Indices

Given,

$$\begin{aligned}\Rightarrow \Big(-\frac{243}{32}\Big)^{-\frac{3}{5}}\\[1em] \Rightarrow \Big(-\frac{3^5}{2^5}\Big)^{-\frac{3}{5}}\\[1em] \Rightarrow \Big[\Big(-\frac{3}{2}\Big)^5\Big]^{-\frac{3}{5}}\\[1em] \Rightarrow \Big(-\frac{3}{2}\Big)^{-\frac{5 \times 3}{5}}\\[1em] \Rightarrow \Big(-\frac{3}{2}\Big)^{-3}\\[1em] \Rightarrow \Big(-\frac{2}{3}\Big)^3\\[1em] \Rightarrow -\frac{8}{27}.\end{aligned}$$

Hence, $\Big(-\frac{243}{32}\Big)^{-\frac{3}{5}} = -\frac{8}{27}$.

(ii) Given,

$$\begin{aligned}\Rightarrow \Big(5\frac{23}{64}\Big)^{-\frac{2}{3}}\\[1em] \Rightarrow \Big(\frac{343}{64}\Big)^{-\frac{2}{3}}\\[1em] \Rightarrow \Big(\frac{7^3}{4^3}\Big)^{-\frac{2}{3}}\\[1em] \Rightarrow \Big[\Big(\frac{7}{4}\Big)^3\Big]^{-\frac{2}{3}}\\[1em] \Rightarrow \Big(\frac{7}{4}\Big)^{-\frac{2 \times 3}{3}}\\[1em] \Rightarrow \Big(\frac{7}{4}\Big)^{-2}\\[1em] \Rightarrow \Big(\frac{4}{7}\Big)^2\\[1em] \Rightarrow \frac{16}{49}.\end{aligned}$$

Hence, $\Big(5\frac{23}{64}\Big)^{-\frac{2}{3}} = \frac{16}{49}$.

Given,

⇒ (2a^-3b²)³ = (2)³(a^-3)³(b²)³

= 8a^-9b⁶.

Hence, (2a^-3b²)³ = 8a^-9b⁶.

Given,

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

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

Given,

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

Hence, $\frac{x^{-1}y^{-1}}{x^{-1} + y^{-1}} = \frac{1}{x + y}$

Given,

$$\begin{aligned}\Rightarrow \frac{(4 \times 10^7)(6 \times 10^{-5})}{8 \times 10^{10}} = \frac{4 \times 6 \times 10^7 \times 10^{-5}}{8 \times 10^{10}} \\[1em] = \frac{24 \times 10^{7 + (-5)}}{8 \times 10^{10}} = \frac{24 \times 10^2}{8 \times 10^{10}} \\[1em] = 3 \times 10^2 \times 10^{-10} = 3 \times 10^{2 + (-10)} \\[1em] = 3 \times 10^{-8}.\end{aligned}$$

Hence, $\frac{(4 \times 10^7)(6 \times 10^{-5})}{8 \times 10^{10}} = 3 \times 10^{-8}.$

Given,

$$\begin{aligned}\Rightarrow \frac{3a}{b^{-1}} + \frac{2b}{a^{-1}} = \frac{3a}{\frac{1}{b}} + \frac{2b}{\frac{1}{a}}\\[1em] = 3ab + 2ab = 5ab.\end{aligned}$$

Hence, $\frac{3a}{b^{-1}} + \frac{2b}{a^{-1}}$ = 5ab.

Given,

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

Hence, $5^0 \times 4^{-1} + 8^{\frac{1}{3}} = 2\frac{1}{4}$.

Given,

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

Hence, $\Big(\frac{8}{125}\Big)^{-\frac{1}{3}} = 2\frac{1}{2}.$

Given,

$$\begin{aligned}\Rightarrow (0.027)^{-\frac{1}{3}} = [(0.3)^3]^{-\frac{1}{3}} \\[1em] = (0.3)^{-1} = \frac{1}{0.3} \\[1em] = \frac{10}{3} = 3\frac{1}{3}. \\[1em]\end{aligned}$$

Hence,$(0.027)^{-\frac{1}{3}} = 3\frac{1}{3}$.

Given,

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

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

Given,

$$\begin{aligned}\Rightarrow (64)^{-\frac{2}{3}} ÷ (9)^{-\frac{3}{2}} = \Big(\frac{1}{64}\Big)^{\frac{2}{3}} ÷ \Big(\frac{1}{9}\Big)^{\frac{3}{2}} \\[1em] = \Big(\frac{1}{2^6}\Big)^{\frac{2}{3}} ÷ \Big(\frac{1}{3^2}\Big)^{\frac{3}{2}} \\[1em] = \frac{(1)^{\frac{2}{3}}}{2^{6 \times \frac{2}{3}}} ÷ \frac{(1)^{\frac{3}{2}}}{3^{2 \times \frac{3}{2}}} \\[1em] = \frac{1}{2^4} ÷ \frac{1}{3^3} \\[1em] = \frac{1}{2^4} \times 3^3 \\[1em] = \frac{27}{16} = 1\frac{11}{16}.\end{aligned}$$

Hence, $(64)^{-\frac{2}{3}} ÷ (9)^{-\frac{3}{2}} = 1\frac{11}{16}$.

Given,

$$\begin{aligned}\Rightarrow \frac{(27)^{\frac{2n}{3}} \times (8)^{-\frac{n}{6}}}{(18)^{-\frac{n}{2}}} \\[1em] = \frac{(3^3)^{\frac{2n}{3}} \times (2^3)^{-\frac{n}{6}}}{(2 \times 3^2)^{-\frac{n}{2}}} \\[1em] = \frac{3^{2n} \times (2)^{-\frac{n}{2}}}{(2)^{-\frac{n}{2}} \times (3^2)^{-\frac{n}{2}}} \\[1em] = \frac{3^{2n}}{3^{-n}} = \frac{3^{2n}}{\frac{1}{3^n}} \\[1em] = 3^{2n} \times 3^n \\[1em] = 3^{(2n + n)} = 3^{3n}. \\[1em]\end{aligned}$$

Hence, $\frac{(27)^{\frac{2n}{3}} \times (8)^{-\frac{n}{6}}}{(18)^{-\frac{n}{2}}} = 3^{3n}$.

