Exponents and Powers

We know that,

For any non-zero rational number ($a$)

$a^m \times a^n = a^{m + n}$

Solving,

$$\begin{aligned}\Rightarrow a^6 \times a^8\\[1em] \Rightarrow a^{6+8}\\[1em] \Rightarrow a^{14}\end{aligned}$$

Hence, $\bm{a^6 \times a^8 = a^{14}}$.

We know that,

For any non-zero rational number ($a$)

$a^m \times a^n = a^{m+n}$

Solving,

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

Hence, $\bm{x^5 \times x^{-3} = x^{2}}$.

We know that,

For any non-zero rational number ($a$)

$a^m \times a^n = a^{m+n}$

Solving,

$$\begin{aligned}\Rightarrow z^9 \times z^3 \times z^{-6}\\[1em] \Rightarrow z^{9+3+(-6)}\\[1em] \Rightarrow z^{6}\end{aligned}$$

Hence, $\bm{z^9 \times z^3 \times z^{-6} = z^{6}}$.

We know that,

For any non-zero rational number ($a$)

$a^m \times a^n = a^{m+n}$

Solving,

$$\begin{aligned}\Rightarrow a^2b^3 \times a^5b^2\\[1em] \Rightarrow a^{2+5} \times b^{3+2}\\[1em] \Rightarrow a^{7}b^{5}\end{aligned}$$

Hence, $\bm{a^2b^3 \times a^5b^2 = a^{7}b^{5}}$.

We know that,

For any non-zero rational number ($a$)

$a^m \times a^n = a^{m+n}$

Solving,

$$\begin{aligned}\Rightarrow 5x^7 \times 3x^4\\[1em] \Rightarrow (5 \times 3) \times x^{7+4}\\[1em] \Rightarrow 15x^{11}\end{aligned}$$

Hence, $\bm{5x^7 \times 3x^4 = 15x^{11}}$.

We know that,

For any non-zero rational number ($a$)

$a^m \times a^n = a^{m+n}$ and $a^{-n} = \frac{1}{a^n}$

Solving,

$$\begin{aligned}\Rightarrow p^3q^4 \times p^5q^{-5}\\[1em] \Rightarrow p^{3+5} \times q^{4+(-5)}\\[1em] \Rightarrow p^{8}q^{-1}\\[1em] \Rightarrow \frac{p^8}{q}\end{aligned}$$

Hence, $\bm{p^3q^4 \times p^5q^{-5} = \frac{p^8}{q}}$.

We know that,

For any non-zero rational number ($a$)

$a^m \times a^n = a^{m+n}$ and $a^{-n} = \frac{1}{a^n}$

Solving,

$$\begin{aligned}\Rightarrow x^7y^{-5} \times x^{-5}y^3\\[1em] \Rightarrow x^{7+(-5)} \times y^{-5+3}\\[1em] \Rightarrow x^{2}y^{-2}\\[1em] \Rightarrow \frac{x^2}{y^2}\end{aligned}$$

Hence, $\bm{x^7y^{-5} \times x^{-5}y^3 = \frac{x^2}{y^2}}$.

We know that,

For any non-zero rational number ($a$)

$a^m \times a^n = a^{m+n}, a^0 = 1$ and $a^{-n} = \frac{1}{a^n}$

Solving,

$$\begin{aligned}\Rightarrow x^{-2}y^5 \times x^0y^{-7}\\[1em] \Rightarrow x^{-2+0} \times y^{5+(-7)}\\[1em] \Rightarrow x^{-2}y^{-2}\\[1em] \Rightarrow \frac{1}{x^2y^2}\end{aligned}$$

Hence, $\bm{x^{-2}y^5 \times x^0y^{-7} = \frac{1}{x^2y^2}}$.

We know that,

For any non-zero rational number ($a$)

$a^m \times a^n = a^{m+n}$ and $a^{-n} = \frac{1}{a^n}$

Solving,

$$\begin{aligned}\Rightarrow x^6y^4z^{-2} \times x^{-3}y^{-5}z^{-1} \times x^2z^4\\[1em] \Rightarrow x^{6+(-3)+2} \times y^{4+(-5)} \times z^{-2+(-1)+4}\\[1em] \Rightarrow x^{5}y^{-1}z^{1}\\[1em] \Rightarrow \frac{x^5z}{y}\end{aligned}$$

Hence, $\bm{x^6y^4z^{-2} \times x^{-3}y^{-5}z^{-1} \times x^2z^4 = \frac{x^5z}{y}}$.

We know that,

For any non-zero rational number ($a$)

$\frac{a^m}{a^n} = a^{m-n}$

Solving,

$$\begin{aligned}\Rightarrow \frac{x^{12}}{x^7}\\[1em] \Rightarrow x^{12-7}\\[1em] \Rightarrow x^{5}\end{aligned}$$

Hence, $\bm{\frac{x^{12}}{x^7} = x^{5}}$.

We know that,

For any non-zero rational number ($a$)

$\frac{a^m}{a^n} = a^{m-n}$

Solving,

$$\begin{aligned}\Rightarrow \frac{z^6}{z^{-3}}\\[1em] \Rightarrow z^{6-(-3)}\\[1em] \Rightarrow z^{9}\end{aligned}$$

Hence, $\bm{\frac{z^6}{z^{-3}} = z^{9}}$.

We know that,

For any non-zero rational number ($a$)

$\frac{a^m}{a^n} = a^{m-n}$

Solving,

$$\begin{aligned}\Rightarrow \frac{m^5n^3}{m^2n^{-4}}\\[1em] \Rightarrow m^{5-2} \times n^{3-(-4)}\\[1em] \Rightarrow m^{3}n^{7}\end{aligned}$$

Hence, $\bm{\frac{m^5n^3}{m^2n^{-4}} = m^{3}n^{7}}$.

We know that,

For any non-zero rational number ($a$)

$\frac{a^m}{a^n} = a^{m-n}$

Solving,

$$\begin{aligned}\Rightarrow 18x^9 \div 6x^7\\[1em] \Rightarrow \frac{18x^9}{6x^7}\\[1em] \Rightarrow \frac{18}{6} \times x^{9-7}\\[1em] \Rightarrow 3x^{2}\end{aligned}$$

Hence, $\bm{18x^9 \div 6x^7 = 3x^{2}}$.

We know that,

For any non-zero rational number ($a$)

$\frac{a^m}{a^n} = a^{m-n}$ and $a^{-n} = \frac{1}{a^n}$

Solving,

$$\begin{aligned}\Rightarrow 7a^{12} \div 56a^{15}\\[1em] \Rightarrow \frac{7a^{12}}{56a^{15}}\\[1em] \Rightarrow \frac{7}{56} \times a^{12-15}\\[1em] \Rightarrow \frac{1}{8} \times a^{-3}\\[1em] \Rightarrow \frac{1}{8a^3}\end{aligned}$$

Hence, $\bm{7a^{12} \div 56a^{15} = \frac{1}{8a^3}}$.

We know that,

For any non-zero rational number ($a$)

$\frac{a^m}{a^n} = a^{m-n}$

Solving,

$$\begin{aligned}\Rightarrow a^{13}b^7 \div a^5b^{-3}\\[1em] \Rightarrow \frac{a^{13}b^7}{a^5b^{-3}}\\[1em] \Rightarrow a^{13-5} \times b^{7-(-3)}\\[1em] \Rightarrow a^{8}b^{10}\end{aligned}$$

Hence, $\bm{a^{13}b^7 \div a^5b^{-3} = a^{8}b^{10}}$.

We know that,

For any non-zero rational number ($a$)

$\frac{a^m}{a^n} = a^{m-n}$

Solving,

$$\begin{aligned}\Rightarrow 7x^{14} \div 21x^{-10}\\[1em] \Rightarrow \frac{7x^{14}}{21x^{-10}}\\[1em] \Rightarrow \frac{7}{21} \times x^{14-(-10)}\\[1em] \Rightarrow \frac{1}{3} \times x^{24}\\[1em] \Rightarrow \frac{x^{24}}{3}\end{aligned}$$

Hence, $\bm{7x^{14} \div 21x^{-10} = \frac{x^{24}}{3}}$.

We know that,

For any non-zero rational number ($a$)

$\frac{a^m}{a^n} = a^{m-n}$ and $a^0 = 1$

Solving,

$$\begin{aligned}\Rightarrow p^{11} \div p^{11}\\[1em] \Rightarrow \frac{p^{11}}{p^{11}}\\[1em] \Rightarrow p^{11-11}\\[1em] \Rightarrow p^{0}\\[1em] \Rightarrow 1\end{aligned}$$

Hence, $\bm{p^{11} \div p^{11} = 1}$.

We know that,

For any non-zero rational number ($a$)

$\frac{a^m}{a^n} = a^{m-n}$ and $a^{-n} = \frac{1}{a^n}$

Solving,

$$\begin{aligned}\Rightarrow \frac{a^7b^5c^4}{a^{-2}b^3c^6}\\[1em] \Rightarrow a^{7-(-2)} \times b^{5-3} \times c^{4-6}\\[1em] \Rightarrow a^{9}b^{2}c^{-2}\\[1em] \Rightarrow \frac{a^9b^2}{c^2}\end{aligned}$$

Hence, $\bm{\frac{a^7b^5c^4}{a^{-2}b^3c^6} = \frac{a^9b^2}{c^2}}$.

