Exponents and Powers

As we know, for any non-zero rational number $a$.

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

Solving,

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

Hence, $\mathbf{\Big(\frac{3}{5}\Big)^{-2} = \frac{25}{9}}$.

As we know, for any non-zero rational number $a$.

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

Solving,

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

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

As we know, for any non-zero rational number $a$.

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

Solving,

$$\begin{aligned}\Rightarrow \Big(\frac{2}{7}\Big)^{-4}\\[1em] \Rightarrow \Big(\frac{7}{2}\Big)^{4}\\[1em] \Rightarrow \Big(\frac{7 \times 7 \times 7 \times 7}{2 \times 2 \times 2 \times 2}\Big)\\[1em] \Rightarrow \frac{2401}{16}\end{aligned}$$

Hence, $\mathbf{\Big(\frac{2}{7}\Big)^{-4} = \frac{2401}{16}}$.

As we know, for any non-zero rational number $a$.

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

Solving,

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

LCM of 2, 4 and 3 is 2 x 2 x 3 = 12

$$\begin{aligned}\Rightarrow \Big[\frac{1 \times 6}{2 \times 6} + \frac{1 \times 3}{4 \times 3} + \frac{1 \times 4}{3 \times 4}\Big]^{-1}\\[1em] \Rightarrow \Big[\frac{6}{12} + \frac{3}{12} + \frac{4}{12}\Big]^{-1}\\[1em] \Rightarrow \Big[\frac{6 + 3 + 4}{12}\Big]^{-1}\\[1em] \Rightarrow \Big[\frac{13}{12}\Big]^{-1}\\[1em] \Rightarrow \frac{12}{13}\end{aligned}$$

Hence, $\mathbf{[(2)^{-1} + (4)^{-1} + (3)^{-1}]^{-1} = \frac{12}{13}}$.

As we know, for any non-zero rational number $a$.

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

Solving,

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

LCM of 4 and 5 is 4 x 5 = 20

$$\begin{aligned}\Rightarrow \Big[\frac{1 \times 5}{4 \times 5} - \frac{1 \times 4}{5 \times 4}\Big]^2 \times \Big(\frac{8}{5}\Big)\\[1em] \Rightarrow \Big[\frac{5}{20} - \frac{4}{20}\Big]^2 \times \Big(\frac{8}{5}\Big)\\[1em] \Rightarrow \Big[\frac{5 - 4}{20}\Big]^2 \times \Big(\frac{8}{5}\Big)\\[1em] \Rightarrow \Big[\frac{1}{20}\Big]^2 \times \Big(\frac{8}{5}\Big)\\[1em] \Rightarrow \frac{1 \times 1 \times 8}{20 \times 20 \times 5}\\[1em] \Rightarrow \frac{8}{2000}\\[1em] \Rightarrow \frac{1}{250}\end{aligned}$$

Hence, $\mathbf{[(4)^{-1}-(5)^{-1}]^2 \times \Big(\frac{5}{8}\Big)^{-1}=\frac{1}{250}}$.

As we know, for any non-zero rational number $a$.

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

Solving,

$$\begin{aligned}\Rightarrow [4^0 + 4^2 - 2^3] \times 3^{-2}\\[1em] \Rightarrow [1 + (4 \times 4) - (2 \times 2 \times 2)] \times \frac{1}{3^2}\\[1em] \Rightarrow [1 + 16 - 8] \times \frac{1}{9}\\[1em] \Rightarrow [9] \times \frac{1}{9}\\[1em] \Rightarrow \frac{9}{9}\\[1em] \Rightarrow 1\end{aligned}$$

Hence, $\mathbf{[4^0 + 4^2 - 2^3] \times 3^{-2} = 1}$.

As we know, for any non-zero rational number $a$.

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

Solving,

$$\begin{aligned}\Rightarrow \Big[(5)^2 - \Big(\frac{1}{4}\Big)^{-2}\Big] \times \Big(\frac{3}{4}\Big)^{-2}\\[1em] \Rightarrow \Big[(5 \times 5) - \Big(\frac{4}{1}\Big)^{2}\Big] \times \Big(\frac{4}{3}\Big)^{2}\\[1em] \Rightarrow \Big[25 - (4 \times 4)\Big] \times \Big(\frac{4 \times 4}{3 \times 3}\Big)\\[1em] \Rightarrow \Big[25 - 16\Big] \times \Big(\frac{16}{9}\Big)\\[1em] \Rightarrow [9] \times \Big(\frac{16}{9}\Big)\\[1em] \Rightarrow \frac{9 \times 16}{9}\\[1em] \Rightarrow 16\end{aligned}$$

Hence, $\mathbf{\Big[(5)^2 - \Big(\frac{1}{4}\Big)^{-2}\Big] \times \Big(\frac{3}{4}\Big)^{-2} = 16}$.

Solving,

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

The multiplicative inverse of a non-zero number is its reciprocal.

Thus, multiplicative inverse of $\frac{8}{27} = \frac{27}{8}$.