Given,

$$\begin{aligned}\Rightarrow \frac{5.(25)^{n + 1} - 25.(5)^{2n}}{5.(5)^{2n + 3} - (25)^{n + 1}} \\[1em] = \frac{5.(5^2)^{n + 1} - (5^2)(5)^{2n}}{5.(5)^{2n + 3} - (5^2)^{n + 1}} \\[1em] = \frac{5.5^{2n + 2} - 5^{2 + 2n}}{5^{2n + 4} - 5^{2n + 2}} \\[1em] = \frac{5^{2n + 3} - 5^{2n + 2}}{5^{2n + 4} - 5^{2n + 2}} \\[1em] = \frac{5^{2n}.5^3 - 5^{2n}.5^2}{5^{2n}.5^4 - 5^{2n}.5^2} \\[1em] = \frac{5^{2n}(5^3 - 5^2)}{5^{2n}(5^4 - 5^2)} \\[1em] = \frac{125 - 25}{625 - 25} \\[1em] = \frac{100}{600} = \frac{1}{6}.\end{aligned}$$

Hence, $\frac{5.(25)^{n + 1} - 25.(5)^{2n}}{5.(5)^{2n + 3} - (25)^{n + 1}} = \frac{1}{6}$.

Given,

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

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

Given,

$$\begin{aligned}\Rightarrow \Big(\frac{27}{8}\Big)^{\frac{2}{3}} - \Big(\frac{1}{4}\Big)^{-2} + 5^0 = \Big(\frac{3^3}{2^3}\Big)^{\frac{2}{3}} - 4^2 + 1 \\[1em] = \frac{3^{3 \times \frac{2}{3}}}{2^{3 \times \frac{2}{3}}} - 16 + 1 \\[1em] = \frac{3^2}{2^2} - 15 \\[1em] = \frac{9}{4} - 15 \\[1em] = \frac{9 - 60}{4} \\[1em] = -\frac{51}{4} = -12\frac{3}{4}.\end{aligned}$$

Hence, $\Big(\frac{27}{8}\Big)^{\frac{2}{3}} - \Big(\frac{1}{4}\Big)^{-2} + 5^0 = -12\frac{3}{4}.$

Given,

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

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

Given,

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

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

Given,

$$\begin{aligned}\Rightarrow (3^2)^0 + 3^{-4} \times 3^6 + \Big(\frac{1}{3}\Big)^{-2} \\[1em] = 9^0 + \frac{1}{3^4} \times 3^6 + (3)^2 \\[1em] = 1 + \frac{1}{3^4} \times 3^6 + 3^2 \\[1em] = 1 + 3^2 + 3^2 \\[1em] = 1 + 9 + 9 = 19.\end{aligned}$$

Hence, $(3^2)^0 + 3^{-4} \times 3^6 ÷ \Big(\frac{1}{3}\Big)^{-2}$ = 19.

Given,

$$\begin{aligned}\Rightarrow (3^2)^{\frac{5}{2}} - 3.1 - \Big(\frac{1}{9^2}\Big)^{-\frac{1}{2}} \\[1em] = 3^{2 \times \frac{5}{2}} - 3 - (9^2)^{\frac{1}{2}} \\[1em] = 3^5 - 3 - 9 \\[1em] = 243 - 12 \\[1em] = 231.\end{aligned}$$

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

Given,

$$\begin{aligned}\Rightarrow 16^{\frac{3}{4}} + 2\Big(\frac{1}{2}\Big)^{-1}(3)^0 = (2^4)^{\frac{3}{4}} + 2.(2)^1 .1 \\[1em] = 2^3 + 4 = 8 + 4 = 12.\end{aligned}$$

Hence, $16^{\frac{3}{4}} + 2\Big(\frac{1}{2}\Big)^{-1}(3)^0$ = 12.

Given,

$$\begin{aligned}\Rightarrow (81)^{\frac{3}{4}} - \Big(\frac{1}{32}\Big)^{-\frac{2}{5}} + (8)^{\frac{1}{3}}\Big(\frac{1}{2}\Big)^{-1}(2)^0 = (3^4)^{\frac{3}{4}} - (32)^{\frac{2}{5}} + (2^3)^{\frac{1}{3}}(2)^1.1 \\[1em] = 3^3 - (2^5)^{\frac{2}{5}} + 2.2 \\[1em] = 27 - 2^2 + 4 \\[1em] = 27 - 4 + 4 = 27.\end{aligned}$$

Hence, $(81)^{\frac{3}{4}} - \Big(\frac{1}{32}\Big)^{-\frac{2}{5}} + (8)^{\frac{1}{3}}\Big(\frac{1}{2}\Big)^{-1}(2)^0$ = 27.

Given,

$$\begin{aligned}\Rightarrow \Big(\frac{64}{125}\Big)^{-\frac{2}{3}} ÷ \frac{1}{\Big(\frac{256}{625}\Big)^{\frac{1}{4}}} + \Big(\frac{\sqrt{25}}{\sqrt[3]{64}}\Big)^0 \\[1em] = \Big(\frac{125}{64}\Big)^{\frac{2}{3}} \times \Big(\frac{256}{625}\Big)^{\frac{1}{4}} + 1 \\[1em] = \Big[\Big(\frac{5}{4}\Big)^3\Big]^{\frac{2}{3}} \times \Big[\Big(\frac{4}{5}\Big)^4\Big]^{\frac{1}{4}} + 1 \\[1em] = \Big(\frac{5}{4}\Big)^2 \times \frac{4}{5} + 1 \\[1em] = \frac{5}{4} + 1 \\[1em] = \frac{9}{4} = 2\frac{1}{4}.\end{aligned}$$

Hence, $\Big(\frac{64}{125}\Big)^{-\frac{2}{3}} ÷ \frac{1}{\Big(\frac{256}{625}\Big)^{\frac{1}{4}}} + \Big(\frac{\sqrt{25}}{\sqrt[3]{64}}\Big)^0 = 2\frac{1}{4}$

Given,

$$\begin{aligned}\Rightarrow \frac{5^{n + 3} - 6 \times 5^{n + 1}}{9 \times 5^n - 2^2 \times 5^n} = \frac{5^n.5^3 - 6.5^n.5^1}{5^n(9 - 2^2)} \\[1em] = \frac{5^n.(5^3 - 30)}{5^n(9 - 4)} \\[1em] = \frac{125 - 30}{5} \\[1em] = \frac{95}{5} = 19.\end{aligned}$$

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

Given,

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

Hence, $\Big[(64)^{-\frac{2}{3}}.2^{-2} ÷ 8^0\Big]^{-\frac{1}{2}} = \frac{1}{2}$.