We know that,

For any non-zero rational number ($a$)

$(a^m)^n = a^{mn}$

Solving,

$$\begin{aligned}\Rightarrow (a^3)^2\\[1em] \Rightarrow a^{3 \times 2}\\[1em] \Rightarrow a^{6}\end{aligned}$$

Hence, $\bm{(a^3)^2 = a^{6}}$.

We know that,

For any non-zero rational numbers ($a$ and $b$)

$(ab)^m = a^mb^m$ and $(a^m)^n = a^{mn}$

Solving,

$$\begin{aligned}\Rightarrow (2x^2y)^4\\[1em] \Rightarrow 2^4 \times x^{2 \times 4} \times y^4\\[1em] \Rightarrow 16x^{8}y^{4}\end{aligned}$$

Hence, $\bm{(2x^2y)^4 = 16x^{8}y^{4}}$.

We know that,

For any non-zero rational numbers ($a$ and $b$)

$(ab)^m = a^mb^m, (a^m)^n = a^{mn}$ and $a^{-n} = \frac{1}{a^n}$

Solving,

$$\begin{aligned}\Rightarrow (m^2n^{-3})^4\\[1em] \Rightarrow m^{2 \times 4} \times n^{-3 \times 4}\\[1em] \Rightarrow m^{8}n^{-12}\\[1em] \Rightarrow \frac{m^8}{n^{12}}\end{aligned}$$

Hence, $\bm{(m^2n^{-3})^4 = \frac{m^8}{n^{12}}}$.

We know that,

For any non-zero rational numbers ($a$ and $b$)

$(ab)^m = a^mb^m, (a^m)^n = a^{mn}$ and $a^{-n} = \frac{1}{a^n}$

Solving,

$$\begin{aligned}\Rightarrow (x^2y^{-3})^{-2}\\[1em] \Rightarrow x^{2 \times (-2)} \times y^{-3 \times (-2)}\\[1em] \Rightarrow x^{-4}y^{6}\\[1em] \Rightarrow \frac{y^6}{x^4}\end{aligned}$$

Hence, $\bm{(x^2y^{-3})^{-2} = \frac{y^6}{x^4}}$.

We know that,

For any non-zero rational numbers ($a$ and $b$)

$(ab)^m = a^mb^m, (a^m)^n = a^{mn}$ and $a^{-n} = \frac{1}{a^n}$

Solving,

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

Hence, $\bm{(3x^3y^{-3})^2 = \frac{9x^6}{y^6}}$.

We know that,

For any non-zero rational numbers ($a$ and $b$)

$(ab)^m = a^mb^m, (a^m)^n = a^{mn}$ and $a^{-n} = \frac{1}{a^n}$

Solving,

$$\begin{aligned}\Rightarrow (5m^4n^{-3})^3\\[1em] \Rightarrow 5^3 \times m^{4 \times 3} \times n^{-3 \times 3}\\[1em] \Rightarrow 125m^{12}n^{-9}\\[1em] \Rightarrow \frac{125m^{12}}{n^9}\end{aligned}$$

Hence, $\bm{(5m^4n^{-3})^3 = \frac{125m^{12}}{n^9}}$.

We know that,

For any non-zero rational number ($a$)

$a^{\frac{m}{n}} = (a^m)^{\frac{1}{n}}$ and $(a^m)^n = a^{mn}$

We know that 36 = 6²

Solving,

$$\begin{aligned}\Rightarrow (36)^{\frac{1}{2}}\\[1em] \Rightarrow (6^2)^{\frac{1}{2}}\\[1em] \Rightarrow 6^{2 \times \frac{1}{2}}\\[1em] \Rightarrow 6^1\\[1em] \Rightarrow 6\end{aligned}$$

Hence, $\mathbf{(36)^{\frac{1}{2}} = 6}$.

We know that,

For any non-zero rational number ($a$)

$a^{\frac{m}{n}} = (a^m)^{\frac{1}{n}}$ and $(a^m)^n = a^{mn}$

We know that 27 = 3³

Solving,

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

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

We know that,

For any non-zero rational number ($a$)

$(a^m)^n = a^{mn}$ and $a^{-n} = \frac{1}{a^n}$

We know that 16 = 2⁴

Solving,

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

Hence, $\mathbf{(16)^{\frac{-3}{4}} = \frac{1}{8}}$.

We know that,

For any non-zero rational number ($a$)

$(a^m)^n = a^{mn}$ and $a^{-n} = \frac{1}{a^n}$

We know that 64 = 4³

Solving,

$$\begin{aligned}\Rightarrow (64)^{\frac{-1}{3}}\\[1em] \Rightarrow (4^3)^{\frac{-1}{3}}\\[1em] \Rightarrow 4^{3 \times \frac{-1}{3}}\\[1em] \Rightarrow 4^{-1}\\[1em] \Rightarrow \frac{1}{4}\end{aligned}$$

Hence, $\mathbf{(64)^{\frac{-1}{3}} = \frac{1}{4}}$.

We know that,

For any non-zero rational number ($a$)

$(a^m)^n = a^{mn}$ and $a^{-n} = \frac{1}{a^n}$

We know that 81 = 3⁴

Solving,

$$\begin{aligned}\Rightarrow (81)^{\frac{-1}{4}}\\[1em] \Rightarrow (3^4)^{\frac{-1}{4}}\\[1em] \Rightarrow 3^{4 \times \frac{-1}{4}}\\[1em] \Rightarrow 3^{-1}\\[1em] \Rightarrow \frac{1}{3}\end{aligned}$$

Hence, $\mathbf{(81)^{\frac{-1}{4}} = \frac{1}{3}}$.

We know that,

For any non-zero rational number ($a$)

$(a^m)^n = a^{mn}$ and $a^{-n} = \frac{1}{a^n}$

We know that 32 = 2⁵

Solving,

$$\begin{aligned}\Rightarrow (32)^{\frac{-4}{5}}\\[1em] \Rightarrow (2^5)^{\frac{-4}{5}}\\[1em] \Rightarrow 2^{5 \times \frac{-4}{5}}\\[1em] \Rightarrow 2^{-4}\\[1em] \Rightarrow \frac{1}{2^4}\\[1em] \Rightarrow \frac{1}{16}\end{aligned}$$

Hence, $\mathbf{(32)^{\frac{-4}{5}} = \frac{1}{16}}$.

We know that,

For any non-zero rational numbers ($a$ and $b$)

$(ab)^m = a^mb^m$ and $(a^m)^n = a^{mn}$

We know that 25 = 5²

Solving,

$$\begin{aligned}\Rightarrow (25a^2)^{\frac{1}{2}}\\[1em] \Rightarrow (5^2)^{\frac{1}{2}} \times a^{2 \times \frac{1}{2}}\\[1em] \Rightarrow 5^1 \times a^1\\[1em] \Rightarrow 5a\end{aligned}$$

Hence, $\bm{(25a^2)^{\frac{1}{2}} = 5a}$.

We know that,

For any non-zero rational numbers ($a$ and $b$)

$(ab)^m = a^mb^m, (a^m)^n = a^{mn}$ and $a^{-n} = \frac{1}{a^n}$

We know that 27 = 3³

Solving,

$$\begin{aligned}\Rightarrow (27x^{-3})^{\frac{1}{3}}\\[1em] \Rightarrow (3^3)^{\frac{1}{3}} \times x^{-3 \times \frac{1}{3}}\\[1em] \Rightarrow 3^1 \times x^{-1}\\[1em] \Rightarrow \frac{3}{x}\end{aligned}$$

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

We know that,

For any non-zero rational numbers ($a$ and $b$)

$(ab)^m = a^mb^m, (a^m)^n = a^{mn}$ and $a^{-n} = \frac{1}{a^n}$

We know that 64 = 4³

Solving,

$$\begin{aligned}\Rightarrow (64m^{-6}n^3)^{\frac{2}{3}}\\[1em] \Rightarrow (4^3)^{\frac{2}{3}} \times m^{-6 \times \frac{2}{3}} \times n^{3 \times \frac{2}{3}}\\[1em] \Rightarrow 4^2 \times m^{-4} \times n^2\\[1em] \Rightarrow 16m^{-4}n^2\\[1em] \Rightarrow \frac{16n^2}{m^4}\end{aligned}$$

Hence, $\bm{(64m^{-6}n^3)^{\frac{2}{3}} = \frac{16n^2}{m^4}}$.

We know that,

For any non-zero rational numbers ($a$ and $b$)

$(ab)^m = a^mb^m, (a^m)^n = a^{mn}$ and $a^{-n} = \frac{1}{a^n}$

We know that 81 = 3⁴

Solving,

$$\begin{aligned}\Rightarrow (81a^4b^8c^{-4})^{\frac{1}{4}}\\[1em] \Rightarrow (3^4)^{\frac{1}{4}} \times a^{4 \times \frac{1}{4}} \times b^{8 \times \frac{1}{4}} \times c^{-4 \times \frac{1}{4}}\\[1em] \Rightarrow 3^1 \times a^1 \times b^2 \times c^{-1}\\[1em] \Rightarrow \frac{3ab^2}{c}\end{aligned}$$

Hence, $\bm{(81a^4b^8c^{-4})^{\frac{1}{4}} = \frac{3ab^2}{c}}$.

