1} = 2^{2 + 3x + 1}

⇒ 2^{5x - 1} = 2^{3x + 3}

⇒ 5x - 1 = 3x + 3

⇒ 5x - 3x = 3 + 1

⇒ 2x = 4

⇒ x = $\frac{4}{2}$ = 2.

Hence, x = 2.

Given,

⇒ 3^{4x + 1} = (27)^{x + 1}

⇒ 3^{4x + 1} = (3³)^{x + 1}

⇒ 3^{4x + 1} = 3^{3(x + 1)}

⇒ 3^{4x + 1} = 3^{3x + 3}

⇒ 4x + 1 = 3x + 3

⇒ 4x - 3x = 3 - 1

⇒ x = 2.

Hence, x = 2.

Given,

⇒ (49)^{x + 4} = 7² × (343)^{x + 1}

⇒ (7²)^{x + 4} = 7² × (7³)^{x + 1}

⇒ 7^{2(x + 4)} = 7² × 7^{3(x + 1)}

⇒ 7^{2x + 8} = 7² × 7^{3x + 3}

⇒ 7^{2x + 8} = 7^{2 + 3x + 3}

⇒ 2x + 8 = 3x + 5

⇒ 3x - 2x = 8 - 5

⇒ x = 3.

Hence, x = 3.

Given,

$$\begin{aligned}\Rightarrow 4^{2x} = \frac{1}{32} \\[1em] \Rightarrow (2^2)^{2x} = \frac{1}{2^5} \\[1em] \Rightarrow 2^{4x} = 2^{-5} \\[1em] \Rightarrow 4x = -5 \\[1em] \Rightarrow x = -\frac{5}{4}.\end{aligned}$$

Hence, x = $-\frac{5}{4}$.

Given,

⇒ $\sqrt{2^{x + 3}} = 16$

Squaring both sides, we get :

⇒ 2^{x + 3} = 16²

⇒ 2^{x + 3} = (2⁴)²

⇒ 2^{x + 3} = 2⁸

⇒ x + 3 = 8

⇒ x = 8 - 3 = 5.

Hence, x = 5.

Given,

$$\begin{aligned}\Rightarrow \Big(\sqrt{\frac{3}{5}}\Big)^{x + 1} = \frac{125}{27} \\[1em] \Rightarrow \Big(\frac{3}{5}^{\frac{1}{2}}\Big)^{x + 1} = \Big(\frac{5}{3}\Big)^3 \\[1em] \Rightarrow \Big(\frac{3}{5}\Big)^{\frac{x + 1}{2}} = \Big(\frac{3}{5}\Big)^{-3} \\[1em] \Rightarrow \frac{x + 1}{2} = -3 \\[1em] \Rightarrow x + 1 = -6 \\[1em] \Rightarrow x = -6 - 1 = -7.\end{aligned}$$

Hence, x = -7.

Given,

$$\begin{aligned}\Rightarrow \Big(\sqrt[3]{\frac{2}{3}}\Big)^{x - 1} = \frac{27}{8} \\[1em] \Rightarrow \Big(\frac{2}{3}^{\frac{1}{3}}\Big)^{x - 1} = \Big(\frac{3}{2}\Big)^3 \\[1em] \Rightarrow \Big(\frac{2}{3}\Big)^{\frac{x - 1}{3}} = \Big(\frac{2}{3}\Big)^{-3} \\[1em] \Rightarrow \frac{x - 1}{3} = -3 \\[1em] \Rightarrow x - 1 = 3 \times -3 \\[1em] \Rightarrow x - 1 = -9 \\[1em] \Rightarrow x = -9 + 1 = -8.\end{aligned}$$

Hence, x = -8.

Given,

⇒ 4^{x - 2} - 2^{x + 1} = 0

⇒ (2²)^{x - 2} - 2^{x + 1} = 0

⇒ 2^{2(x - 2)} = 2^{x + 1}

⇒ 2^{2x - 4} = 2^{x + 1}

⇒ 2x - 4 = x + 1

⇒ 2x - x = 1 + 4

⇒ x = 5.

Hence, x = 5.

Given,

⇒ 3^x2 : 3^x = 9 : 1

$$\begin{aligned}\Rightarrow \frac{3^{x^2}}{3^x} = \frac{9}{1} \\[1em] \Rightarrow \frac{3^{x^2}}{3^x} = \frac{3^2}{3^0} \\[1em] \Rightarrow 3^{x^2 - x} = 3^{2- 0} \\[1em] \Rightarrow x^2 - x = 2 - 0 \\[1em] \Rightarrow x^2 - x = 2 \\[1em] \Rightarrow x^2 - x - 2 = 0 \\[1em] \Rightarrow x^2 - 2x + x - 2 = 0 \\[1em] \Rightarrow x(x - 2) + 1(x - 2) = 0 \\[1em] \Rightarrow (x + 1)(x - 2) = 0 \\[1em] \Rightarrow x + 1 = 0 \text{ or } x - 2 = 0 \\[1em] \Rightarrow x = -1 \text{ or } x = 2.\end{aligned}$$

Hence, x = -1 or x = 2.

Given,

⇒ 8 × 2^2x + 4 × 2^{x + 1} = 1 + 2^x

⇒ 8 × 2^(x)(2) + 4 × 2^x.2¹ = 1 + 2^x

Substituting 2^x = a, in above equation, we get :

⇒ 8 × a² + 4a × 2 = 1 + a

⇒ 8a² + 8a = 1 + a

⇒ 8a² + 8a - a - 1 = 0

⇒ 8a(a + 1) - 1(a + 1) = 0

⇒ (8a - 1)(a + 1) = 0

⇒ 8a - 1 = 0 or a + 1 = 0

⇒ 8a = 1 or a = -1

a cannot be negative as 2^x, for any value of x is greater than 0.

⇒ 8(2^x) = 1

⇒ 2^x = $\frac{1}{8}$

⇒ $2^x = \frac{1}{2^3}$

⇒ 2^x = 2^-3

⇒ x = -3.