Given,

$$\begin{aligned}\Rightarrow 3^n \times 9^{n + 1} ÷ (3^{n - 1} \times 9^{n - 1}) \\[1em] = 3^n \times 9^{n + 1} \times \frac{1}{3^{n - 1} \times 9^{n - 1}} \\[1em] = \frac{3^n \times 9^n.9^1}{3^n.3^{-1} \times 9^n.9^{-1}} \\[1em] = \frac{9}{3^{-1} \times 9^{-1}} \\[1em] = 9 \times 3 \times 9 \\[1em] = 243.\end{aligned}$$

Hence, $3^n \times 9^{n + 1} ÷ (3^{n - 1} \times 9^{n - 1})$ = 243.

Given,

$$\begin{aligned}\Rightarrow \frac{\sqrt{2^2} \times \sqrt[4]{256}}{\sqrt[3]{64}} - \Big(\frac{1}{2}\Big)^{-2} \\[1em] \Rightarrow \frac{\sqrt{2^2} \times \sqrt[4]{2^8}}{\sqrt[3]{2^6}} - (2)^{2} \\[1em] \Rightarrow \frac{(2^2)^{\frac{1}{2}} \times (2^8)^{\frac{1}{4}}}{(2^6)^{\frac{1}{3}}} - 4 \\[1em] \Rightarrow \frac{(2)^{2 \times \frac{1}{2}} \times (2)^{8 \times \frac{1}{4}}}{(2)^{6 \times \frac{1}{3}}} - 4 \\[1em] \Rightarrow \frac{2 \times 2^2}{2^2} - 4 \\[1em] \Rightarrow 2 - 4 \\[1em] \Rightarrow -2.\end{aligned}$$

Hence, $\frac{\sqrt{2^2} \times \sqrt[4]{256}}{\sqrt[3]{64}} - \Big(\frac{1}{2}\Big)^{-2}$ = -2.

Given,

$$\begin{aligned}\Rightarrow \frac{3^{-\frac{6}{7}} \times 4^{-\frac{3}{7}} \times 9^{\frac{3}{7}} \times 2^{\frac{6}{7}}}{2^2 + 2^0 + 2^{-2}} \\[1em] \Rightarrow \frac{3^{-\frac{6}{7}} \times (2^2)^{-\frac{3}{7}} \times (3^2)^{\frac{3}{7}} \times 2^{\frac{6}{7}}}{4 + 1 + \Big(\frac{1}{2}\Big)^{2}} \\[1em] \Rightarrow \frac{3^{-\frac{6}{7}} \times (2)^{-\frac{6}{7}} \times (3)^{\frac{6}{7}} \times 2^{\frac{6}{7}}}{4 + 1 + \Big(\frac{1}{4}\Big)} \\[1em] \Rightarrow \frac{3^{-\frac{6}{7} + \frac{6}{7}} \times (2)^{-\frac{6}{7} + \frac{6}{7}}}{\Big(\frac{16 + 4 + 1}{4}\Big)} \\[1em] \Rightarrow \frac{3^0 \times 2^0}{\frac{21}{4}} \\[1em] \Rightarrow \frac{1}{\frac{21}{4}} \\[1em] \Rightarrow \frac{4}{21}.\end{aligned}$$

Hence, $\frac{3^{-\frac{6}{7}} \times 4^{-\frac{3}{7}} \times 9^{\frac{3}{7}} \times 2^{\frac{6}{7}}}{2^2 + 2^0 + 2^{-2}} = \frac{4}{21}$.

Given,

$$\begin{aligned}\Rightarrow \frac{(32)^{\frac{2}{5}} \times (4)^{-\frac{1}{2}} \times (8)^{\frac{1}{3}}}{2^{-2} ÷ (64)^{-\frac{1}{3}}} \\[1em] \Rightarrow \frac{(2^5)^{\frac{2}{5}} \times (2^2)^{-\frac{1}{2}} \times (2^3)^{\frac{1}{3}}}{2^{-2} ÷ (2^6)^{-\frac{1}{3}}} \\[1em] \Rightarrow \frac{2^2 \times 2^{-1} \times 2^1}{2^{-2} ÷ 2^{-2}} \\[1em] \Rightarrow \frac{2^{2 - 1 + 1}}{\frac{2^{-2}}{2^{-2}}} \\[1em] \Rightarrow \frac{2^2}{1} = 2^2 = 4.\end{aligned}$$

Hence, $\frac{(32)^{\frac{2}{5}} \times (4)^{-\frac{1}{2}} \times (8)^{\frac{1}{3}}}{2^{-2} ÷ (64)^{-\frac{1}{3}}}$ = 4.

Given,

$$\begin{aligned}\Rightarrow \frac{5^{2(x + 6)} \times (25)^{-7 + 2x}}{(125)^{2x}} \\[1em] \Rightarrow \frac{5^{2x + 12} \times (5^2)^{-7 + 2x}}{(5^3)^{2x}} \\[1em] \Rightarrow \frac{5^{2x + 12} \times 5^{-14 + 4x}}{5^{6x}} \\[1em] \Rightarrow \frac{5^{2x + 12 + (-14 + 4x)}}{5^{6x}} \\[1em] \Rightarrow 5^{6x - 2}.5^{-6x} \\[1em] \Rightarrow 5^{6x - 2 + (-6x)} \\[1em] \Rightarrow 5^{-2} = \frac{1}{5^2} = \frac{1}{25}.\end{aligned}$$

Hence, $\frac{5^{2(x + 6)} \times (25)^{-7 + 2x}}{(125)^{2x}} = \frac{1}{25}$.

Given,

$$\begin{aligned}\Rightarrow \frac{7^{2n + 3} - (49)^{n + 2}}{((343)^{n + 1})^{\frac{2}{3}}} \\[1em] \Rightarrow \frac{7^{2n + 3} - (7^2)^{n + 2}}{((7^3)^{n + 1})^{\frac{2}{3}}} \\[1em] \Rightarrow \frac{7^{2n + 3} - 7^{2n + 4}}{7^{3 \times (n + 1) \times \frac{2}{3}}} \\[1em] \Rightarrow \frac{7^{2n}.7^3 - 7^{2n}.7^{4}}{7^{2(n + 1)}} \\[1em] \Rightarrow \frac{343.7^{2n} - 2401.7^{2n}}{7^{2n + 2}} \\[1em] \Rightarrow \frac{7^{2n}(343 - 2401)}{7^{2n}.7^2} \\[1em] \Rightarrow \frac{-2058}{49} = -42.\end{aligned}$$

Hence, $\frac{7^{2n + 3} - (49)^{n + 2}}{((343)^{n + 1})^{\frac{2}{3}}}$ = -42.