We know that,

For any non-zero rational numbers ($a$ and $b$)

$(ab)^m = a^mb^m, (a^m)^n = a^{mn}$ and $a^{-n} = \frac{1}{a^n}$

Solving,

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

Hence, $\bm{(3x^{-3}y^3)^{-2} = \frac{x^6}{9y^6}}$.

We know that,

For any non-zero rational numbers ($a$ and $b$)

$(ab)^m = a^mb^m, (a^m)^n = a^{mn}$ and $a^{-n} = \frac{1}{a^n}$

Solving,

$$\begin{aligned}\Rightarrow (6ab^2c^{-3})^{-1}\\[1em] \Rightarrow 6^{-1} \times a^{-1} \times b^{2 \times (-1)} \times c^{-3 \times (-1)}\\[1em] \Rightarrow \frac{1}{6} \times a^{-1} \times b^{-2} \times c^3\\[1em] \Rightarrow \frac{c^3}{6ab^2}\end{aligned}$$

Hence, $\bm{(6ab^2c^{-3})^{-1} = \frac{c^3}{6ab^2}}$.

We know that,

For any non-zero rational numbers ($a$ and $b$)

$(ab)^m = a^mb^m, (a^m)^n = a^{mn}$ and $a^{-n} = \frac{1}{a^n}$

Solving,

$$\begin{aligned}\Rightarrow \left(-3a^{3/4}b^{-1/4}\right)^4\\[1em] \Rightarrow (-3)^4 \times a^{(3/4) \times 4} \times b^{(-1/4) \times 4}\\[1em] \Rightarrow 81 \times a^3 \times b^{-1}\\[1em] \Rightarrow \frac{81a^3}{b}\end{aligned}$$

Hence, $\bm{\left(-3a^{3/4}b^{-1/4}\right)^4 = \frac{81a^3}{b}}$.

We know that,

For any non-zero rational numbers ($a$ and $b$)

$(ab)^m = a^mb^m, (a^m)^n = a^{mn}$ and $a^{-n} = \frac{1}{a^n}$

We know that 32 = 2⁵

Solving,

$$\begin{aligned}\Rightarrow (32a^{10}b^{-5})^{\frac{1}{5}}\\[1em] \Rightarrow (2^5)^{\frac{1}{5}} \times a^{10 \times \frac{1}{5}} \times b^{-5 \times \frac{1}{5}}\\[1em] \Rightarrow 2^1 \times a^2 \times b^{-1}\\[1em] \Rightarrow \frac{2a^2}{b}\end{aligned}$$

Hence, $\bm{(32a^{10}b^{-5})^{\frac{1}{5}} = \frac{2a^2}{b}}$.

We know that,

For any non-zero rational number ($a$)

$\sqrt[n]{a} = a^{\frac{1}{n}}, (a^m)^n = a^{mn}$ and $a^{-n} = \frac{1}{a^n}$

Solving,

$$\begin{aligned}\Rightarrow \sqrt[3]{x^{18}y^{-12}z^3}\\[1em] \Rightarrow \left(x^{18}y^{-12}z^3\right)^{\frac{1}{3}}\\[1em] \Rightarrow x^{18 \times \frac{1}{3}} \times y^{-12 \times \frac{1}{3}} \times z^{3 \times \frac{1}{3}}\\[1em] \Rightarrow x^6 \times y^{-4} \times z^1\\[1em] \Rightarrow \frac{x^6z}{y^4}\end{aligned}$$

Hence, $\bm{\sqrt[3]{x^{18}y^{-12}z^3} = \frac{x^6z}{y^4}}$.

We know that,

For any non-zero rational number ($a$)

$a^m \times a^n = a^{m+n}, \frac{a^m}{a^n} = a^{m-n}, (a^m)^n = a^{mn}$ and $a^0 = 1$

Solving L.H.S.,

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

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

Hence, $\bm{\frac{x^{m+n} \times x^{n+l} \times x^{l+m}}{\left(x^m \times x^n \times x^l\right)^2} = 1}$.

We know that,

For any non-zero rational number ($a$)

$\sqrt{a} = a^{\frac{1}{2}}, (a^m)^n = a^{mn}, a^m \times a^n = a^{m+n}$ and $a^0 = 1$

Solving L.H.S.,

$$\begin{aligned}\Rightarrow \sqrt{x^{p-q}} \times \sqrt{x^{q-r}} \times \sqrt{x^{r-p}}\\[1em] \Rightarrow \left(x^{p-q}\right)^{\frac{1}{2}} \times \left(x^{q-r}\right)^{\frac{1}{2}} \times \left(x^{r-p}\right)^{\frac{1}{2}}\\[1em] \Rightarrow x^{\frac{p-q}{2}} \times x^{\frac{q-r}{2}} \times x^{\frac{r-p}{2}}\\[1em] \Rightarrow x^{\frac{p-q}{2}+\frac{q-r}{2}+\frac{r-p}{2}}\\[1em] \Rightarrow x^{\frac{p-q+q-r+r-p}{2}}\\[1em] \Rightarrow x^{\frac{0}{2}}\\[1em] \Rightarrow x^0\\[1em] \Rightarrow 1\end{aligned}$$

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

Hence, $\bm{\sqrt{x^{p-q}} \times \sqrt{x^{q-r}} \times \sqrt{x^{r-p}} = 1}$.

We know that,

For any non-zero rational number ($a$)

$\frac{a^m}{a^n} = a^{m-n}, (a^m)^n = a^{mn}, a^m \times a^n = a^{m+n}$ and $a^0 = 1$

Solving L.H.S.,

$$\begin{aligned}\Rightarrow \left(\frac{x^p}{x^q}\right)^r \times \left(\frac{x^q}{x^r}\right)^p \times \left(\frac{x^r}{x^p}\right)^q\\[1em] \Rightarrow \left(x^{p-q}\right)^r \times \left(x^{q-r}\right)^p \times \left(x^{r-p}\right)^q\\[1em] \Rightarrow x^{pr-qr} \times x^{pq-pr} \times x^{qr-pq}\\[1em] \Rightarrow x^{(pr-qr)+(pq-pr)+(qr-pq)}\\[1em] \Rightarrow x^{pr - qr + pq - pr + qr - pq}\\[1em] \Rightarrow x^0\\[1em] \Rightarrow 1\end{aligned}$$

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

Hence, $\bm{\left(\frac{x^p}{x^q}\right)^r \times \left(\frac{x^q}{x^r}\right)^p \times \left(\frac{x^r}{x^p}\right)^q = 1}$.

We know that,

For any non-zero rational number ($a$)

$(a^m)^n = a^{mn}, a^m \times a^n = a^{m+n}, a^0 = 1$ and $(a-b)(a+b) = a^2 - b^2$

Solving L.H.S.,

$$\begin{aligned}\Rightarrow \left(x^{a+b}\right)^{a-b} \times \left(x^{b+c}\right)^{b-c} \times \left(x^{c+a}\right)^{c-a}\\[1em] \Rightarrow x^{(a+b)(a-b)} \times x^{(b+c)(b-c)} \times x^{(c+a)(c-a)}\\[1em] \Rightarrow x^{a^2-b^2} \times x^{b^2-c^2} \times x^{c^2-a^2}\\[1em] \Rightarrow x^{(a^2-b^2)+(b^2-c^2)+(c^2-a^2)}\\[1em] \Rightarrow x^{a^2-b^2+b^2-c^2+c^2-a^2}\\[1em] \Rightarrow x^0\\[1em] \Rightarrow 1\end{aligned}$$

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

Hence, $\bm{\left(x^{a+b}\right)^{a-b} \times \left(x^{b+c}\right)^{b-c} \times \left(x^{c+a}\right)^{c-a} = 1}$.

We know that,

For any non-zero rational number ($a$)

$\frac{a^m}{a^n} = a^{m-n}, (a^m)^n = a^{mn}, a^m \times a^n = a^{m+n}, a^0 = 1$ and $(a-b)(a+b) = a^2 - b^2$

Solving L.H.S.,

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

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

Hence, $\bm{\left(\frac{x^a}{x^{-b}}\right)^{a-b} \times \left(\frac{x^b}{x^{-c}}\right)^{b-c} \times \left(\frac{x^c}{x^{-a}}\right)^{c-a} = 1}$.

We know that,

For any non-zero rational number ($a$)

$\frac{a^m}{a^n} = a^{m-n}, (a^m)^n = a^{mn}, a^m \times a^n = a^{m+n}, a^0 = 1$ and $(a-b)(a+b) = a^2 - b^2$

Solving L.H.S.,

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

Adding the powers,

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

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

Hence, $\bm{\left(\frac{x^{a+b}}{x^c}\right)^{a-b} \times \left(\frac{x^{b+c}}{x^a}\right)^{b-c} \times \left(\frac{x^{c+a}}{x^b}\right)^{c-a} = 1}$.

We know that,

For any non-zero rational number ($a$)

$\frac{a^m}{a^n} = a^{m-n}, (a^m)^n = a^{mn}, a^m \times a^n = a^{m+n}, a^0 = 1$ and $a^2 - b^2 = (a-b)(a+b)$

Solving L.H.S.,

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

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

Hence, $\bm{\left(\frac{x^{a^2}}{x^{b^2}}\right)^{\frac{1}{a+b}} \times \left(\frac{x^{b^2}}{x^{c^2}}\right)^{\frac{1}{b+c}} \times \left(\frac{x^{c^2}}{x^{a^2}}\right)^{\frac{1}{c+a}} = 1}$.