Hence, x = -3.

Given,

⇒ 2^2x + 2^{x + 2} - 4 × 2³ = 0

⇒ 2^(x)(2) + 2^x.2² - 4 × 8 = 0

Substituting 2^x = a, we get :

⇒ a² + 4a - 32 = 0

⇒ a² + 8a - 4a - 32 = 0

⇒ a(a + 8) - 4(a + 8) = 0

⇒ (a - 4)(a + 8) = 0

⇒ a - 4 = 0 or a + 8 = 0

⇒ a = 4 or a = -8

a cannot be negative as 2^x, for any value of x is greater than 0.

⇒ a = 4

⇒ 2^x = 4

⇒ 2^x = 2²

⇒ x = 2.

Hence, x = 2.

Given,

$$\begin{aligned}\Rightarrow (\sqrt{3})^{x - 3} = (\sqrt[4]{3})^{x + 1} \\[1em] \Rightarrow (3^{\frac{1}{2}})^{x - 3} = (3^{\frac{1}{4}})^{x + 1} \\[1em] \Rightarrow 3^{\frac{x - 3}{2}} = 3^{\frac{x + 1}{4}} \\[1em] \Rightarrow \frac{x - 3}{2} = \frac{x + 1}{4}\\[1em] \Rightarrow 4(x - 3) = 2(x + 1) \\[1em] \Rightarrow 4x - 12 = 2x + 2 \\[1em] \Rightarrow 4x - 2x = 2 + 12 \\[1em] \Rightarrow 2x = 14 \\[1em] \Rightarrow x = \frac{14}{2} = 7.\end{aligned}$$

Hence, x = 7.

Given,

4^2m = $(\sqrt[3]{16})^{-\frac{6}{n}} = (\sqrt{8})^2$

Considering,

$$\begin{aligned}\Rightarrow 4^{2m} = (\sqrt{8})^2 \\[1em] \Rightarrow (2^2)^{2m} = (\sqrt{8})^2 \\[1em] \Rightarrow 2^{4m} = 8 \\[1em] \Rightarrow 2^{4m} = 2^3 \\[1em] \Rightarrow 4m = 3 \\[1em] \Rightarrow m = \frac{3}{4}.\end{aligned}$$

Considering,

$$\begin{aligned}\Rightarrow (\sqrt[3]{16})^{-\frac{6}{n}} = (\sqrt{8})^2 \\[1em] \Rightarrow (16)^{\frac{1}{3} \times \Big(-\frac{6}{n}\Big)} = 8 \\[1em] \Rightarrow (16)^{-\frac{2}{n}} = 8 \\[1em] \Rightarrow (2^4)^{-\frac{2}{n}} = 2^3 \\[1em] \Rightarrow (2)^{4 \times -\frac{2}{n}} = 2^3 \\[1em] \Rightarrow (2)^{-\frac{8}{n}} = 2^3 \\[1em] \Rightarrow -\frac{8}{n} = 3 \\[1em] \Rightarrow n = -\frac{8}{3}.\end{aligned}$$

Hence, $m = \frac{3}{4} \text{ and } n = -\frac{8}{3}$.

Given,

$$\begin{aligned}\Rightarrow (\sqrt{32})^x ÷ 2^{y + 1} = 1 \\[1em] \Rightarrow \frac{(\sqrt{32})^x}{2^{y + 1}} = 1 \\[1em] \Rightarrow \frac{(\sqrt{2^5})^x}{2^{y + 1}} = 1 \\[1em] \Rightarrow [(2^5)^{\frac{1}{2}}]^x = 2^{y + 1} \\[1em] \Rightarrow 2^{5 \times \frac{1}{2} \times x} = 2^{y + 1} \\[1em] \Rightarrow \frac{5x}{2} = y + 1 \\[1em] \Rightarrow 5x = 2y + 2 \\[1em] \Rightarrow x = \frac{2y + 2}{5} \text{ ......(1)}\end{aligned}$$

Given,

$$\begin{aligned}\Rightarrow 8^y - 16^{4 - \frac{x}{2}} = 0 \\[1em] \Rightarrow 8^y = 16^{4 - \frac{x}{2}} \\[1em] \Rightarrow (2^3)^y = (2^4)^{4 - \frac{x}{2}} \\[1em] \Rightarrow 2^{3y} = (2)^{4\Big(4 - \frac{x}{2}\Big)} \\[1em] \Rightarrow 2^{3y} = 2^{16 - 2x} \\[1em] \Rightarrow 3y = 16 - 2x \\[1em] \Rightarrow 2x = 16 - 3y \text{ ........(2)}\end{aligned}$$

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

$$\begin{aligned}\Rightarrow 2\Big(\frac{2y + 2}{5}\Big) = 16 - 3y \\[1em] \Rightarrow \frac{4y + 4}{5} = 16 - 3y \\[1em] \Rightarrow 4y + 4 = 5(16 - 3y) \\[1em] \Rightarrow 4y + 4 = 80 - 15y \\[1em] \Rightarrow 4y + 15y = 80 - 4 \\[1em] \Rightarrow 19y = 76 \\[1em] \Rightarrow y = \frac{76}{19} = 4.\end{aligned}$$

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

$$\begin{aligned}\Rightarrow x = \frac{2y + 2}{5} = \frac{2 \times 4 + 2}{5} \\[1em] = \frac{8 + 2}{5} \\[1em] = \frac{10}{5} = 2.\end{aligned}$$

Hence, x = 2 and y = 4.