Given,

$$\begin{aligned}\Rightarrow (27)^{\frac{4}{3}} + (32)^{0.8} + (0.8)^{-1} \\[1em] \Rightarrow (3^3)^{\frac{4}{3}} + (2^5)^{\frac{8}{10}} + \Big(\frac{1}{0.8}\Big)^{1} \\[1em] \Rightarrow (3)^{3 \times \frac{4}{3}} + (2)^{5 \times \frac{8}{10}} + \Big(\frac{10}{8}\Big) \\[1em] \Rightarrow (3)^4 + 2^4 + \frac{5}{4} \\[1em] \Rightarrow 81 + 16 + \frac{5}{4} \\[1em] \Rightarrow 97 + \frac{5}{4} \\[1em] \Rightarrow \frac{388 + 5}{4} \\[1em] \Rightarrow \frac{393}{4} \\[1em] \Rightarrow 98.25\end{aligned}$$

Hence, $(27)^{\frac{4}{3}} + (32)^{0.8} + (0.8)^{-1}$ = 98.25

Given,

$$\begin{aligned}\Rightarrow (\sqrt{32} - \sqrt{5})^{\frac{1}{3}}(\sqrt{32} + \sqrt{5})^{\frac{1}{3}} \\[1em] \Rightarrow [(\sqrt{32} - \sqrt{5})(\sqrt{32} + \sqrt{5})]^{\frac{1}{3}}\end{aligned}$$

As (a - b)(a + b) = a² - b² we get,

$$\begin{aligned}\Rightarrow [(\sqrt{32})^2 - (\sqrt{5}^2)]^{\frac{1}{3}} \\[1em] \Rightarrow [32 - 5]^{\frac{1}{3}} \\[1em] \Rightarrow (27)^{\frac{1}{3}} \\[1em] \Rightarrow (3^3)^{\frac{1}{3}} \\[1em] \Rightarrow 3.\end{aligned}$$

Hence, $(\sqrt{32} - \sqrt{5})^{\frac{1}{3}}(\sqrt{32} + \sqrt{5})^{\frac{1}{3}}$ = 3.

The above equation can be written as,

$\Rightarrow \Big(x^{\frac{1}{3}} - x^{-\frac{1}{3}}\Big)\Big[\Big(x^{\frac{1}{3}}\Big)^2 + x^{\frac{1}{3}}.x^{-\frac{1}{3}} + \Big(x^{-\frac{1}{3}}\Big)^2\Big)$

We know that,

a³ - b³ = (a - b)(a² + ab + b²)

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

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

Given,

$$\begin{aligned}\Rightarrow \Big(\frac{x^m}{x^n}\Big)^l.\Big(\frac{x^n}{x^l}\Big)^m.\Big(\frac{x^l}{x^m}\Big)^n = (x^m.x^{-n})^l.(x^n.x^{-l})^m.(x^l.x^{-m})^n \\[1em] = (x^{m - n})^l.(x^{n - l})^m.(x^{l - m})^n \\[1em] = x^{ml - nl}.x^{nm - lm}.x^{ln - mn} \\[1em] = x^{ml - nl + nm - lm + ln - mn} \\[1em] = x^0 = 1.\end{aligned}$$

Hence, $\Big(\frac{x^m}{x^n}\Big)^l.\Big(\frac{x^n}{x^l}\Big)^m.\Big(\frac{x^l}{x^m}\Big)^n$ = 1.

Given,

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

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

Given,

$$\begin{aligned}\Rightarrow \sqrt[lm]{\frac{x^l}{x^m}}.\sqrt[mn]{\frac{x^m}{x^n}}.\sqrt[nl]{\frac{x^n}{x^l}} \\[1em] = \sqrt[lm]{x^{l - m}}.\sqrt[mn]{x^{m - n}}.\sqrt[nl]{x^{n - l}} \\[1em] = (x)^{\frac{l - m}{lm}}.(x )^{\frac{m - n}{mn}}.(x)^{\frac{n - l}{nl}} \\[1em] = (x)^{\frac{l - m}{lm} + \frac{m - n}{mn} + \frac{n - l}{nl}} \\[1em] = (x)^{\frac{(l - m)n + (m - n)l + (n - l)m}{lmn}} \\[1em] = (x)^{\frac{ln - mn + ml - nl + nm - lm}{lmn}} \\[1em] = (x)^0 \\[1em] = 1.\end{aligned}$$

Hence, $\sqrt[lm]{\frac{x^l}{x^m}}.\sqrt[mn]{\frac{x^m}{x^n}}.\sqrt[nl]{\frac{x^n}{x^l}}$ = 1.

Given,

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

By formula,

(x³ - y³) = (x - y)(x² + y² + xy) we get,

$$\begin{aligned}\Rightarrow (x)^{a^3 - b^3}.(x)^{b^3 - c^3}.(x)^{c^3 - a^3} \\[1em] \Rightarrow x^{a^3 - b^3 + b^3 - c^3 + c^3 - a^3} \\[1em] \Rightarrow x^{0} = 1.\end{aligned}$$

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

Given,

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

By formula,

(x³ + y³) = (x + y)(x² + y² - xy) we get,

$$\begin{aligned}\Rightarrow (x)^{a^3 + b^3}.(x)^{b^3 + c^3}.(x)^{c^3 + a^3} \\[1em] \Rightarrow x^{a^3 + b^3 + b^3 + c^3 + c^3 + a^3} \\[1em] \Rightarrow x^{2(a^3 + b^3 + c^3)}.\end{aligned}$$

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

Given,

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

Hence, (a^-1 + b^-1) ÷ (a^-2 - b^-2) = $\frac{ab}{b - a}.$

Given,

$$\begin{aligned}\Rightarrow \frac{1}{1 + a^{m - n}} + \frac{1}{1 + a^{n - m}} \\[1em] \Rightarrow \frac{1}{1 + a^m.a^{-n}} + \frac{1}{1 + a^n.a^{-m}} \\[1em] \Rightarrow \frac{1}{1 + \frac{a^m}{a^n}} + \frac{1}{1 + \frac{a^n}{a^m}} \\[1em] \Rightarrow \frac{1}{\frac{a^n + a^m}{a^n}} + \frac{1}{\frac{a^m + a^n}{a^m}} \\[1em] \Rightarrow \frac{a^n}{a^n + a^m} + \frac{a^m}{a^m + a^n} \\[1em] \Rightarrow \frac{a^n + a^m}{a^n + a^m} \\[1em] \Rightarrow 1.\end{aligned}$$

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

Given,

$\Rightarrow (a + b)^{-1}(a^{-1} + b^{-1}) = \frac{1}{ab}$

Solving L.H.S. of above equation,

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

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

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

Given,

$\frac{x + y + z}{x^{-1}y^{-1} + y^{-1}z^{-1} + z^{-1}x^{-1}} = xyz$

Solving L.H.S. of above equation,

$$\begin{aligned}\Rightarrow \frac{x + y + z}{\frac{1}{xy} + \frac{1}{yz} + \frac{1}{zx}} \\[1em] \Rightarrow \frac{x + y + z}{\frac{z + x + y}{xyz}} \\[1em] \Rightarrow \frac{xyz(x + y + z)}{x + y + z} \\[1em] \Rightarrow xyz.\end{aligned}$$

Hence, proved that $\frac{x + y + z}{x^{-1}y^{-1} + y^{-1}z^{-1} + z^{-1}x^{-1}}$ = xyz.