We know that,

For any non-zero rational number ($a$)

$\frac{a^m}{a^n} = a^{m-n}, (a^m)^n = a^{mn}, a^m \times a^n = a^{m+n}, a^0 = 1$ and $(a-b)(a^2+ab+b^2) = a^3 - b^3$

Solving L.H.S.,

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

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

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

We know that,

For any non-zero rational number ($a$)

$\frac{a^m}{a^n} = a^{m-n}, (a^m)^n = a^{mn}, a^m \times a^n = a^{m+n}$ and $a^0 = 1$

Solving,

$$\begin{aligned}\Rightarrow \left(\frac{x^a}{x^b}\right)^{\frac{1}{ab}} \times \left(\frac{x^b}{x^c}\right)^{\frac{1}{bc}} \times \left(\frac{x^c}{x^a}\right)^{\frac{1}{ca}}\\[1em] \Rightarrow \left(x^{a-b}\right)^{\frac{1}{ab}} \times \left(x^{b-c}\right)^{\frac{1}{bc}} \times \left(x^{c-a}\right)^{\frac{1}{ca}}\\[1em] \Rightarrow x^{\frac{a-b}{ab}} \times x^{\frac{b-c}{bc}} \times x^{\frac{c-a}{ca}}\\[1em] \Rightarrow x^{\left(\frac{1}{b}-\frac{1}{a}\right)} \times x^{\left(\frac{1}{c}-\frac{1}{b}\right)} \times x^{\left(\frac{1}{a}-\frac{1}{c}\right)}\\[1em] \Rightarrow x^{\frac{1}{b}-\frac{1}{a}+\frac{1}{c}-\frac{1}{b}+\frac{1}{a}-\frac{1}{c}}\\[1em] \Rightarrow x^0\\[1em] \Rightarrow 1\end{aligned}$$

Hence, $\bm{\left(\frac{x^a}{x^b}\right)^{\frac{1}{ab}} \times \left(\frac{x^b}{x^c}\right)^{\frac{1}{bc}} \times \left(\frac{x^c}{x^a}\right)^{\frac{1}{ca}} = 1}$.

We know that,

For any non-zero rational number ($a$)

$a^{m-n} = \frac{a^m}{a^n}$

Solving,

$$\begin{aligned}\Rightarrow \frac{1}{1+x^{a-b}} + \frac{1}{1+x^{b-a}}\\[1em] \Rightarrow \frac{1}{1+\frac{x^a}{x^b}} + \frac{1}{1+\frac{x^b}{x^a}}\\[1em] \Rightarrow \frac{1}{\frac{x^b+x^a}{x^b}} + \frac{1}{\frac{x^a+x^b}{x^a}}\\[1em] \Rightarrow \frac{x^b}{x^b+x^a} + \frac{x^a}{x^a+x^b}\\[1em] \Rightarrow \frac{x^b+x^a}{x^a+x^b}\\[1em] \Rightarrow 1\end{aligned}$$

Hence, $\bm{\frac{1}{1+x^{a-b}} + \frac{1}{1+x^{b-a}} = 1}$.

We know that,

For any non-zero rational numbers ($a$ and $b$)

$(ab)^m = a^mb^m$ and $a^m \times a^n = a^{m+n}$

Solving,

$$\begin{aligned}\Rightarrow (ab)^{y-z} \cdot (bc)^{z-x} \cdot (ca)^{x-y}\\[1em] \Rightarrow a^{y-z}b^{y-z} \cdot b^{z-x}c^{z-x} \cdot c^{x-y}a^{x-y}\end{aligned}$$

Collecting the powers of $a, b$ and $c$ separately,

$$\begin{aligned}\Rightarrow a^{(y-z)+(x-y)} \cdot b^{(y-z)+(z-x)} \cdot c^{(z-x)+(x-y)}\\[1em] \Rightarrow a^{y-z+x-y} \cdot b^{y-z+z-x} \cdot c^{z-x+x-y}\\[1em] \Rightarrow a^{x-z} \cdot b^{y-x} \cdot c^{z-y}\end{aligned}$$

Hence, $\bm{(ab)^{y-z} \cdot (bc)^{z-x} \cdot (ca)^{x-y} = a^{x-z}b^{y-x}c^{z-y}}$.

We know that,

For any non-zero rational number ($a$)

$a^m \times a^n = a^{m+n}, \frac{a^m}{a^n} = a^{m-n}$ and $(a^m)^n = a^{mn}$

Solving the numerator,

$$\begin{aligned}\Rightarrow x^{2n+3} \cdot x^{(2n+1)(n+2)}\\[1em] \Rightarrow x^{(2n+3)+(2n^2+4n+n+2)}\\[1em] \Rightarrow x^{2n+3+2n^2+5n+2}\\[1em] \Rightarrow x^{2n^2+7n+5}\end{aligned}$$

Solving the denominator,

$$\begin{aligned}\Rightarrow \left(x^3\right)^{2n+1} \cdot x^{n(2n+1)}\\[1em] \Rightarrow x^{3(2n+1)} \cdot x^{2n^2+n}\\[1em] \Rightarrow x^{(6n+3)+(2n^2+n)}\\[1em] \Rightarrow x^{2n^2+7n+3}\end{aligned}$$

So,

$$\begin{aligned}\Rightarrow \frac{x^{2n^2+7n+5}}{x^{2n^2+7n+3}}\\[1em] \Rightarrow x^{(2n^2+7n+5)-(2n^2+7n+3)}\\[1em] \Rightarrow x^{2}\end{aligned}$$

Hence, $\bm{\frac{x^{2n+3} \cdot x^{(2n+1)(n+2)}}{\left(x^3\right)^{2n+1} \cdot x^{n(2n+1)}} = x^2}$.

We know that,

For any non-zero rational number ($a$)

$a^m \times a^n = a^{m+n}, \frac{a^m}{a^n} = a^{m-n}, (a^m)^n = a^{mn}$ and $a^{-n} = \frac{1}{a^n}$

Solving,

$$\begin{aligned}\Rightarrow \frac{a^{7+2n} \cdot \left(a^2\right)^{3n+2}}{\left(a^4\right)^{2n+3}}\\[1em] \Rightarrow \frac{a^{7+2n} \cdot a^{2(3n+2)}}{a^{4(2n+3)}}\\[1em] \Rightarrow \frac{a^{7+2n} \cdot a^{6n+4}}{a^{8n+12}}\\[1em] \Rightarrow \frac{a^{(7+2n)+(6n+4)}}{a^{8n+12}}\\[1em] \Rightarrow \frac{a^{8n+11}}{a^{8n+12}}\\[1em] \Rightarrow a^{(8n+11)-(8n+12)}\\[1em] \Rightarrow a^{-1}\\[1em] \Rightarrow \frac{1}{a}\end{aligned}$$

Hence, $\bm{\frac{a^{7+2n} \cdot \left(a^2\right)^{3n+2}}{\left(a^4\right)^{2n+3}} = \frac{1}{a}}$.

We know that,

For any non-zero rational number ($a$)

$\Big(\frac{a}{b}\Big)^m = \frac{a^m}{b^m}$ and $(a^m)^n = a^{mn}$

We know that 16 = 2⁴ and 625 = 5⁴

Solving,

$$\begin{aligned}\Rightarrow \Big(\frac{16}{625}\Big)^{\frac{1}{4}}\\[1em] \Rightarrow \Big(\frac{2^4}{5^4}\Big)^{\frac{1}{4}}\\[1em] \Rightarrow \Big(\frac{2}{5}\Big)^{4 \times \frac{1}{4}}\\[1em] \Rightarrow \Big(\frac{2}{5}\Big)^{1}\\[1em] \Rightarrow \frac{2}{5}\end{aligned}$$

Hence, $\mathbf{\Big(\frac{16}{625}\Big)^{\frac{1}{4}} = \frac{2}{5}}$.

We know that,

For any non-zero rational number ($a$)

$\Big(\frac{a}{b}\Big)^{-m} = \Big(\frac{b}{a}\Big)^{m}$ and $(a^m)^n = a^{mn}$

We know that 81 = 3⁴ and 16 = 2⁴

Solving,

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

Hence, $\mathbf{\Big(\frac{81}{16}\Big)^{\frac{-1}{4}} = \frac{2}{3}}$.