Solving L.H.S. of the given equation :

$$\begin{aligned}\Rightarrow \Big(\frac{x^a}{x^b}\Big)^{a + b - c} \Big(\frac{x^b}{x^c}\Big)^{b + c - a} \Big(\frac{x^c}{x^a}\Big)^{c + a - b} \\[1em] = (x^{a - b})^{a + b - c}.(x^{b - c})^{b + c - a}.(x^{c - a})^{c + a - b} \\[1em] = x^{(a - b)(a + b - c)}.x^{(b - c)(b + c - a)}.x^{(c - a)(c + a - b)} \\[1em] = x^{a^2 + ab - ac - ab - b^2 + bc}.x^{b^2 + bc - ab - bc - c^2 + ac}.x^{c^2 + ac - bc - ac - a^2 + ab} \\[1em] = x^{a^2 - b^2 - ac + bc}.x^{b^2 - c^2 - ab + ac}.x^{c^2 - a^2 - bc + ab} \\[1em] = x^{a^2 - b^2 - ac + bc + b^2 - c^2 - ab + ac + c^2 - a^2 - bc + ab} \\[1em] = x^{a^2 - a^2 - b^2 + b^2 - c^2 + c^2 - ac + ac + bc - bc - ab + ab} \\[1em] = x^0 \\[1em] = 1.\end{aligned}$$

Since, L.H.S. = R.H.S. = 1.

Hence, proved that $\Big(\frac{x^a}{x^b}\Big)^{a + b - c} \Big(\frac{x^b}{x^c}\Big)^{b + c - a} \Big(\frac{x^c}{x^a}\Big)^{c + a - b}$ = 1.

Solving L.H.S. of the given equation :

$$\begin{aligned}\Rightarrow \frac{x^{a(b - c)}}{x^{b(a - c)}} ÷ \Big(\frac{x^b}{x^a}\Big)^c = \frac{x^{ab - ac}}{x^{ab - bc}} ÷ \frac{x^{bc}}{x^{ac}} \\[1em] = x^{ab - ac - (ab - bc)} ÷ x^{bc - ac} \\[1em] = x^{ab - ab - ac + bc} ÷ x^{bc - ac} \\[1em] = x^{bc - ac} ÷ x^{bc - ac} \\[1em] = \frac{x^{bc - ac}}{x^{bc - ac}} \\[1em] = 1.\end{aligned}$$

Since, L.H.S. = R.H.S. = 1.

Hence, proved that $\frac{x^{a(b - c)}}{x^{b(a - c)}} ÷ \Big(\frac{x^b}{x^a}\Big)^c = 1$.

Given,

⇒ a = c^z .......(1)

⇒ c = b^y .......(2)

⇒ b = a^x ........(3)

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

⇒ a = (b^y)^z

⇒ a = b^yz

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

⇒ a = (a^x)^yz

⇒ a = a^xyz

⇒ xyz = 1.

Hence, proved that xyz = 1.

Given,

⇒ a^x = b^y = c^z = k (let)

⇒ a^x = k

⇒ a = $k^{\frac{1}{x}}$ ........(1)

⇒ b^y = k

⇒ b = $k^{\frac{1}{y}}$ ........(2)

⇒ c^z = k

⇒ c = $k^{\frac{1}{z}}$ ........(3)

Given,

⇒ b² = ac

Substituting values of a, b and c in above equation, we get :

$$\begin{aligned}\Rightarrow (k^{\frac{1}{y}})^2 = k^{\frac{1}{x}} \times k^{\frac{1}{z}} \\[1em] \Rightarrow k^{\frac{2}{y}} = k^{\frac{1}{x} + \frac{1}{z}} \\[1em] \Rightarrow \frac{2}{y} = \frac{1}{x} + \frac{1}{z} \\[1em] \Rightarrow \frac{2}{y} = \frac{z + x}{xz} \\[1em] \Rightarrow y = \frac{2xz}{x + z}.\end{aligned}$$

Hence, proved that $y = \frac{2xz}{x + z}.$

Given,

⇒ 5^-p = 4^-q = 20^r = k (let)

⇒ 5^-p = k

⇒ 5 = $k^{-\frac{1}{p}}$ .......(1)

⇒ 4^-q = k

⇒ 4 = $k^{-\frac{1}{q}}$ ......(2)

⇒ 20^r = k

⇒ 20 = $k^{\frac{1}{r}}$ ............(3)

We know that,

⇒ 5 × 4 = 20

From equations (1), (2) and (3), we get :

$$\begin{aligned}\Rightarrow k^{-\frac{1}{p}} \times k^{-\frac{1}{q}} = k^{\frac{1}{r}} \\[1em] \Rightarrow k^{-\frac{1}{p} + \Big(-\frac{1}{q}\Big)} = k^{\frac{1}{r}} \\[1em] \Rightarrow k^{-\frac{1}{p} - \frac{1}{q}} = k^{\frac{1}{r}} \\[1em] \Rightarrow -\frac{1}{p} - \frac{1}{q} = \frac{1}{r} \\[1em] \Rightarrow \frac{1}{p} + \frac{1}{q} + \frac{1}{r} = 0.\end{aligned}$$

Hence, proved that $\frac{1}{p} + \frac{1}{q} + \frac{1}{r} = 0.$.

Given,

$$\begin{aligned}\Rightarrow (m + n)^{-1}(m^{-1} + n^{-1}) = m^xn^y \\[1em] \Rightarrow \frac{1}{(m + n)} \times \Big(\frac{1}{m} + \frac{1}{n}\Big) = m^xn^y \\[1em] \Rightarrow \frac{1}{(m + n)} \times \Big(\frac{n + m}{mn}\Big) = m^xn^y \\[1em] \Rightarrow \frac{1}{mn} = m^xn^y \\[1em] \Rightarrow m^{-1}n^{-1} = m^xn^y \\[1em] \Rightarrow x = -1 \text{ and } y = -1.\end{aligned}$$

Substituting value in L.H.S. of equation x + y + 2 = 0, we get :

⇒ x + y + 2 = (-1) + (-1) + 2 = -2 + 2 = 0.

Since, L.H.S. = R.H.S. = 0.

Hence, proved that x + y + 2 = 0.

Given,

⇒ 5^{x + 1} = 25^{x - 2}

⇒ 5^{x + 1} = (5²)^{x - 2}

⇒ 5^{x + 1} = 5^{2(x - 2)}

⇒ x + 1 = 2(x - 2)

⇒ x + 1 = 2x - 4

⇒ 2x - x = 1 + 4

⇒ x = 5.