Given,

a = c^z, b = a^x and c = b^y

Substituting value of c in a = c^z we get,

⇒ a = (b^y)^z = b^yz

Substituting value of b in above equation we get,

⇒ a = (a^x)^yz
⇒ a = a^xyz

∴ xyz = 1.

Hence, proved that xyz = 1.

Given,

a = xy^{p - 1}, b = xy^{q - 1} and c = xy^{r - 1}.

Substituting value of a, b and c in L.H.S. of a^{q - r}.b^{r - p}.c^{p - q} = 1 we get,

⇒ (xy^{p - 1})^{q - r}.(xy^{q - 1})^{r - p}.(xy^{r - 1})^{p - q}

⇒ (xy)^{(p - 1)(q - r)}.(xy)^{(q - 1)(r - p)}.(xy)^{(r - 1)(p - q)}

⇒ (xy)^{(p - 1)(q - r) + (q - 1)(r - p) + (r - 1)(p - q)}

⇒ (xy)^{pq - pr - q + r + qr - qp - r + p + rp - rq - p + q}

⇒ (xy)^{p - p - q + q + r - r + pq - qp - pr + rp + qr - rq}

⇒ (xy)⁰

⇒ 1.

Hence,prove that a^{q - r}.b^{r - p}.c^{p - q} = 1.

Let 2^x = 3^y = 6^-z = k.

$\therefore 2^x = k \text{ or } 2 = k^{\frac{1}{x}}$ ......(i)

$\therefore 3^y = k \text{ or } 3 = k^{\frac{1}{y}}$ ......(ii)

$\therefore 6^{-z} = k \text{ or } 6 = k^{-\frac{1}{z}}$ ......(iii)

We know that,

2 × 3 = 6

Substituting value of 2, 3 and 6 from (i), (ii) and (iii) in above equation we get,

$$\begin{aligned}\Rightarrow k^{\frac{1}{x}} \times k^{\frac{1}{y}} = k^{-\frac{1}{z}} \\[1em] \Rightarrow k^{\frac{1}{x} + \frac{1}{y}} = k^{-\frac{1}{z}} \\[1em] \therefore \frac{1}{x} + \frac{1}{y} = -\frac{1}{z} \\[1em] \Rightarrow \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0.\end{aligned}$$

Hence, proved that $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$.

Let 2^x = 3^y = 12^z = k.

$\therefore 2^x = k \text{ or } 2 = k^{\frac{1}{x}}$ ......(i)

$\therefore 3^y = k \text{ or } 3 = k^{\frac{1}{y}}$ ......(ii)

$\therefore 12^{z} = k \text{ or } 12 = k^{\frac{1}{z}}$ ......(iii)

We know that,

2² × 3 = 12

Substituting value of 2, 3 and 12 from (i), (ii) and (iii) in above equation we get,

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

Hence, proved that $x = \frac{2yz}{y - z}.$

(3x²)⁰ = 1,

$(xy)^{-2} = \Big(\frac{1}{xy}\Big)^2 = \frac{1}{x^2y^2}$

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

(i) Given, a = 3 and b = -2.

Substituting values of a and b in a^a + b^b we get,

= 3³ + (-2)^-2

= 27 + $\Big(-\frac{1}{2}\Big)^2$

= 27 + $\frac{1}{4} = 27\frac{1}{4}$.

Hence, $a^a + b^b = 27\frac{1}{4}.$

(ii) Given, a = 3 and b = -2.

Substituting values of a and b in a^b + b^a we get,

$$\begin{aligned}= (3)^{-2} + (-2)^3 \\[1em] = \Big(\frac{1}{3}\Big)^2 + (-8) \\[1em] = \frac{1}{9} - 8 \\[1em] = \frac{1 - 72}{9} \\[1em] = -\frac{71}{9} = -7\frac{8}{9}.\end{aligned}$$

Hence, $a^a + b^b = -7\frac{8}{9}.$

Substituting value of x and y in $\sqrt{\frac{x}{y}}$ we get,

$$\begin{aligned}\Rightarrow \sqrt{\frac{x}{y}} = \sqrt{\frac{10^3 \times 0.0099}{10^{-2} \times 110}} \\[1em] = \sqrt{\frac{9.9}{10^{-2} \times 110}} \\[1em] = \sqrt{\frac{9.9 \times 10^2}{110}} \\[1em] = \sqrt{\frac{990}{110}} \\[1em] = \sqrt{9} = 3.\end{aligned}$$

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

Substituting value of x, y and z in $x^{\frac{1}{2}}.y^{-1}.z^{\frac{2}{3}}$ we get,

$$\begin{aligned}\Rightarrow x^{\frac{1}{2}}.y^{-1}.z^{\frac{2}{3}} = (9)^{\frac{1}{2}}.(2)^{-1}.(8)^{\frac{2}{3}} \\[1em] = (3^2)^{\frac{1}{2}} \times \frac{1}{2} \times (2^3)^{\frac{2}{3}} \\[1em] = 3 \times \frac{1}{2} \times 2^2 \\[1em] = \frac{12}{2} = 6.\end{aligned}$$

Hence, $x^{\frac{1}{2}}.y^{-1}.z^{\frac{2}{3}}$ = 6.

On prime factorization of 49392 we get,

49392 = 2⁴3²7³

∴ x⁴y²z³ = 2⁴3²7³

On comparing we get,

x = 2, y = 3 and z = 7.

Hence, x = 2, y = 3 and z = 7.