We know that,

For any non-zero rational number ($a$)

$(a^m)^n = a^{mn}, a^0 = 1, \frac{1}{a^{-m}} = a^m$ and $\Big(\frac{a}{b}\Big)^{-m} = \Big(\frac{b}{a}\Big)^{m}$

We know that 64 = 4³, 125 = 5³, 27 = 3³, 25 = 5² and 9 = 3²

Solving each term separately,

$$\begin{aligned}\Rightarrow (64)^{\frac{2}{3}} = (4^3)^{\frac{2}{3}} = 4^2 = 16\\[1em] \Rightarrow \sqrt[3]{125} = (5^3)^{\frac{1}{3}} = 5\\[1em] \Rightarrow 3^0 = 1\\[1em] \Rightarrow \frac{1}{2^{-5}} = 2^5 = 32\\[1em] \Rightarrow (27)^{\frac{-2}{3}} = (3^3)^{\frac{-2}{3}} = 3^{-2} = \frac{1}{9}\\[1em] \Rightarrow \Big(\frac{25}{9}\Big)^{\frac{-1}{2}} = \Big(\frac{9}{25}\Big)^{\frac{1}{2}} = \Big(\frac{3^2}{5^2}\Big)^{\frac{1}{2}} = \frac{3}{5}\end{aligned}$$

Substituting the values,

$$\begin{aligned}\Rightarrow 16 + 5 + 1 + 32 + \frac{1}{9} \times \frac{3}{5}\\[1em] \Rightarrow 54 + \frac{3}{45}\\[1em] \Rightarrow 54 + \frac{1}{15}\\[1em] \Rightarrow \frac{810 + 1}{15}\\[1em] \Rightarrow \frac{811}{15}\\[1em] \Rightarrow 54\frac{1}{15}\end{aligned}$$

Hence, the value of the given expression is $\mathbf{54\frac{1}{15}}$.

We know that,

For any non-zero rational number ($a$)

$a^m \times a^n = a^{m+n}, (a^m)^n = a^{mn}, a^{-n} = \frac{1}{a^n}$ and $a^0 = 1$

We know that 81 = 3⁴, 64 = 2⁶ and $\sqrt[3]{3} = 3^{\frac{1}{3}}$

Solving,

$$\begin{aligned}\Rightarrow (81)^{-1} \times 3^{-5} \times 3^9 \times (64)^{\frac{5}{6}} \times \left(\sqrt[3]{3}\right)^6\\[1em] \Rightarrow (3^4)^{-1} \times 3^{-5} \times 3^9 \times (2^6)^{\frac{5}{6}} \times \left(3^{\frac{1}{3}}\right)^6\\[1em] \Rightarrow 3^{-4} \times 3^{-5} \times 3^9 \times 2^5 \times 3^2\\[1em] \Rightarrow 3^{-4-5+9+2} \times 32\\[1em] \Rightarrow 3^{2} \times 32\\[1em] \Rightarrow 9 \times 32\\[1em] \Rightarrow 288\end{aligned}$$

Hence, the value of the given expression is 288.

We know that,

For any non-zero rational numbers ($a$ and $b$)

$\sqrt{a} \times \sqrt{b} = \sqrt{ab}$

Solving,

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

Hence, $\bm{\sqrt{\frac{y^3}{x}} \times \sqrt{\frac{y}{x}} = \frac{y^2}{x}}$.

We know that,

For any non-zero rational number ($a$)

$\frac{a^m}{a^n} = a^{m-n}, a^m \times a^n = a^{m+n}$ and $(a^m)^n = a^{mn}$

Solving L.H.S.,

$$\begin{aligned}\Rightarrow \Big(\frac{-3}{11}\Big)^{x+5} \div \Big(\frac{-3}{11}\Big)^{-2x+3}\\[1em] \Rightarrow \Big(\frac{-3}{11}\Big)^{(x+5)-(-2x+3)}\\[1em] \Rightarrow \Big(\frac{-3}{11}\Big)^{x+5+2x-3}\\[1em] \Rightarrow \Big(\frac{-3}{11}\Big)^{3x+2}\end{aligned}$$

Solving R.H.S.,

$$\begin{aligned}\Rightarrow \Big(\frac{-3}{11}\Big)^{2x-5} \times \Big[\Big(\frac{-3}{11}\Big)^{-2}\Big]^{(x+4)}\\[1em] \Rightarrow \Big(\frac{-3}{11}\Big)^{2x-5} \times \Big(\frac{-3}{11}\Big)^{-2(x+4)}\\[1em] \Rightarrow \Big(\frac{-3}{11}\Big)^{(2x-5)+(-2x-8)}\\[1em] \Rightarrow \Big(\frac{-3}{11}\Big)^{-13}\end{aligned}$$

So,

$\Rightarrow \Big(\frac{-3}{11}\Big)^{3x+2} = \Big(\frac{-3}{11}\Big)^{-13}$

Using $a^m = a^n \Rightarrow m = n$

$$\begin{aligned}\Rightarrow 3x + 2 = -13\\[1em] \Rightarrow 3x = -13 - 2\\[1em] \Rightarrow 3x = -15\\[1em] \Rightarrow x = \frac{-15}{3}\\[1em] \Rightarrow x = -5\end{aligned}$$

Hence, the value of x is -5.

We know that,

For any non-zero rational number ($a$)

$(a^m)^n = a^{mn}$

Solving L.H.S.,

$$\begin{aligned}\Rightarrow \Big[\Big\lbrace\Big(\frac{2}{5}\Big)^2\Big\rbrace^4\Big]^{x+2}\\[1em] \Rightarrow \Big[\Big(\frac{2}{5}\Big)^{2 \times 4}\Big]^{x+2}\\[1em] \Rightarrow \Big(\frac{2}{5}\Big)^{8(x+2)}\\[1em] \Rightarrow \Big(\frac{2}{5}\Big)^{8x+16}\end{aligned}$$

Solving R.H.S.,

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

So,

$\Rightarrow \Big(\frac{2}{5}\Big)^{8x+16} = \Big(\frac{2}{5}\Big)^{-6+6x}$

Using $a^m = a^n \Rightarrow m = n$

$$\begin{aligned}\Rightarrow 8x + 16 = -6 + 6x\\[1em] \Rightarrow 8x - 6x = -6 - 16\\[1em] \Rightarrow 2x = -22\\[1em] \Rightarrow x = \frac{-22}{2}\\[1em] \Rightarrow x = -11\end{aligned}$$

Hence, the value of $\bm{x}$ is -11.

We know that,

For any non-zero rational numbers ($a$ and $b$)

$(a^m)^n = a^{mn}, (ab)^m = a^mb^m$ and $a^{-n} = \frac{1}{a^n}$

Solving,

$$\begin{aligned}\Rightarrow \Big[\Big\lbrace 2p^{-1}q^2r \Big\rbrace^3\Big]^{-2}\\[1em] \Rightarrow \Big(2p^{-1}q^2r\Big)^{3 \times (-2)}\\[1em] \Rightarrow \Big(2p^{-1}q^2r\Big)^{-6}\\[1em] \Rightarrow 2^{-6} \times p^{-1 \times (-6)} \times q^{2 \times (-6)} \times r^{-6}\\[1em] \Rightarrow \frac{1}{2^6} \times p^{6} \times q^{-12} \times r^{-6}\\[1em] \Rightarrow \frac{p^6}{64q^{12}r^6}\end{aligned}$$

Hence, $\bm{\Big[\Big\lbrace 2p^{-1}q^2r \Big\rbrace^3\Big]^{-2} = \frac{p^6}{64q^{12}r^6}}$.

We know that,

For any non-zero rational numbers ($a$ and $b$)

$\frac{a^m}{a^n} = a^{m-n}, (a^m)^n = a^{mn}, (ab)^m = a^mb^m$ and $a^{-n} = \frac{1}{a^n}$

Solving,

$$\begin{aligned}\Rightarrow \Big(\frac{3p^2qr^{-2}}{2p^{-1}q^3}\Big)^2 \div \Big(2p^3r\Big)^{-1}\\[1em] \Rightarrow \Big(\frac{3^2p^{2 \times 2}q^2r^{-2 \times 2}}{2^2p^{-1 \times 2}q^{3 \times 2}}\Big) \div 2^{-1}p^{-3}r^{-1}\\[1em] \Rightarrow \Big(\frac{9p^4q^2r^{-4}}{4p^{-2}q^6}\Big) \times \frac{1}{2^{-1}p^{-3}r^{-1}}\\[1em] \Rightarrow \Big(\frac{9}{4}p^{4+2}q^{2-6}r^{-4}\Big) \times 2p^3r\\[1em] \Rightarrow \frac{9}{4}p^6q^{-4}r^{-4} \times 2p^3r\\[1em] \Rightarrow \frac{9 \times 2}{4}p^{6+3}q^{-4}r^{-4+1}\\[1em] \Rightarrow \frac{9}{2}p^9q^{-4}r^{-3}\\[1em] \Rightarrow \frac{9p^9}{2q^4r^3}\end{aligned}$$

Hence, $\bm{\Big(\frac{3p^2qr^{-2}}{2p^{-1}q^3}\Big)^2 \div \Big(2p^3r\Big)^{-1} = \frac{9p^9}{2q^4r^3}}$.

We know that,

$a^0 = 1$ and $a^1 = a$

Solving each value,

$$\begin{aligned}\Rightarrow 1^5 = 1 \times 1 \times 1 \times 1 \times 1 = 1\\[1em] \Rightarrow 5^0 = 1\\[1em] \Rightarrow 0^5 = 0 \times 0 \times 0 \times 0 \times 0 = 0\\[1em] \Rightarrow 5^1 = 5\end{aligned}$$

Since, I = II = 1

Hence, option 1 is the correct option.

We know that 3125 = 5 × 5 × 5 × 5 × 5 = 5⁵

$$\begin{aligned}\Rightarrow 5^x = 3125\\[1em] \Rightarrow 5^x = 5^5\end{aligned}$$

Using $a^m = a^n \Rightarrow m = n$

$\Rightarrow x = 5$

Substituting the value of $x$,

$$\begin{aligned}\Rightarrow 5^{x-3}\\[1em] \Rightarrow 5^{5-3}\\[1em] \Rightarrow 5^2\\[1em] \Rightarrow 25\end{aligned}$$

Hence, option 1 is the correct option.