Substituting value of x in 3^{x - 3} × 2^{3 - x}, we get :

⇒ 3^{x - 3} × 2^{3 - x} = 3^{5 - 3} × 2^{3 - 5}

= 3² × 2^(-2)

= $9 \times \frac{1}{2^2} = \frac{9}{4} = 2\frac{1}{4}$.

Hence, 3^{x - 3} × 2^{3 - x} = $2\frac{1}{4}$.

Given,

⇒ 4^{x + 3} = 112 + 8 × 4^x

⇒ 4^x.4³ = 8(14 + 4^x)

⇒ 64.4^x = 8(14 + 4^x)

⇒ $\frac{64.4^x}{8}$ = 14 + 4^x

⇒ 8.4^x = 14 + 4^x

⇒ 8.4^x - 4^x = 14

⇒ 4^x(8 - 1) = 14

⇒ 7.4^x = 14

⇒ 4^x = $\frac{14}{7}$

⇒ (2²)^x = 2

⇒ 2^2x = 2¹

⇒ 2x = 1

⇒ x = $\frac{1}{2}$.

Substituting value of x in (18x)^3x, we get :

$$\begin{aligned}\Rightarrow (18x)^{3x} = \Big(18 \times \frac{1}{2}\Big)^{3 \times \frac{1}{2}} \\[1em] = 9^{\frac{3}{2}} \\[1em] = (3^2)^{\frac{3}{2}} \\[1em] = 3^{2 \times \frac{3}{2}} \\[1em] = 3^3 \\[1em] = 27.\end{aligned}$$

Hence, (18x)^3x = 27.

Given,

$$\begin{aligned}\Rightarrow 4^{x - 1} \times (0.5)^{3 - 2x} = \Big(\frac{1}{8}\Big)^{-x} \\[1em] \Rightarrow (2^2)^{x - 1} \times \Big(\frac{5}{10}\Big)^{3 - 2x} = \Big(\frac{1}{2^3}\Big)^{-x} \\[1em] \Rightarrow 2^{2(x - 1)} \times \Big(\frac{1}{2}\Big)^{3 - 2x} = (2^{-3})^{-x} \\[1em] \Rightarrow 2^{2x - 2} \times (2^{-1})^{3 - 2x} = 2^{3x} \\[1em] \Rightarrow 2^{2x - 2} \times 2^{-1(3 - 2x)} = 2^{3x} \\[1em] \Rightarrow 2^{(2x - 2) + [-1(3 - 2x)]} = 2^{3x} \\[1em] \Rightarrow 2^{2x - 2 - 3 + 2x} = 2^{3x} \\[1em] \Rightarrow 2^{4x - 5} = 2^{3x} \\[1em] \Rightarrow 4x - 5 = 3x \\[1em] \Rightarrow 4x - 3x = 5 \\[1em] \Rightarrow x = 5.\end{aligned}$$

Hence, x = 5.

Given,

⇒ (a^{3x + 5})².(a^x)⁴ = a^{8x + 12}

⇒ a^{2(3x + 5)}.a^4x = a^{8x + 12}

⇒ a^{6x + 10}.a^4x = a^{8x + 12}

⇒ a^{6x + 10 + 4x} = a^{8x + 12}

⇒ a^{10x + 10} = a^{8x + 12}

⇒ 10x + 10 = 8x + 12

⇒ 10x - 8x = 12 - 10

⇒ 2x = 2

⇒ x = $\frac{2}{2}$ = 1.

Hence, x = 1.

Given,

$$\begin{aligned}\Rightarrow (81)^{\frac{3}{4}} - \Big(\frac{1}{32}\Big)^{-\frac{2}{5}} + x\Big(\frac{1}{2}\Big)^{-1}.2^0 = 27 \\[1em] \Rightarrow (3^4)^{\frac{3}{4}} - \Big(\frac{1}{2^5}\Big)^{-\frac{2}{5}} + (2^{-1})^{-1}x.1 = 27 \\[1em] \Rightarrow 3^{4 \times \frac{3}{4}} - (2^{-5})^{-\frac{2}{5}} + 2^{-1\times -1}.x = 27 \\[1em] \Rightarrow 3^3 - 2^{-5 \times -\frac{2}{5}} + 2x = 27 \\[1em] \Rightarrow 27 - 2^2 + 2x = 27 \\[1em] \Rightarrow 27 - 4 + 2x = 27 \\[1em] \Rightarrow 23 + 2x = 27 \\[1em] \Rightarrow 2x = 27 - 23 \\[1em] \Rightarrow 2x = 4 \\[1em] \Rightarrow x = \frac{4}{2} = 2.\end{aligned}$$

Hence, x = 2.

Given,

⇒ 2^{3x + 3} = 2^{3x + 1} + 48

⇒ 2^3x.2³ = 2^3x.2¹ + 48

⇒ 8.2^3x = 2.2^3x + 48

⇒ 8.2^3x - 2.2^3x = 48

⇒ 6.2^3x = 48

⇒ 2^3x = $\frac{48}{6}$

⇒ 2^3x = 8

⇒ 2^3x = 2³

⇒ 3x = 3

⇒ x = $\frac{3}{3}$ = 1.

Hence, x = 1.

Given,

⇒ 3(2^x + 1) - 2^{x + 2} + 5 = 0

⇒ 3.2^x + 3 - 2^x.2² + 5 = 0

⇒ 3.2^x + 3 - 4.2^x + 5 = 0

⇒ -2^x + 8 = 0

⇒ 2^x = 8

⇒ 2^x = 2³

⇒ x = 3.

Hence, x = 3.

Given,

⇒ 9^{x + 2} = 720 + 9^x

⇒ 9^x.9² = 720 + 9^x

⇒ 81.9^x - 9^x = 720

⇒ 9^x(81 - 1) = 720

⇒ 9^x.80 = 720

⇒ 9^x = $\frac{720}{80}$

⇒ 9^x = 9

⇒ x = 1.

Hence, x = 1.