Given,

$$\begin{aligned}\Rightarrow \sqrt[3]{a^6b^{-4}} = a^x.b^{2y} \\[1em] \Rightarrow (a^6b^{-4})^{\frac{1}{3}} = a^x.b^{2y} \\[1em] \Rightarrow a^{\frac{6}{3}}.b^{-\frac{4}{3}} = a^x.b^{2y} \\[1em] \Rightarrow a^2.b^{-\frac{4}{3}} = a^x.b^{2y} \\[1em] \therefore x = 2 \text{ and } 2y = -\frac{4}{3} \\[1em] \Rightarrow x = 2 \text{ and } y = -\frac{2}{3}.\end{aligned}$$

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

Given,

$$\begin{aligned}\Rightarrow (p + q)^{-1}(p^{-1} + q^{-1}) = p^aq^b \\[1em] \Rightarrow \frac{1}{(p + q)}\Big(\frac{1}{p} + \frac{1}{q}\Big) = p^aq^b \\[1em] \Rightarrow \frac{1}{(p + q)} \times \frac{p + q}{pq} = p^aq^b \\[1em] \Rightarrow \frac{1}{pq} = p^aq^b \\[1em] \Rightarrow p^{-1}q^{-1} = p^aq^b \\[1em] \therefore a = -1 \text{ and } b = -1.\end{aligned}$$

Substituting values of a and b in L.H.S. of a + b + 2 = 0 we get,

a + b + 2 = -1 + (-1) + 2 = -2 + 2 = 0.

Hence, proved that a + b + 2 = 0.

Given,

$$\begin{aligned}\Rightarrow \Big(\frac{p^{-1}q^{2}}{p^2q^{-4}}\Big)^7 ÷ \Big(\frac{p^3q^{-5}}{p^{-2}q^{3}}\Big)^{-5} = p^xq^y \\[1em] \Rightarrow \Big(\frac{p^{-1}q^{2}}{p^2q^{-4}}\Big)^7 ÷ \Big(\frac{p^{-2}q^3}{p^3q^{-5}}\Big)^5 = p^xq^y \\[1em] \Rightarrow \frac{p^{-7}q^{14}}{p^{14}q^{-28}} \times \Big(\frac{p^3q^{-5}}{p^{-2}q^3}\Big)^5 = p^xq^y \\[1em] \Rightarrow \frac{q^{14}.q^{28}}{p^{14}.p^7} \times \frac{p^{15}q^{-25}}{p^{-10}q^{15}} = p^xq^y \\[1em] \Rightarrow \frac{q^{14 +28}}{p^{14 + 7}} \times \frac{p^{15}.p^{10}}{q^{25}.q^{15}} = p^xq^y \\[1em] \Rightarrow \frac{q^{42}}{p^{21}} \times \frac{p^{25}}{q^{40}} = p^xq^y \\[1em] \Rightarrow p^{4}q^{2} = p^xq^y \\[1em] \therefore x = 4 \text{ and } y = 2.\end{aligned}$$

x + y = 4 + 2 = 6.

Hence, x + y = 6.

Given,

⇒ 5^{2x + 3} = 1

⇒ 5^{2x + 3} = 5⁰

⇒ 2x + 3 = 0

⇒ 2x = -3

⇒ x = -$\frac{3}{2}$.

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

Given,

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

Hence, x = 4.

Given,

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

Hence, x = -7.

Given,

$$\begin{aligned}\Rightarrow (\sqrt[3]{4})^{2x + \frac{1}{2}} = \frac{1}{32} \\[1em] \Rightarrow [(2^2)^{\frac{1}{3}}]^{2x + \frac{1}{2}} = \frac{1}{2^5} \\[1em] \Rightarrow (2^2)^{\frac{2x}{3} + \frac{1}{6}} = 2^{-5} \\[1em] \Rightarrow (2)^{\frac{4x}{3} + \frac{1}{3}} = 2^{-5} \\[1em] \Rightarrow \frac{4x + 1}{3} = -5 \\[1em] \Rightarrow 4x + 1 = -15 \\[1em] \Rightarrow 4x = -16 \\[1em] \Rightarrow x = -4.\end{aligned}$$

Hence, x = -4.

Given,

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

Hence, x = $\frac{3}{4}$.

Given,

$$\begin{aligned}\Rightarrow 4^{x - 1} \times (0.5)^{3 - 2x} = \Big(\frac{1}{8}\Big)^x \\[1em] \Rightarrow (2^2)^{x - 1} \times \Big(\frac{5}{10}\Big)^{3 - 2x} = \Big(\frac{1}{2^3}\Big)^x \\[1em] \Rightarrow 2^{2x - 2} \times \Big(\frac{1}{2}\Big)^{3 - 2x} = (2^{-3})^x \\[1em] \Rightarrow 2^{2x - 2} \times (2^{-1})^{3 - 2x} = 2^{-3x} \\[1em] \Rightarrow 2^{2x - 2} \times 2^{2x - 3} = 2^{-3x} \\[1em] \Rightarrow 2^{2x - 2 + 2x - 3} = 2^{-3x} \\[1em] \Rightarrow 2^{4x - 5} = 2^{-3x} \\[1em] \Rightarrow 4x - 5 = -3x \\[1em] \Rightarrow 4x + 3x = 5 \\[1em] \Rightarrow 7x = 5 \\[1em] \Rightarrow x = \frac{5}{7}.\end{aligned}$$

Hence, x = $\frac{5}{7}$.

Given,

⇒ 5^3x = 125

⇒ 5^3x = 5³

⇒ 3x = 3

⇒ x = 1.

Given,

$$\begin{aligned}\Rightarrow 10^y = 0.001 \\[1em] \Rightarrow 10^y = \frac{1}{1000} \\[1em] \Rightarrow 10^y = \frac{1}{10^3} \\[1em] \Rightarrow 10^y = 10^{-3} \\[1em] \Rightarrow y = -3.\end{aligned}$$

Hence, x = 1 and y = -3.

Given,

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

Hence, proved that m = 1 + n.

Given,

⇒ 3^4x = 81^-1

⇒ 3^4x = (3⁴)^-1

⇒ 3^4x = (3^-4)

⇒ 4x = -4

⇒ x = -1.

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,

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

Hence, 2^-x.(16)^y = 1.

Given,

⇒ 3^{x + 1} = 9^{x - 2}

⇒ 3^{x + 1} = (3²)^{x - 2}

⇒ 3^{x + 1} = 3^{2(x - 2)}

⇒ 3^{x + 1} = 3^{2x - 4}

⇒ x + 1 = 2x - 4

⇒ 2x - x = 1 + 4

⇒ x = 5.

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

⇒ 2^{1 + x} = 2^{1 + 5} = 2⁶ = 64.

Hence, 2^{1 + x} = 64.

(i) Given,

⇒ 3(2^x + 1) - 2^{x + 2} + 5 = 0

⇒ 3.2^x + 3 - 2^x.2² + 5 = 0

⇒ 3.2^x - 4.2^x + 3 + 5 = 0

⇒ -2^x + 8 = 0

⇒ 2^x = 8

⇒ 2^x = 2³

⇒ x = 3.