Hence, the multiplicative inverse of $\mathbf{\Big(\frac{81}{16}\Big)^{\frac{-3}{4}}}$ is $\mathbf{\frac{27}{8}}$.

Solving,

$$\begin{aligned}\Rightarrow \Big\lbrace\Big(\frac{-3}{2}\Big)^{-4}\Big\rbrace^{\frac{1}{2}}\\[1em] \Rightarrow \Big(\frac{-3}{2}\Big)^{-4 \times \frac{1}{2}}\\[1em] \Rightarrow \Big(\frac{-3}{2}\Big)^{-2}\\[1em] \Rightarrow \Big(\frac{2}{-3}\Big)^{2}\\[1em] \Rightarrow \frac{2 \times 2}{(-3) \times (-3)}\\[1em] \Rightarrow \frac{4}{9}\end{aligned}$$

Hence, the multiplicative inverse of $\mathbf{\Big\lbrace\Big(\frac{-3}{2}\Big)^{-4}\Big\rbrace^{\frac{1}{2}}}$ is $\mathbf{\frac{9}{4}}$.

According to the laws of exponents,

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

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

Hence, the multiplicative inverse of $\mathbf{\Big(\frac{5}{7}\Big)^{-2} \times \Big(\frac{5}{7}\Big)^{4} \div \Big(\frac{5}{7}\Big)^{3}}$ is $\mathbf{\frac{5}{7}}$.

We know that,

16 = 2 x 2 x 2 x 2 = 2⁴

Solving,

⇒ 16^-2

⇒ (2⁴)^-2

⇒ 2^{4 × (-2)}

⇒ 2^-8

Hence, 16^-2 = 2^-8.

We know that 125 = 5 x 5 x 5 = 5³

⇒ 125^-4

⇒ (5³)^-4

⇒ 5^{3 × (-4)}

⇒ 5^-12

Hence, 125^-4 = 5^-12.

Solving,

$$\begin{aligned}\Rightarrow \Big[\Big\lbrace\Big(\frac{5}{7}\Big)^2\Big\rbrace^{-1}\Big]^{-3}\\[1em] \Rightarrow \Big(\frac{5}{7}\Big)^{2 \times (-1) \times (-3)}\\[1em] \Rightarrow \Big(\frac{5}{7}\Big)^{6}\end{aligned}$$

Hence, $\mathbf{\Big[\Big\lbrace\Big(\frac{5}{7}\Big)^2\Big\rbrace^{-1}\Big]^{-3} = \Big(\frac{5}{7}\Big)^{6}}$.

Solving,

$$\begin{aligned}\Rightarrow \Big(\frac{2}{7}\Big)^2 \times \Big(\frac{7}{2}\Big)^{-3} \div \Big\lbrace\Big(\frac{7}{5}\Big)^{-2}\Big\rbrace^{-4}\\[1em] \Rightarrow \Big(\frac{2}{7}\Big)^2 \times \Big(\frac{2}{7}\Big)^{3} \div \Big(\frac{7}{5}\Big)^{8}\\[1em] \Rightarrow \Big(\frac{2}{7}\Big)^{2 + 3} \div \Big(\frac{7}{5}\Big)^{8}\\[1em] \Rightarrow \Big(\frac{2}{7}\Big)^{5} \times \Big(\frac{5}{7}\Big)^{8}\\[1em] \Rightarrow \frac{2^5}{7^5} \times \frac{5^8}{7^8}\\[1em] \Rightarrow \frac{2^5 \times 5^8}{7^{5+8}}\\[1em] \Rightarrow \frac{2^5 \times 5^8}{7^{13}}\end{aligned}$$

Hence, $\mathbf{\Big(\frac{2}{7}\Big)^2 \times \Big(\frac{7}{2}\Big)^{-3} \div \Big\lbrace\Big(\frac{7}{5}\Big)^{-2}\Big\rbrace^{-4} = \frac{2^5 \times 5^8}{7^{13}}}$.

Solving,

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

Hence, $\mathbf{\Big(\frac{4}{5}\Big)^2 \times 5^4 \times \Big(\frac{2}{5}\Big)^{-2} \div \Big(\frac{5}{2}\Big)^{-3} = \frac{5^7}{2^1}}$.

Solving,

We know that 8 = 2³

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

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

Simplifying,

$$\begin{aligned}\Rightarrow ((-2)^3)^2 + 5^{-3} \div 5^{-5} - \Big(-\frac{1}{2}\Big)^0\\[1em] \Rightarrow (-2)^{3 \times 2} + 5^{-3 - (-5)} - 1\\[1em] \Rightarrow (-2)^{6} + 5^{-3 + 5} - 1\\[1em] \Rightarrow 64 + 5^{2} - 1\\[1em] \Rightarrow 64 + 25 - 1\\[1em] \Rightarrow 88\\[1em] \Rightarrow 2 \times 2 \times 2 \times 11\\[1em] \Rightarrow 2^{3} \times 11\end{aligned}$$

Hence, $\mathbf{((-2)^3)^2 + 5^{-3} \div 5^{-5} - \Big(-\frac{1}{2}\Big)^0 = 2^3 \times 11}$.