Given, (20⁰ - 18⁰) x 7⁰

As we know that a⁰ = 1

⇒ (1 - 1) x 1

⇒ 0 x 1

⇒ 0

Hence, option 1 is the correct option.

Given,

⇒ (-2)^-1 ÷ (-2)^-4

⇒ (-2)^-1 ÷ $\frac{1}{(-2)^4}$

⇒ $\frac{1}{(-2)} ÷ \frac{1}{(-2)^4}$

⇒ $\frac{1}{(-2)} \times (-2)^4$

⇒ (-2)³

⇒ -8.

Hence, option 3 is the correct option.

Given, $\frac{a}{b} = \frac{2}{3} ÷ \Big(-\frac{2}{3}\Big)^0$

As we know that a⁰ = 1

$$\begin{aligned}\Rightarrow \frac{a}{b} = \frac{2}{3} ÷ 1 \\[1em] \Rightarrow \frac{a}{b} = \frac{2}{3}\\[1em] \Rightarrow \Big(\frac{a}{b}\Big)^{-2} = \Big(\frac{2}{3}\Big)^{-2}\\[1em] \Rightarrow \Big(\frac{a}{b}\Big)^{-2} = \Big(\frac{3}{2}\Big)^2 \\[1em] \Rightarrow \Big(\frac{a}{b}\Big)^{-2} = \frac{9}{4}.\end{aligned}$$

Hence, option 3 is the correct option.

Given, 5^{2x + 3} = 1

As we know that a⁰ = 1,

⇒ 5^{2x + 3} = 5⁰

⇒ 2x + 3 = 0

⇒ 2x = -3

⇒ x = $-\frac{3}{2}$

Hence, option 2 is the correct option.

Given,

⇒ (2 + 3)^-1 x (2^-1 + 3^-1)

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

Hence, option 3 is the correct option.

Given,

$$\begin{aligned}\Rightarrow \Big(\frac{3}{4}\Big)^{-4} \times \Big(\frac{3}{4}\Big)^{-5} = \Big(\frac{3}{4}\Big)^{3x}\\[1em] \Rightarrow \Big(\frac{3}{4}\Big)^{-4 + (-5)} = \Big(\frac{3}{4}\Big)^{3x}\\[1em] \Rightarrow \Big(\frac{3}{4}\Big)^{-4 - 5} = \Big(\frac{3}{4}\Big)^{3x}\\[1em] \Rightarrow \Big(\frac{3}{4}\Big)^{-9} = \Big(\frac{3}{4}\Big)^{3x}\\[1em] \Rightarrow 3x = -9\\[1em] \Rightarrow x = -\frac{9}{3}\\[1em] \Rightarrow x = -3.\end{aligned}$$

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

Hence, option 4 is the correct option.

Given,

$$\begin{aligned}\Rightarrow \Big(\frac{5}{8}\Big)^{-7} \times \Big(\frac{8}{5}\Big)^{-4} = x\\[1em] \Rightarrow \Big(\frac{5}{8}\Big)^{-7} \times \Big(\frac{5}{8}\Big)^{4} = x\\[1em] \Rightarrow \Big(\frac{5}{8}\Big)^{-7 + 4} = x\\[1em] \Rightarrow \Big(\frac{5}{8}\Big)^{-3} = x\\[1em] \Rightarrow x = \Big(\frac{8}{5}\Big)^{3}\end{aligned}$$

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

Hence, option 4 is the correct option.

Given,

(3^-7 ÷ 3^-10) x 3^-5

$$\begin{aligned}\Rightarrow \Big[\Big(\frac{1}{3}\Big)^7 ÷ \Big(\frac{1}{3}\Big)^{10}\Big] \times \Big(\frac{1}{3}\Big)^5 \\[1em] \Rightarrow \Big[\Big(\frac{1}{3}\Big)^7 \times 3^{10}\Big] \times \frac{1}{3^5} \\[1em] \Rightarrow \frac{3^{10}}{3^7 \times 3^5} \\[1em] \Rightarrow \frac{3^{10}}{3^{12}} \\[1em] \Rightarrow \frac{1}{3^2} \\[1em] \Rightarrow \frac{1}{9}.\end{aligned}$$

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

Hence, option 3 is the correct option.

Given, x = (1³ + 2³ + 3³)$^\frac{1}{2}$

$$\begin{aligned}\Rightarrow (1 + 8 + 27)^\frac{1}{2}\\[1em] \Rightarrow (36)^\frac{1}{2}\\[1em] \Rightarrow \sqrt{36}\\[1em] \Rightarrow 6.\end{aligned}$$

∴ A is false, but R is true.

Hence, option 2 is the correct option.

Simplifying the expression :

$$\begin{aligned}\Rightarrow 9^{\frac{5}{2}} - 3 \times 8^0 - \Big(\frac{1}{81}\Big)^{-\frac{1}{2}} = (3^2)^{\frac{5}{2}} - 3 \times 1 - \Big(\frac{1}{3^4}\Big)^{-\frac{1}{2}} \\[1em] = (3)^{2 \times \frac{5}{2}} - 3 - (3^{-4})^{-\frac{1}{2}} \\[1em] = 3^5 - 3 - 3^{-4 \times -\frac{1}{2}} \\[1em] = 243 - 3 - 3^2 \\[1em] = 243 - 3 - 9 \\[1em] = 231.\end{aligned}$$

Hence, $9^{\frac{5}{2}} - 3 \times 8^0 - \Big(\frac{1}{81}\Big)^{-\frac{1}{2}}$ = 231.