Hence, x = 3.

(ii) Given,

⇒ 3^x = 9.3^y

⇒ 3^x = 3².3^y

⇒ 3^x = 3^{2 + y}

⇒ x = y + 2 ......(i)

Given,

⇒ 8.2^y = 4^x

⇒ 2³.2^y = (2²)^x

⇒ 2^{3 + y} = 2^2x

⇒ 2x = y + 3 .......(ii)

Subtracting (i) from (ii) we get,

⇒ 2x - x = (y + 3) - (y + 2)

⇒ x = 1.

Substituting value of x in (i) we get,

⇒ 1 = y + 2

⇒ y = -1.

Hence, x = 1 and y = -1.

Given,

$$\begin{aligned}\Rightarrow \Big(5\frac{1}{16}\Big)^{-\frac{3}{4}} = \Big(\frac{81}{16}\Big)^{-\frac{3}{4}} \\[1em] = \Big(\frac{16}{81}\Big)^{\frac{3}{4}} \\[1em] = \Big(\frac{2^4}{3^4}\Big)^{\frac{3}{4}} = \frac{2^{4 \times \frac{3}{4}}}{3^{4 \times \frac{3}{4}}} \\[1em] = \frac{2^3}{3^3} = \frac{8}{27}.\end{aligned}$$

Hence, Option 4 is the correct option.

Given,

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

Hence, Option 3 is the correct option.

Given,

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

Hence, Option 2 is the correct option.

Given,

$$\begin{aligned}\Rightarrow \sqrt[4]{(81)^{-2}} = \Big[\Big(\frac{1}{81}\Big)^{2}\Big]^{\frac{1}{4}} \\[1em] = \Big(\frac{1}{81}\Big)^{\frac{1}{2}} = \frac{1}{9}.\end{aligned}$$

Hence, Option 1 is the correct option.

Given,

$$\begin{aligned}\Rightarrow (256)^{0.16} \times (256)^{0.09} = (256)^{\frac{16}{100}} \times (256)^{\frac{9}{100}} \\[1em] = (256)^{\frac{16 + 9}{100}} = (256)^{\frac{25}{100}}\\[1em] = (256)^{\frac{1}{4}} = (4^4)^{\frac{1}{4}} \\[1em] = 4.\end{aligned}$$

Hence, Option 1 is the correct option.

1. $x^{\frac{12}{7}} - x^{\frac{5}{7}}$

2. $\sqrt[12]{(x^4)^{\frac{1}{3}}}$

3. $(\sqrt{x^3})^{\frac{2}{3}}$

4. $x^{\frac{12}{7}} \times x^{\frac{7}{12}}$

On solving $(\sqrt{x^3})^{\frac{2}{3}}$ we get,

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

Hence, Option 3 is the correct option.

According to statement 1 :

3^m + 2ⁿ = 5^{m + n}

Lets take some examples

Let m = 1, n = 1

Taking L.H.S.: 3^m + 2ⁿ

= 3¹ + 2¹

= 3 + 2

= 5

Taking R.H.S.: 5^{m + n}

= 5^{1 + 1}

= 5²

= 25

As L.H.S. ≠ R.H.S.

We can conclude that 3^m + 2ⁿ ≠ 5^{m + n}

∴ Statement 1 is false.

According to statement 2; a^m + bⁿ = (a + b)^{m + n}

Lets take some examples

Let a = 1, b = 1, m = 1, n = 1

Taking L.H.S.: a^m + bⁿ

= 1¹ + 1¹

= 1 + 1

= 2.

Taking R.H.S.: (a + b)^{m + n}

= (1 + 1)^{1 + 1}

= 2²

= 4.

As L.H.S. ≠ R.H.S.

We can conclude that a^m + bⁿ ≠ (a + b)^{m + n}

∴ Statement 2 is false.

∴ Both the statements are false.

Hence, option 2 is the correct option.

If p, q are different prime numbers, then p^m.qⁿ = p^l.q^k ⇒ m = l and n = k.

This statement is true. This is a direct consequence of the Fundamental Theorem of Arithmetic.

∴ Reason (R) is true.

Given, 2^x.3^y.5^z = 16200

⇒ 2^x.3^y.5^z = 2³.3⁴.5²

⇒ x = 3, y = 4 and z = 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.

According to Assertion: If x = 9 and y = 2, then x^y = y^x.

Taking L.H.S. = x^y

= 9²

= 81.

Taking R.H.S. = y^x

= 2⁹

= 512.

∴ L.H.S. ≠ R.H.S.

So, x^y ≠ y^x

∴ Assertion (A) is false.

a^-n = $\Big(\frac{1}{a}\Big)^n$ when a is a real positive number and n is a rational number.

This statement is a fundamental property of exponents. It correctly defines how to handle negative exponents. For example, 2^-3 = $\Big(\frac{1}{2}\Big)^3$

∴ Reason (R) is true.

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

Hence, option 2 is the correct option.

If a is a real positive number and n is positive integer then $\sqrt[n]{a}$ is also written as $a^\frac{1}{n}$.

This statement is a fundamental property of exponents and radicals. It defines the relationship between roots and fractional exponents. For example, $\sqrt{a} = a^\frac{1}{2}, \sqrt[3]{a} = a^\frac{1}{3}$.

∴ Reason (R) is true.

Given,

$$\begin{aligned}\Rightarrow 3^x = 9 \sqrt{3}\\[1em] \Rightarrow 3^x = 3^2 . 3^\frac{1}{2}\\[1em] \Rightarrow 3^x = (3)^{2 + \frac{1}{2}}\\[1em] \Rightarrow 3^x = (3)^{\frac{4 + 1}{2}}\\[1em] \Rightarrow 3^x = (3)^{\frac{5}{2}}\\[1em] \Rightarrow x = \frac{5}{2}.\end{aligned}$$

∴ 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.

2160 = 2⁴.3³.5¹.

∴ 2⁴.3³.5¹ = 2^x.3^y.5^z.

∴ x = 4, y = 3 and z = 1.

Substituting values of x, y and z in 3^x.2^-y.5^-z

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

Hence, x = 4, y = 3 and z = 1 and 3^x.2^-y.5^-z = $2\frac{1}{40}.$

(i) x^x + y^y = 2² + (-3)^-3

= 4 + $\Big(-\frac{1}{3}\Big)^3$

= 4 + $-\frac{1}{27}$

= 4 - $\frac{1}{27}$

= $\frac{108 - 1}{27} = \frac{107}{27} = 3\frac{26}{27}$.

Hence, x^x + y^y = $3\frac{26}{27}$.