We know that,

For any non-zero rational number ($a$)

$(a^m)^n = a^{mn}$

We know that 256 = 4⁴

Solving,

$$\begin{aligned}\Rightarrow (256)^{\frac{5}{4}}\\[1em] \Rightarrow (4^4)^{\frac{5}{4}}\\[1em] \Rightarrow 4^{4 \times \frac{5}{4}}\\[1em] \Rightarrow 4^5\\[1em] \Rightarrow 4 \times 4 \times 4 \times 4 \times 4\\[1em] \Rightarrow 1024\end{aligned}$$

Hence, option 3 is the correct option.

We know that,

For any non-zero rational number ($a$)

$a^m \times a^n = a^{m+n}$

Solving,

$$\begin{aligned}\Rightarrow 2^{\frac{5}{4}}\\[1em] \Rightarrow 2^{1 + \frac{1}{4}}\\[1em] \Rightarrow 2^1 \times 2^{\frac{1}{4}}\\[1em] \Rightarrow 2 \times \sqrt[4]{2}\end{aligned}$$

Since $\sqrt[4]{2}$ lies between 1 and 2, the value of $2 \times \sqrt[4]{2}$ lies between 2 and 4.

More precisely, $\sqrt[4]{2} \approx 1.189$, so

$$\begin{aligned}\Rightarrow 2 \times 1.189\\[1em] \Rightarrow 2.378\end{aligned}$$

Since, 2 < 2.378 < 3

Hence, option 3 is the correct option.

We know that,

For any non-zero rational number ($a$)

$\Big(\frac{a}{b}\Big)^{-m} = \Big(\frac{b}{a}\Big)^{m}$ and $(a^m)^n = a^{mn}$

We know that 32 = 2⁵ and 243 = 3⁵

Solving,

$$\begin{aligned}\Rightarrow \Big(\frac{32}{243}\Big)^{\frac{-4}{5}}\\[1em] \Rightarrow \Big(\frac{243}{32}\Big)^{\frac{4}{5}}\\[1em] \Rightarrow \Big(\frac{3^5}{2^5}\Big)^{\frac{4}{5}}\\[1em] \Rightarrow \Big(\frac{3}{2}\Big)^{5 \times \frac{4}{5}}\\[1em] \Rightarrow \Big(\frac{3}{2}\Big)^{4}\\[1em] \Rightarrow \frac{81}{16}\end{aligned}$$

Hence, option 4 is the correct option.

We know that,

For any non-zero rational number ($a$)

$a^{-n} = \frac{1}{a^n}$

Solving,

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

Hence, option 3 is the correct option.

We know that,

For any non-zero rational number ($a$)

$a^{-n} = \frac{1}{a^n}$

Solving the first bracket,

$$\begin{aligned}\Rightarrow (8^{-1} - 9^{-1})^{-1}\\[1em] \Rightarrow \Big(\frac{1}{8} - \frac{1}{9}\Big)^{-1}\\[1em] \Rightarrow \Big(\frac{9 - 8}{72}\Big)^{-1}\\[1em] \Rightarrow \Big(\frac{1}{72}\Big)^{-1}\\[1em] \Rightarrow 72\end{aligned}$$

Solving the second bracket,

$$\begin{aligned}\Rightarrow (4^{-1} - 9^{-1})^{-1}\\[1em] \Rightarrow \Big(\frac{1}{4} - \frac{1}{9}\Big)^{-1}\\[1em] \Rightarrow \Big(\frac{9 - 4}{36}\Big)^{-1}\\[1em] \Rightarrow \Big(\frac{5}{36}\Big)^{-1}\\[1em] \Rightarrow \frac{36}{5}\end{aligned}$$

So,

$$\begin{aligned}\Rightarrow 72 \div \frac{36}{5}\\[1em] \Rightarrow 72 \times \frac{5}{36}\\[1em] \Rightarrow \frac{360}{36}\\[1em] \Rightarrow 10\end{aligned}$$

Hence, option 2 is the correct option.

We know that,

For any non-zero rational number ($a$)

$(a^m)^n = a^{mn}$ and $a^{-n} = \frac{1}{a^n}$

We know that 64 = 8² and −32 = (−2)⁵

Solving the first term,

$$\begin{aligned}\Rightarrow (64)^{\frac{-1}{2}}\\[1em] \Rightarrow (8^2)^{\frac{-1}{2}}\\[1em] \Rightarrow 8^{-1}\\[1em] \Rightarrow \frac{1}{8}\end{aligned}$$

Solving the second term,

$$\begin{aligned}\Rightarrow (-32)^{\frac{-4}{5}}\\[1em] \Rightarrow ((-2)^5)^{\frac{-4}{5}}\\[1em] \Rightarrow (-2)^{-4}\\[1em] \Rightarrow \frac{1}{(-2)^4}\\[1em] \Rightarrow \frac{1}{16}\end{aligned}$$

So,

$$\begin{aligned}\Rightarrow \frac{1}{8} - \frac{1}{16}\\[1em] \Rightarrow \frac{2 - 1}{16}\\[1em] \Rightarrow \frac{1}{16}\end{aligned}$$

Hence, option 3 is the correct option.

We know that,

For any non-zero rational number ($a$)

$\Big(\frac{p}{q}\Big)^{-m} = \Big(\frac{q}{p}\Big)^{m}$

Solving,

$$\begin{aligned}\Rightarrow \Big(\frac{-5}{7}\Big)^{-7}\\[1em] \Rightarrow \Big(\frac{7}{-5}\Big)^{7}\\[1em] \Rightarrow \Big(\frac{-7}{5}\Big)^{7}\end{aligned}$$

Hence, option 4 is the correct option.

We know that 81 = 3⁴

Solving the first equation,

$$\begin{aligned}\Rightarrow 3^{x+y} = 81\\[1em] \Rightarrow 3^{x+y} = 3^4\end{aligned}$$

Using $a^m = a^n \Rightarrow m = n$

$\Rightarrow x + y = 4$    ..... (1)

Solving the second equation,

$$\begin{aligned}\Rightarrow 81^{x-y} = 3^8\\[1em] \Rightarrow (3^4)^{x-y} = 3^8\\[1em] \Rightarrow 3^{4(x-y)} = 3^8\end{aligned}$$

Using $a^m = a^n \Rightarrow m = n$

$$\begin{aligned}\Rightarrow 4(x - y) = 8\\[1em] \Rightarrow x - y = 2\end{aligned}$$

   ..... (2)

Adding equations (1) and (2),

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

Substituting $x = 3$ in equation (1),

$$\begin{aligned}\Rightarrow 3 + y = 4\\[1em] \Rightarrow y = 1\end{aligned}$$

Hence, option 4 is the correct option.

We know that,

For any non-zero rational number ($a$)

$(a^m)^n = a^{mn}$

Solving each value,

$$\begin{aligned}\Rightarrow (0.008)^{\frac{1}{3}} = ((0.2)^3)^{\frac{1}{3}} = 0.2\\[1em] \Rightarrow (0.01)^{\frac{1}{2}} = ((0.1)^2)^{\frac{1}{2}} = 0.1\\[1em] \Rightarrow (0.2)^2 = 0.2 \times 0.2 = 0.04\\[1em] \Rightarrow \frac{1}{100} = 0.01\end{aligned}$$

Since, 0.01 < 0.04 < 0.1 < 0.2

So, the greatest value is 0.2.

Hence, option 1 is the correct option.

We know that,

For any non-zero rational number ($a$)

$(a^m)^n = a^{mn}$ and $\frac{a^m}{a^n} = a^{m-n}$

We know that 1000 = 10³

Solving,

$$\begin{aligned}\Rightarrow (1000)^{12} \div (10)^{30}\\[1em] \Rightarrow (10^3)^{12} \div 10^{30}\\[1em] \Rightarrow 10^{36} \div 10^{30}\\[1em] \Rightarrow 10^{36-30}\\[1em] \Rightarrow 10^6\end{aligned}$$

Also, (1000)² = (10³)² = 10⁶

Hence, option 1 is the correct option.

We know that,

For any non-zero rational number ($a$)

$(a^m)^n = a^{mn}, \frac{a^m}{a^n} = a^{m-n}$ and $a^m \times a^n = a^{m+n}$

We know that 0.04 = (0.2)² and 0.008 = (0.2)³

Solving L.H.S.,

$$\begin{aligned}\Rightarrow (0.04)^2 \div (0.008) \times (0.2)^6\\[1em] \Rightarrow ((0.2)^2)^2 \div (0.2)^3 \times (0.2)^6\\[1em] \Rightarrow (0.2)^4 \div (0.2)^3 \times (0.2)^6\\[1em] \Rightarrow (0.2)^{4-3+6}\\[1em] \Rightarrow (0.2)^7\end{aligned}$$

So,

$\Rightarrow (0.2)^7 = (0.2)^x$

Using $a^m = a^n \Rightarrow m = n$

$\Rightarrow x = 7$

Hence, option 4 is the correct option.