Simplifying the expression :

$$\begin{aligned}\Rightarrow (64)^{\frac{2}{3}} - \sqrt[3]{125} - \frac{1}{2^{-5}} + (27)^{-\frac{2}{3}} \times \Big(\frac{25}{9}\Big)^{-\frac{1}{2}} \\[1em] = (2^6)^{\frac{2}{3}} - (5^3)^{\frac{1}{3}} - 2^5 + (3^3)^{-\frac{2}{3}} \times \Big[\Big(\frac{5}{3}\Big)^2\Big]^{-\frac{1}{2}}\\[1em] = (2)^{6 \times \frac{2}{3}} - (5)^{3 \times \frac{1}{3}} - 32 + (3)^{3 \times -\frac{2}{3}} \times \Big(\frac{5}{3}\Big)^{2 \times -\frac{1}{2}} \\[1em] = 2^4 - 5^1 - 32 + 3^{-2} \times \Big(\frac{5}{3}\Big)^{-1} \\[1em] = 16 - 5 - 32 + \frac{1}{3^2} \times \frac{3}{5} \\[1em] = -21 + \frac{1}{9} \times \frac{3}{5} \\[1em] = -21 + \frac{3}{45} \\[1em] = -21 + \frac{1}{15} \\[1em] = \frac{-315 + 1}{15} \\[1em] = \frac{-314}{15} \\[1em] = -20\frac{14}{15}.\end{aligned}$$

Hence, $(64)^{\frac{2}{3}} - \sqrt[3]{125} - \frac{1}{2^{-5}} + (27)^{-\frac{2}{3}} \times \Big(\frac{25}{9}\Big)^{-\frac{1}{2}} = -20\frac{14}{15}.$

Simplify the expression :

$$\begin{aligned}\Rightarrow \Big[\Big(-\frac{2}{3}\Big)^{-2}\Big]^3 \times \Big(\frac{1}{3}\Big)^{-4} \times 3^{-1} \times \frac{1}{6} \\[1em] = \Big(-\frac{2}{3}\Big)^{-6} \times (3^{-1})^{-4} \times \frac{1}{3} \times \frac{1}{6} \\[1em] = \Big(\frac{3}{2}\Big)^6 \times 3^4 \times \frac{1}{18} \\[1em] = \frac{3^6 \times 3^4}{2^6 \times 18} \\[1em] = \frac{3^{6 + 4}}{2^6 \times (2 \times 3 \times 3)} \\[1em] = \frac{3^{10}}{2^{6 + 1} \times 3^2} \\[1em] = \frac{3^{10 - 2}}{2^7} \\[1em] = \frac{3^8}{2^7} \\[1em] = 3^8 ÷ 2^7.\end{aligned}$$

Hence, $\Big[\Big(-\frac{2}{3}\Big)^{-2}\Big]^3 \times \Big(\frac{1}{3}\Big)^{-4} \times 3^{-1} \times \frac{1}{6} = 3^8 ÷ 2^7.$

Simplify the expression :

$$\begin{aligned}\Rightarrow \frac{3 \times 9^{n + 1} - 9 \times 3^{2n}}{3 \times 3^{2n + 3} - 9^{n + 1}} = \frac{3 \times (3^2)^{n + 1} - 9 \times 3^{2n}}{3 \times 3^{2n}.3^3 - (3^2)^{n + 1}} \\[1em] = \frac{3 \times 3^{2(n + 1)} - 9 \times 3^{2n}}{81.3^{2n} - 3^{2(n + 1)}} \\[1em] = \frac{3 \times 3^{2n + 2} - 9 \times 3^{2n}}{81.3^{2n} - 3^{2n + 2}} \\[1em] = \frac{3 \times 3^{2n} \times 3^2 - 9 \times 3^{2n}}{81.3^{2n} - 3^{2n}.3^2} \\[1em] = \frac{3^{2n}(3 \times 3^2 - 9)}{3^{2n}(81 - 3^2)} \\[1em] = \frac{27 - 9}{81 - 9} \\[1em] = \frac{18}{72} \\[1em] = \frac{1}{4}.\end{aligned}$$

Hence, $\frac{3 \times 9^{n + 1} - 3 \times 3^{2n}}{3 \times 3^{2n + 3} - 9^{n + 1}} = \frac{1}{4}$.

Solving the expression :

$$\begin{aligned}\Rightarrow 3^{x - 1} \times 5^{2y - 3} = 225 \\[1em] \Rightarrow 3^x.3^{-1} \times 5^{2y}.5^{-3} = 3^2 \times 5^2 \\[1em] \Rightarrow \frac{3^x}{3^1} \times \frac{5^{2y}}{5^3} = 3^2 \times 5^2 \\[1em] \Rightarrow 3^x \times 5^{2y} = 3^2 \times 5^2 \times 3^1 \times 5^3 \\[1em] \Rightarrow 3^x \times 5^{2y} = 3^{2 + 1} \times 5^{2 + 3} \\[1em] \Rightarrow 3^x \times 5^{2y} = 3^3 \times 5^5 \\[1em] \Rightarrow x = 3 \text{ and } 2y = 5 \\[1em] \Rightarrow x = 3 \text{ and } y = \frac{5}{2} = 2\frac{1}{2}.\end{aligned}$$

Hence, x = 3 and y = $2\frac{1}{2}$.

Given,

$$\begin{aligned}\Rightarrow \Big(\frac{a^{-1}b^2}{a^2b^{-4}}\Big)^7 ÷ \Big(\frac{a^3b^{-5}}{a^{-2}b^3}\Big)^{-5} = a^x.b^y \\[1em] \Rightarrow (a^{-1 - 2}b^{2 - (-4)})^7 ÷ (a^{3 - (-2)}b^{-5 - 3})^{-5} = a^x.b^y \\[1em] \Rightarrow (a^{-3}b^6)^7 ÷ (a^5b^{-8})^{-5} = a^x.b^y \\[1em] \Rightarrow (a^{-3 \times 7}.b^{6 \times 7}) ÷ (a^{5 \times -5}.b^{-8 \times -5}) = a^x.b^y \\[1em] \Rightarrow (a^{-21}.b^{42}) ÷ (a^{-25}.b^{40}) = a^x.b^y \\[1em] \Rightarrow \frac{a^{-21}.b^{42}}{a^{-25}.b^{40}} = a^xb^y \\[1em] \Rightarrow a^{-21 - (-25)}.b^{42 - 40} = a^xb^y \\[1em] \Rightarrow a^{-21 + 25}.b^{2} = a^x.b^y \\[1em] \Rightarrow a^4.b^2 = a^x.b^y \\[1em] \Rightarrow x = 4 \text{ and } y = 2.\end{aligned}$$

x + y = 4 + 2 = 6.