(ii) x^y + y^x

$$\begin{aligned}x^y + y^x = 2^{-3} + (-3)^{2} \\[1em] = \Big(\frac{1}{2}\Big)^3 + 9 \\[1em] = \frac{1}{8} + 9 \\[1em] = \frac{1 + 72}{8} \\[1em] = \frac{73}{8} = 9\frac{1}{8}.\end{aligned}$$

Hence, x^y + y^x = $9\frac{1}{8}$.

Substituting value of p, q and r in p^{m - n}.q^{n - l}.r^{l - m} = 1.

= (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.

Hence, proved that, p^{m - n}.q^{n - l}.r^{l - m} = 1.

Substituting values of x, y and z in L.H.S. of x^myⁿz^l = xⁿy^lz^m we get,

x^myⁿz^l = (a^{m + n})^m.(a^{n + l})ⁿ.(a^{l + m})^l

= a^{m2 + mn}.a^{n2 + ln}.a^{lm + l2}

= a^{m2 + n2 + l2 + mn + lm + ln} .......(i)

Substituting values of x, y and z in R.H.S. of x^myⁿz^l = xⁿy^lz^m we get,

xⁿy^lz^m = (a^{m + n})ⁿ.(a^{n + l})^l.(a^{l + m})^m

= a^{mn + n2}.a^{nl + l2}.a^{lm + m2}

= a^{m2 + n2 + l2 + mn + lm + ln} .........(ii)

Since (i) = (ii),

Hence, proved that x^myⁿz^l = xⁿy^lz^m.

Given,

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

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

Given,

$$\begin{aligned}\Rightarrow \frac{(x^{(a + b)})^2(x^{(b + c)})^2(x^{(c + a)})^2}{(x^ax^bx^c)^4} = \frac{x^{2a + 2b}.x^{2b +2c}.x^{2c + 2a}}{x^{(a + b + c)4}} \\[1em] = \frac{x^{2a + 2b + 2b + 2c + 2c + 2a}}{x^{4a + 4b + 4c}} \\[1em] = \frac{x^{4a + 4b + 4c}}{x^{4a + 4b + 4c}} \\[1em] = 1.\end{aligned}$$

Hence, $\frac{(x^{(a + b)})^2(x^{(b + c)})^2(x^{(c + a)})^2}{(x^ax^bx^c)^4}$ = 1.

Given,

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

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

Given,

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

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

Given,

$\frac{1}{1 + a^{y - x} + a^{z - x}} + \frac{1}{1 + a^{z - y} + a^{x - y}} + \frac{1}{1 + a^{x - z} + a^{y - z}}$ = 1.

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

$$\begin{aligned}\Rightarrow \frac{1}{1 + a^{y - x} + a^{z - x}} + \frac{1}{1 + a^{z - y} + a^{x - y}} + \frac{1}{1 + a^{x - z} + a^{y - z}} \\[1em] \Rightarrow \frac{1}{1 + a^ya^{-x} + a^{z}.a^{-x}} + \frac{1}{1 + a^z.a^{-y} + a^x.a^{-y}} + \frac{1}{1 + a^x.a^{-z} + a^y.a^{-z}} \\[1em] \Rightarrow \frac{1}{1 + \frac{a^y}{a^x} + \frac{a^z}{a^x}} + \frac{1}{1 + \frac{a^z}{a^y} + \frac{a^x}{a^y}} + \frac{1}{1 + \frac{a^x}{a^z} + \frac{a^y}{a^z}} \\[1em] \Rightarrow \frac{1}{\frac{a^x + a^y + a^z}{a^x}} + \frac{1}{\frac{a^y + a^z + a^x}{a^y}} + \frac{1}{\frac{a^z + a^x + a^y}{a^z}} \\[1em] \Rightarrow \frac{a^x}{a^x + a^y + a^z} + \frac{a^y}{a^x + a^y + a^z} + \frac{a^z}{a^x + a^y + a^z} \\[1em] \Rightarrow \frac{a^x + a^y + a^z}{a^x + a^y + a^z} \\[1em] \Rightarrow 1.\end{aligned}$$

Hence, proved that,

$\frac{1}{1 + a^{y - x} + a^{z - x}} + \frac{1}{1 + a^{z - y} + a^{x - y}} + \frac{1}{1 + a^{x - z} + a^{y - z}}$ = 1.

Let (3)^x = (5)^y = (75)^z = k

∴ (3)^x = k

⇒ 3 = $k^{\frac{1}{x}}$ ......(i)

∴ (5)^y = k

⇒ 5 = $k^{\frac{1}{y}}$ .......(ii)

∴ (75)^z = k

⇒ 75 = $k^{\frac{1}{z}}$

⇒ 3.5² = $k^{\frac{1}{z}}$

Substituting value of x and y from (i) and (ii) in above equation we get,

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

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

(i) Solving,

$$\begin{aligned}\Rightarrow \sqrt{3^{x + 1}} = 27 \\[1em] \Rightarrow \sqrt{3^{x + 1}} = 3^3 \\[1em] \Rightarrow \sqrt{3^{x + 1}} = \sqrt{3^6} \\[1em] \Rightarrow x + 1 = 6 \\[1em] \Rightarrow x = 6 - 1 = 5.\end{aligned}$$

Hence, x = 5.

Given,

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

Similarly,

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

Hence, x = $\frac{3}{4} \text{ and } y = -\frac{8}{3}$.

Given,

⇒ 3^{x - 1} × 5^{2y - 3} = 225

⇒ 3^{x - 1} × 5^{2y - 3} = 3².5²

∴ x - 1 = 2 and 2y - 3 = 2

⇒ x = 2 + 1 and 2y = 2 + 3

⇒ x = 3 and y = $\frac{5}{2}$.

Hence, x = 3 and y = $\frac{5}{2}$.

Given,

⇒ 8^{x + 1} = 16^{y + 2}

⇒ (2³)^{x + 1} = (2⁴)^{y + 2}

⇒ 2^{3x + 3} = 2^{4y + 8}

⇒ 3x + 3 = 4y + 8

⇒ 3x - 4y = 5 .......(i)

Given,

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

Multiplying (ii) by 3 we get,

⇒ 18y - 3x = 9 .......(iii)

Adding (i) and (iii) we get,

⇒ 3x - 4y + (18y - 3x) = 5 + 9
⇒ 14y = 14
⇒ y = 1.

Substituting value of y in (i) we get,

⇒ 3x - 4(1) = 5
⇒ 3x - 4 = 5
⇒ 3x = 9
⇒ x = 3.

Hence, x = 3 and y = 1.