We know that,

For any non-zero rational number ($a$)

$(a^m)^n = a^{mn}, a^m \times a^n = a^{m+n}$ and $\frac{a^m}{a^n} = a^{m-n}$

We know that 25 = 5² and 125 = 5³

Solving L.H.S.,

$$\begin{aligned}\Rightarrow (25)^{7.5} \times (5)^{2.5} \div (125)^{1.5}\\[1em] \Rightarrow (5^2)^{7.5} \times 5^{2.5} \div (5^3)^{1.5}\\[1em] \Rightarrow 5^{15} \times 5^{2.5} \div 5^{4.5}\\[1em] \Rightarrow 5^{15 + 2.5 - 4.5}\\[1em] \Rightarrow 5^{13}\end{aligned}$$

So,

$\Rightarrow 5^{13} = 5^x$

Using $a^m = a^n \Rightarrow m = n$

$\Rightarrow x = 13$

Hence, option 2 is the correct option.

We know that,

For any non-zero rational number ($a$)

$(a^m)^n = a^{mn}$

We know that 8 = 2³

Solving,

$$\begin{aligned}\Rightarrow 2^{x+1} = 8^x\\[1em] \Rightarrow 2^{x+1} = (2^3)^x\\[1em] \Rightarrow 2^{x+1} = 2^{3x}\end{aligned}$$

Using $a^m = a^n \Rightarrow m = n$

$$\begin{aligned}\Rightarrow x + 1 = 3x\\[1em] \Rightarrow 1 = 3x - x\\[1em] \Rightarrow 2x = 1\\[1em] \Rightarrow x = \frac{1}{2}\end{aligned}$$

Hence, option 2 is the correct option.

We know that,

For any non-zero rational number ($a$)

$(a^m)^n = a^{mn}$ and $a^m \times a^n = a^{m+n}$

We know that 9 = 3²

Solving L.H.S.,

$$\begin{aligned}\Rightarrow 9^n + 9^n + 9^n\\[1em] \Rightarrow 3 \times 9^n\\[1em] \Rightarrow 3 \times (3^2)^n\\[1em] \Rightarrow 3^1 \times 3^{2n}\\[1em] \Rightarrow 3^{2n+1}\end{aligned}$$

So,

$\Rightarrow 3^{2n+1} = 3^{2013}$

Using $a^m = a^n \Rightarrow m = n$

$$\begin{aligned}\Rightarrow 2n + 1 = 2013\\[1em] \Rightarrow 2n = 2012\\[1em] \Rightarrow n = \frac{2012}{2}\\[1em] \Rightarrow n = 1006\end{aligned}$$

Hence, option 2 is the correct option.

We know that,

For any non-zero rational number ($a$)

$a^m \times a^n = a^{m+n}$ and $a^{-n} = \frac{1}{a^n}$

Solving L.H.S.,

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

So,

$$\begin{aligned}\Rightarrow 3^x \times \frac{10}{3} = 90\\[1em] \Rightarrow 3^x = 90 \times \frac{3}{10}\\[1em] \Rightarrow 3^x = 27\\[1em] \Rightarrow 3^x = 3^3\end{aligned}$$

Using $a^m = a^n \Rightarrow m = n$

$\Rightarrow x = 3$

Hence, option 4 is the correct option.

We know that,

For any non-zero rational number ($a$)

$\Big(\frac{p}{q}\Big)^{-m} = \Big(\frac{q}{p}\Big)^{m}$

Solving R.H.S.,

$$\begin{aligned}\Rightarrow \Big(\frac{y}{x}\Big)^{n-3}\\[1em] \Rightarrow \Big(\frac{x}{y}\Big)^{-(n-3)}\\[1em] \Rightarrow \Big(\frac{x}{y}\Big)^{3-n}\end{aligned}$$

So,

$\Rightarrow \Big(\frac{x}{y}\Big)^{n-1} = \Big(\frac{x}{y}\Big)^{3-n}$

Using $a^m = a^n \Rightarrow m = n$

$$\begin{aligned}\Rightarrow n - 1 = 3 - n\\[1em] \Rightarrow n + n = 3 + 1\\[1em] \Rightarrow 2n = 4\\[1em] \Rightarrow n = 2\end{aligned}$$

Hence, option 3 is the correct option.

We know that,

For any non-zero rational number ($a$)

$(a^m)^n = a^{mn}$

We know that 25 = 5² and 125 = 5³

Solving,

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

Using $a^m = a^n \Rightarrow m = n$

$$\begin{aligned}\Rightarrow 2x = 3y\\[1em] \Rightarrow \frac{x}{y} = \frac{3}{2}\\[1em] \Rightarrow x : y = 3 : 2\end{aligned}$$

Hence, option 4 is the correct option.

We know that,

For any non-zero rational number ($a$)

$a^m \times a^n = a^{m + n}$ and $\Big(\frac{1}{a}\Big)^{m} = a^{-m}$

Solving the numerator,

$$\begin{aligned}\Rightarrow 2 \cdot 3^{n+1} + 7 \cdot 3^{n-1}\\[1em] \Rightarrow 2 \cdot 3^{n-1} \cdot 3^2 + 7 \cdot 3^{n-1}\\[1em] \Rightarrow 3^{n-1}(2 \times 9 + 7)\\[1em] \Rightarrow 3^{n-1}(18 + 7)\\[1em] \Rightarrow 25 \cdot 3^{n-1}\end{aligned}$$

Solving the denominator,

$$\begin{aligned}\Rightarrow 3^{n+2} - 2\Big(\frac{1}{3}\Big)^{1-n}\\[1em] \Rightarrow 3^{n+2} - 2 \cdot 3^{-(1-n)}\\[1em] \Rightarrow 3^{n+2} - 2 \cdot 3^{n-1}\\[1em] \Rightarrow 3^{n-1} \cdot 3^3 - 2 \cdot 3^{n-1}\\[1em] \Rightarrow 3^{n-1}(27 - 2)\\[1em] \Rightarrow 25 \cdot 3^{n-1}\end{aligned}$$

Substituting the values of numerator and denominator in the given expression,

$$\begin{aligned}\Rightarrow \frac{2 \cdot 3^{n + 1} + 7 \cdot 3^{n - 1}}{3^{n + 2} - 2\Big(\frac{1}{3}\Big)^{1 - n}}\\[1em] \Rightarrow \frac{25 \cdot 3^{n - 1}}{25 \cdot 3^{n - 1}}\\[1em] \Rightarrow 1\end{aligned}$$

Hence, option 3 is the correct option.

(i) We know that,

For any non-zero rational number ($a$)

$a^{-n} = \frac{1}{a^n}$

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

Hence, the value of (-5)^-3 is $\mathbf{-\frac{1}{125}}$.

(ii) We know that 512 = 8 × 8 × 8 = 8³

$$\begin{aligned}\Rightarrow \frac{1}{512}\\[1em] \Rightarrow \frac{1}{8^3}\\[1em] \Rightarrow 8^{-3}\end{aligned}$$

Hence, $\mathbf{\frac{1}{512}}$ expressed as a power with base 8 is $\mathbf{8^{-3}}$.

(iii) We know that,

For any non-zero rational number ($a$)

$\Big(\frac{p}{q}\Big)^{-m} = \Big(\frac{q}{p}\Big)^{m}$

$$\begin{aligned}\Rightarrow \Big(\frac{1}{2}\Big)^{-2} + \Big(\frac{1}{3}\Big)^{-2} + \Big(\frac{1}{4}\Big)^{-2}\\[1em] \Rightarrow \Big(\frac{2}{1}\Big)^{2} + \Big(\frac{3}{1}\Big)^{2} + \Big(\frac{4}{1}\Big)^{2}\\[1em] \Rightarrow 4 + 9 + 16\\[1em] \Rightarrow 29\end{aligned}$$

Hence, the value is 29.

(iv) Solving,

$$\begin{aligned}\Rightarrow (-3)^4 \div 3^4\\[1em] \Rightarrow \frac{(-3) \times (-3) \times (-3) \times (-3)}{3 \times 3 \times 3 \times 3}\\[1em] \Rightarrow \frac{81}{81}\\[1em] \Rightarrow 1\end{aligned}$$

Hence, (-3)⁴ ÷ 3⁴ is equal to 1.

(v) We know that,

For any non-zero rational number ($a$)

$a^m \times a^n = a^{m+n}$

$$\begin{aligned}\Rightarrow 3 \cdot 3^n = 3\\[1em] \Rightarrow 3^{1+n} = 3^1\end{aligned}$$

Using $a^m = a^n \Rightarrow m = n$

$$\begin{aligned}\Rightarrow 1 + n = 1\\[1em] \Rightarrow n = 0\end{aligned}$$

Hence, the value of $\bm{n}$ is 0.

(vi) As the multiplicative inverse of $a^m = a^{-m}$

$$\begin{aligned}\Rightarrow 10^{-10}\\[1em] \Rightarrow \frac{1}{10^{10}}\end{aligned}$$

Hence, the multiplicative inverse of 10¹⁰ is $\mathbf{10^{-10}}$.

(vii) We know that $a^0 = 1$

⇒ 5⁰ + 6⁰ + 7⁰

⇒ 1 + 1 + 1

⇒ 3

Hence, (5⁰ + 6⁰ + 7⁰) = 3.

(i) The law $a^m \times b^n$ can be combined into a single power only when the bases are the same or the exponents are the same. The correct law is $a^m \times b^m = (ab)^m$.