Hence, x + y = 6.

Given,

⇒ 3^{x + 1} = 9^{x - 3}

⇒ 3^{x + 1} = (3²)^{x - 3}

⇒ 3^{x + 1} = 3^{2(x - 3)}

⇒ 3^{x + 1} = 3^{2x - 6}

⇒ x + 1 = 2x - 6

⇒ 2x - x = 1 + 6

⇒ x = 7.

Substituting value of x in 2^{1 + x}, we get :

⇒ 2^{1 + 7} = 2⁸ = 256.

Hence, 2^{1 + x} = 256.

Given,

⇒ 2^x = 4^y = 8^z

⇒ 2^x = (2²)^y = (2³)^z

⇒ 2^x = 2^2y = 2^3z

⇒ x = 2y = 3z

⇒ x = 2y and x = 3z

$\Rightarrow y = \frac{x}{2} \text{ and } z = \frac{x}{3}$.

Substituting value of y and z in $\frac{1}{2x} + \frac{1}{4y} + \frac{1}{8z} = 4$, we get :

$$\begin{aligned}\Rightarrow \frac{1}{2x} + \frac{1}{4 \times \frac{x}{2}} + \frac{1}{8 \times \frac{x}{3}} = 4 \\[1em] \Rightarrow \frac{1}{2x} + \frac{1}{2x} + \frac{3}{8x} = 4 \\[1em] \Rightarrow \frac{2}{2x} + \frac{3}{8x} = 4 \\[1em] \Rightarrow \frac{8 + 3}{8x} = 4 \\[1em] \Rightarrow \frac{11}{8x} = 4 \\[1em] \Rightarrow x = \frac{11}{8 \times 4} = \frac{11}{32}.\end{aligned}$$

Hence, x = $\frac{11}{32}$.

Given,

$$\begin{aligned}\Rightarrow \frac{9^n.3^2.3^n - (27)^n}{(3^m.2)^3} = 3^{-3} \\[1em] \Rightarrow \frac{(3^2)^n.3^2.3^n - (3^3)^n}{(3^m)^3.(2)^3} = 3^{-3} \\[1em] \Rightarrow 3^{2n}.3^2.3^n - 3^{3n} = 3^{-3}.(3^m)^3.(2)^3 \\[1em] \Rightarrow 9.3^{2n + n} - 3^{3n} = 3^{-3}.3^{3m}.8 \\[1em] \Rightarrow 9.3^{3n} - 3^{3n} = 8.3^{3m - 3} \\[1em] \Rightarrow 3^{3n}(9 - 1) = 8.3^{3m - 3} \\[1em] \Rightarrow 8.3^{3n} = 8.3^{3(m - 1)} \\[1em] \Rightarrow 3^{3n} = 3^{3(m - 1)} \\[1em] \Rightarrow 3n = 3(m - 1) \\[1em] \Rightarrow n = m - 1 \\[1em] \Rightarrow m - n = 1.\end{aligned}$$

Hence, proved that m - n = 1.

Given,

$$\begin{aligned}\Rightarrow (13)^{\sqrt{x}} = 4^4 - 3^4 - 6 \\[1em] \Rightarrow (13)^{\sqrt{x}} = 256 - 81 - 6 \\[1em] \Rightarrow (13)^{\sqrt{x}} = 169 \\[1em] \Rightarrow (13)^{\sqrt{x}} = 13^2 \\[1em] \Rightarrow \sqrt{x} = 2\end{aligned}$$

Squaring both sides of the above equation, we get :

$$\begin{aligned}\Rightarrow (\sqrt{x})^2 = 2^2 \\[1em] \Rightarrow x = 4.\end{aligned}$$

Hence, x = 4.

Given,

$$\begin{aligned}\Rightarrow 3^{4x} = (3^4)^{-1} \\[1em] \Rightarrow 3^{4x} = 3^{-4} \\[1em] \Rightarrow 4x = -4 \\[1em] \Rightarrow x = -\frac{4}{4} \\[1em] \Rightarrow x = -1.\end{aligned}$$

Given,

$$\begin{aligned}\Rightarrow (10)^{\frac{1}{y}} = 0.0001 \\[1em] \Rightarrow (10)^{\frac{1}{y}} = \frac{1}{10^4} \\[1em] \Rightarrow (10)^{\frac{1}{y}} = 10^{-4} \\[1em] \Rightarrow \frac{1}{y} = -4 \\[1em] \Rightarrow y = -\frac{1}{4}.\end{aligned}$$

Substituting value of x and y in 2^-x × 16^y, we get :

$$\begin{aligned}\Rightarrow 2^{-x} \times 16^y = 2^{-(-1)} \times 16^{-\frac{1}{4}} \\[1em] = 2^1 \times (2^4)^{-\frac{1}{4}} \\[1em] = 2 \times 2^{4 \times -\frac{1}{4}} \\[1em] = 2 \times 2^{-1} \\[1em] = 2 \times \frac{1}{2} \\[1em] = 1.\end{aligned}$$

Hence, 2^-x × 16^y = 1.

Given,

⇒ (a^m)ⁿ = a^m.aⁿ

⇒ a^mn = a^{m + n}

⇒ mn = m + n

⇒ mn - m = n

⇒ m(n - 1) = n

⇒ m = $\frac{n}{n - 1}$ .......(1)

Substituting value of m from equation (1) in m(n - 1) - (n - 1), we get :

$$\begin{aligned}\Rightarrow m(n - 1) - (n - 1) = \frac{n}{n - 1} \times (n - 1) - (n - 1) \\[1em] = n - (n - 1) \\[1em] = n - n + 1 \\[1em] = 1.\end{aligned}$$

Hence, m(n - 1) - (n - 1) = 1.

Given,

$$\begin{aligned}\Rightarrow m = \sqrt[3]{15} \text{ and } n = \sqrt[3]{14} \\[1em] \Rightarrow m = (15)^{\frac{1}{3}} \text{ and } n = (14)^{\frac{1}{3}}\end{aligned}$$

Cubing both sides, we get :

$$\begin{aligned}\Rightarrow m^3 = [(15)^{\frac{1}{3}}]^3 \text{ and } n^3 = [(14)^{\frac{1}{3}}]^3 \\[1em] \Rightarrow m^3 = (15)^{\frac{1}{3} \times 3} \text{ and } n^3 = (14)^{\frac{1}{3} \times 3} \\[1em] \Rightarrow m^3 = 15 \text{ and } n^3 = 14.\end{aligned}$$

Simplifying the expression $m - n - \frac{1}{m^2 + mn + n^2}$, we get :

$$\begin{aligned}\Rightarrow \frac{m(m^2 + mn + n^2) - n(m^2 + mn + n^2) - 1}{m^2 + mn + n^2} \\[1em] \Rightarrow \frac{m^3 + m^2n + mn^2 - nm^2 - mn^2 - n^3 - 1}{m^2 + mn + n^2} \\[1em] \Rightarrow \frac{m^3 - n^3 - 1}{m^2 + mn + n^2} \\[1em]\end{aligned}$$

Substituting value of m³ and n³ in above equation, we get :

$$\begin{aligned}\Rightarrow \frac{15 - 14 - 1}{m^2 + mn + n^2} \\[1em] \Rightarrow \frac{1 - 1}{m^2 + mn + n^2} \\[1em] \Rightarrow \frac{0}{m^2 + mn + n^2} \\[1em] \Rightarrow 0.\end{aligned}$$

Hence, $m - n - \frac{1}{m^2 + mn + n^2} = 0$

Simplifying the expression :

$$\begin{aligned}\Rightarrow \Big(\frac{x^q}{x^r}\Big)^{\frac{1}{qr}} \times \Big(\frac{x^r}{x^p}\Big)^{\frac{1}{rp}} \times \Big(\frac{x^p}{x^q}\Big)^{\frac{1}{pq}} \\[1em] = (x^{q - r})^{\frac{1}{qr}} \times (x^{r - p})^{\frac{1}{rp}} \times (x^{p - q})^{\frac{1}{pq}} \\[1em] = (x)^{\frac{q - r}{qr}} \times (x)^{\frac{r - p}{rp}} \times (x)^{\frac{p - q}{pq}} \\[1em] = x^{\Big(\frac{q - r}{qr} + \frac{r - p}{rp} + \frac{p - q}{pq}\Big)} \\[1em] = x^{\Big(\frac{p(q - r) + q(r - p) + r(p - q)}{pqr}\Big)} \\[1em] = x^{\Big(\frac{pq - pr + qr - qp + rp - rq}{pqr}\Big)} \\[1em] = x^{\Big(\frac{0}{pqr}\Big)} \\[1em] = x^0 \\[1em] = 1.\end{aligned}$$

Hence, $\Big(\frac{x^q}{x^r}\Big)^{\frac{1}{qr}} \times \Big(\frac{x^r}{x^p}\Big)^{\frac{1}{rp}} \times \Big(\frac{x^p}{x^q}\Big)^{\frac{1}{pq}} = 1$.

To prove:

$\frac{a^{-1}}{a^{-1} + b^{-1}} + \frac{a^{-1}}{a^{-1} - b^{-1}} = \frac{2b^2}{b^2 - a^2}$

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

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

Since, L.H.S. = R.H.S. = $\frac{2b^2}{b^2 - a^2}$

Hence, proved that $\frac{a^{-1}}{a^{-1} + b^{-1}} + \frac{a^{-1}}{a^{-1} - b^{-1}} = \frac{2b^2}{b^2 - a^2}$.

To prove:

$\frac{a + b + c}{a^{-1} b^{-1} + b^{-1}c^{-1} + c^{-1}a^{-1}} = abc$

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

$$\begin{aligned}\Rightarrow \frac{a + b + c}{a^{-1} b^{-1} + b^{-1}c^{-1} + c^{-1}a^{-1}} = \frac{a + b + c}{\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}} \\[1em] = \frac{a + b + c}{\frac{c + a + b}{abc}} \\[1em] = \frac{abc(a + b + c)}{a + b + c} \\[1em] = abc.\end{aligned}$$

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

Hence, proved that $\frac{a + b + c}{a^{-1} b^{-1} + b^{-1}c^{-1} + c^{-1}a^{-1}} = abc$.

We know the algebraic identity:

(a - b) (a + b) = a² - b²

The given expression can also be written as,

(4 + 3) (4² + 3²) (4⁴ + 3⁴) (4⁸ + 3⁸) (4¹⁶ + 3¹⁶) (4³² + 3³²) = (4^x - 3^x)

To use our identity, we need a (4 - 3) term at the beginning. Since 4 - 3 = 1, multiplying the left side by (4 - 3) does not change its value:

∴ Given expression can be re-written as:

(4 - 3)(4 + 3) (4² + 3²) (4⁴ + 3⁴) (4⁸ + 3⁸) (4¹⁶ + 3¹⁶) (4³² + 3³²) = (4^x - 3^x)

From the algebraic identity,

(4 - 3) (4 + 3) = (4² - 3²)

(4² - 3²) (4² + 3²) = (4⁴ - 3⁴)

(4⁴ - 3⁴) (4⁴ + 3⁴) = (4⁸ - 3⁸)

(4⁸ - 3⁸) (4⁸ + 3⁸) = (4¹⁶ - 3¹⁶)

(4¹⁶ - 3¹⁶)(4¹⁶ + 3¹⁶) = (4³² - 3³²)

(4³² - 3³²)(4³² + 3³²) = (4⁶⁴ - 3⁶⁴)

4⁶⁴ - 3⁶⁴ = 4^x - 3^x

∴ x = 64

Hence, x = 64.