Taking an example with $a = 2, b = 3, m = 2$ and $n = 3$,

$$\begin{aligned}\Rightarrow a^m \times b^n = 2^2 \times 3^3 = 4 \times 27 = 108\\[1em] \Rightarrow (ab)^{mn} = (2 \times 3)^{2 \times 3} = 6^6 = 46656\end{aligned}$$

Since, L.H.S. ≠ R.H.S.

Hence, the given statement is False.

(ii) We know that 100 = 10²

Solving R.H.S.,

$$\begin{aligned}\Rightarrow (100)^{-10}\\[1em] \Rightarrow (10^2)^{-10}\\[1em] \Rightarrow 10^{-20}\end{aligned}$$

Since (10)^-100 ≠ 10^-20,

Hence, the given statement is False.

(iii) As the reciprocal of $a^{-m} = a^m$

Solving,

$$\begin{aligned}\Rightarrow \Big(\frac{3}{7}\Big)^{-2}\\[1em] \Rightarrow \Big(\frac{7}{3}\Big)^{2}\end{aligned}$$

The reciprocal of $\Big(\frac{7}{3}\Big)^{2}$ is $\Big(\frac{3}{7}\Big)^{2}$ and not $\Big(\frac{7}{3}\Big)^{2}$.

Hence, the given statement is False.

(iv) Solving L.H.S.,

$$\begin{aligned}\Rightarrow (-a)^{-m}\\[1em] \Rightarrow \frac{1}{(-a)^m}\end{aligned}$$

When $m$ is odd, $(-a)^m = -a^m$ and so $(-a)^{-m} = -\frac{1}{a^m}$, which is not equal to $\frac{1}{a^m}$.

For example, taking $a = 2$ and $m = 3$,

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

Since, L.H.S. ≠ R.H.S.

Hence, the given statement is False.

(v) We know that,

$a^m \times a^n = a^{m+n}$

$$\begin{aligned}\Rightarrow 10^5 \times 10^5\\[1em] \Rightarrow 10^{5+5}\\[1em] \Rightarrow 10^{10}\end{aligned}$$

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

Hence, the given statement is True.

(vi) We know that,

$a^m \times a^n = a^{m+n}$ and $a^0 = 1$

$$\begin{aligned}\Rightarrow p^{\frac{-2}{5}} \times p^{\frac{2}{5}}\\[1em] \Rightarrow p^{\frac{-2}{5}+\frac{2}{5}}\\[1em] \Rightarrow p^{0}\\[1em] \Rightarrow 1\end{aligned}$$

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

Hence, the given statement is True.

(vii) We know that,

$a^m \times a^n = a^{m+n}$

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

Since (-8)^-7 ≠ (-8)^-12,

Hence, the given statement is False.

We know that,

$\frac{a^m}{b^m} = \Big(\frac{a}{b}\Big)^m$ and $\Big(\frac{p}{q}\Big)^{-m} = \Big(\frac{q}{p}\Big)^{m}$

Solving,

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

So, Assertion (A) is true.

The correct law of exponents is $\left(x^m\right)^n = x^{mn}$ and not $x^{m + n}$.

So, Reason (R) is false.

Hence, option 3 is the correct option.

The reciprocal of $a^m = a^{-m}$

Solving,

$$\begin{aligned}\Rightarrow \Big(\frac{4}{5}\Big)^{3} \times \Big(\frac{4}{5}\Big)^{-3}\\[1em] \Rightarrow \Big(\frac{4}{5}\Big)^{3 + (-3)}\\[1em] \Rightarrow \Big(\frac{4}{5}\Big)^{0}\\[1em] \Rightarrow 1\end{aligned}$$

So, Assertion (A) is true.

If $p = 0$, then $\frac{p}{q} = 0$. Since the product of 0 with any number is 0 and never 1, the reciprocal of 0 does not exist.

So, Reason (R) is also true.

However, Assertion (A) follows from the law of reciprocals of powers and not from the fact that 0 has no reciprocal.

So, Reason (R) is not the correct explanation of Assertion (A).

Hence, option 2 is the correct option.

We know that,

For any non-zero rational number ($a$)

$a^{-n} = \frac{1}{a^n}$

Solving,

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

Hence, option 2 is the correct option.

Let $a^x = b^y = c^z = k$. So,

$\Rightarrow a = k^{\frac{1}{x}}, \quad b = k^{\frac{1}{y}}, \quad c = k^{\frac{1}{z}}$

Given, $b^2 = ac$

Substituting the values,

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

Using $a^m = a^n \Rightarrow m = n$

$$\begin{aligned}\Rightarrow \frac{2}{y} = \frac{1}{x} + \frac{1}{z}\\[1em] \Rightarrow \frac{2}{y} = \frac{z + x}{xz}\\[1em] \Rightarrow y = \frac{2xz}{x + z}\end{aligned}$$

Hence, option 1 is the correct option.

We know that,

For any non-zero rational numbers ($a$ and $b$)

$(ab)^m = a^mb^m$ and $(a^m)^n = a^{mn}$

We know that 4 = 2² and 18 = 2 × 3²

Solving R.H.S.,

$$\begin{aligned}\Rightarrow (18)^{10}\\[1em] \Rightarrow (2 \times 3^2)^{10}\\[1em] \Rightarrow 2^{10} \times 3^{20}\end{aligned}$$

Solving L.H.S.,

$$\begin{aligned}\Rightarrow 3^a \cdot 4^5\\[1em] \Rightarrow 3^a \cdot (2^2)^5\\[1em] \Rightarrow 3^a \cdot 2^{10}\end{aligned}$$

So,

$$\begin{aligned}\Rightarrow 3^a \cdot 2^{10} = 2^{10} \times 3^{20}\\[1em] \Rightarrow 3^a = 3^{20}\end{aligned}$$

Using $a^m = a^n \Rightarrow m = n$

$\Rightarrow a = 20$

Hence, option 3 is the correct option.

We know that,

For any non-zero rational number ($a$)

$a^m \times a^n = a^{m+n}$

Solving L.H.S.,

$$\begin{aligned}\Rightarrow 2^{12} - 2^{10}\\[1em] \Rightarrow 2^{10} \times 2^2 - 2^{10}\\[1em] \Rightarrow 2^{10}(4 - 1)\\[1em] \Rightarrow 3 \times 2^{10}\end{aligned}$$

So,

$$\begin{aligned}\Rightarrow 3 \times 2^{10} = 3 \times 2^x\\[1em] \Rightarrow 2^{10} = 2^x\end{aligned}$$

Using $a^m = a^n \Rightarrow m = n$

$\Rightarrow x = 10$

Hence, option 1 is the correct option.

We know that,

For any non-zero rational number ($a$)

$a^{-n} = \frac{1}{a^n}$

Solving the first bracket,

$$\begin{aligned}\Rightarrow (7^{-1} - 8^{-1})^{-1}\\[1em] \Rightarrow \Big(\frac{1}{7} - \frac{1}{8}\Big)^{-1}\\[1em] \Rightarrow \Big(\frac{8 - 7}{56}\Big)^{-1}\\[1em] \Rightarrow \Big(\frac{1}{56}\Big)^{-1}\\[1em] \Rightarrow 56\end{aligned}$$

Solving the second bracket,

$$\begin{aligned}\Rightarrow (3^{-1} - 4^{-1})^{-1}\\[1em] \Rightarrow \Big(\frac{1}{3} - \frac{1}{4}\Big)^{-1}\\[1em] \Rightarrow \Big(\frac{4 - 3}{12}\Big)^{-1}\\[1em] \Rightarrow \Big(\frac{1}{12}\Big)^{-1}\\[1em] \Rightarrow 12\end{aligned}$$

So,

$$\begin{aligned}\Rightarrow 56 - 12\\[1em] \Rightarrow 44\end{aligned}$$

Hence, option 2 is the correct option.

Let the original number of cells be $N$.

Since the cells double after every one hour,

$$\begin{aligned}\Rightarrow \text{After 1 hour} = N \times 2 = 2^1N\\[1em] \Rightarrow \text{After 2 hours} = 2^1N \times 2 = 2^2N\\[1em] \Rightarrow \text{After 3 hours} = 2^2N \times 2 = 2^3N\end{aligned}$$

Continuing this pattern, after 8 hours the number of cells

$\Rightarrow 2^8N$

So, there will be 2⁸ times the original number of cells.

Hence, option 3 is the correct option.

We know that,

For any non-zero rational number ($a$)

$a^{-n} = \frac{1}{a^n}$

Solving one pair of terms,

$$\begin{aligned}\Rightarrow 2^1 - 2^{-1}\\[1em] \Rightarrow 2 - \frac{1}{2}\\[1em] \Rightarrow \frac{4 - 1}{2}\\[1em] \Rightarrow \frac{3}{2}\end{aligned}$$

Since there are 100 terms in all, the number of such pairs

$$\begin{aligned}\Rightarrow \frac{100}{2}\\[1em] \Rightarrow 50\end{aligned}$$

So, the value of the given expression

$$\begin{aligned}\Rightarrow 50 \times \frac{3}{2}\\[1em] \Rightarrow \frac{150}{2}\\[1em] \Rightarrow 75\end{aligned}$$

Hence, option 4 is the correct option.