Simplifying,

$$\begin{aligned}\Rightarrow 3^{-5} \times 3^2 \div 3^{-6} + (2^2 \times 3)^2 + \Big(\frac{2}{3}\Big)^{-1} + 2^{-1} + \Big(\frac{1}{19}\Big)^{-1}\\[1em] \Rightarrow 3^{-5 + 2 - (-6)} + (2^2)^2 \times 3^2 + \frac{3}{2} + \frac{1}{2} + 19\\[1em] \Rightarrow 3^{-5 + 2 + 6} + 2^4 \times 3^2 + \frac{3}{2} + \frac{1}{2} + 19\\[1em] \Rightarrow 3^{3} + (16 \times 9) + \frac{3 + 1}{2} + 19\\[1em] \Rightarrow 27 + 144 + \frac{4}{2} + 19\\[1em] \Rightarrow 27 + 144 + 2 + 19\\[1em] \Rightarrow 192\\[1em] \Rightarrow 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3\\[1em] \Rightarrow 2^6 \times 3^1\end{aligned}$$

Hence, $\mathbf{3^{-5} \times 3^2 \div 3^{-6} + (2^2 \times 3)^2 + \Big(\frac{2}{3}\Big)^{-1} + 2^{-1} + \Big(\frac{1}{19}\Big)^{-1} = 2^6 \times 3^1}$.

Simplifying,

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

Hence, $\mathbf{5^3 \times \Big(\frac{4}{5}\Big)^3 = \Big(\frac{1}{4}\Big)^{-3}}$.

Simplifying,

$$\begin{aligned}\Rightarrow \Big[\Big(\frac{3}{7}\Big)^{-2}\Big]^{-3}\\[1em] \Rightarrow \Big(\frac{3}{7}\Big)^{(-2) \times (-3)}\\[1em] \Rightarrow \Big(\frac{3}{7}\Big)^{6}\\[1em] \Rightarrow \Big(\frac{7}{3}\Big)^{-6}\end{aligned}$$

Hence, $\mathbf{\Big[\Big(\frac{3}{7}\Big)^{-2}\Big]^{-3} = \Big(\frac{7}{3}\Big)^{-6}}$.

Simplifying,

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

Hence, $\mathbf{\Big(\frac{5}{9}\Big)^{-2} \times \Big(\frac{5}{3}\Big)^2 \div \Big(\frac{1}{5}\Big)^{-2} = \Big(\frac{5}{3}\Big)^{-2}}$.

Simplifying,

$$\begin{aligned}\Rightarrow 2^{-1}\Big[\Big(\frac{5}{3}\Big)^4 + \Big(\frac{3}{5}\Big)^{-2}\Big] \div \frac{17}{9}\\[1em] \Rightarrow \frac{1}{2}\Big[\Big(\frac{5}{3}\Big)^4 + \Big(\frac{5}{3}\Big)^{2}\Big] \div \frac{17}{9}\\[1em] \Rightarrow \frac{1}{2}\Big[\frac{625}{81} + \frac{25}{9}\Big] \div \frac{17}{9}\end{aligned}$$

LCM of 81 and 9 is 81

$$\begin{aligned}\Rightarrow \frac{1}{2}\Big[\frac{625}{81} + \frac{25 \times 9}{9 \times 9}\Big] \div \frac{17}{9}\\[1em] \Rightarrow \frac{1}{2}\Big[\frac{625}{81} + \frac{225}{81}\Big] \div \frac{17}{9}\\[1em] \Rightarrow \frac{1}{2}\Big[\frac{625 + 225}{81}\Big] \div \frac{17}{9}\\[1em] \Rightarrow \frac{1}{2} \times \frac{850}{81} \times \frac{9}{17}\\[1em] \Rightarrow \frac{1 \times 850 \times 9}{2 \times 81 \times 17}\\[1em] \Rightarrow \frac{7650}{2754}\\[1em] \Rightarrow \frac{25}{9}\\[1em] \Rightarrow \Big(\frac{5}{3}\Big)^{2}\\[1em] \Rightarrow \Big(\frac{3}{5}\Big)^{-2}\end{aligned}$$

Hence, $\mathbf{2^{-1}\Big[\Big(\frac{5}{3}\Big)^4 + \Big(\frac{3}{5}\Big)^{-2}\Big] \div \frac{17}{9} = \Big(\frac{3}{5}\Big)^{-2}}$.

Simplifying,

$$\begin{aligned}\Rightarrow (-7)^3 \times \Big(\frac{1}{-7}\Big)^{-9} \div (-7)^{10}\\[1em] \Rightarrow (-7)^{3} \times (-7)^{9} \div (-7)^{10}\\[1em] \Rightarrow (-7)^{3 + 9 - 10}\\[1em] \Rightarrow (-7)^{2}\\[1em] \Rightarrow 7^{2}\\[1em] \Rightarrow \Big(\frac{1}{7}\Big)^{-2}\end{aligned}$$

Hence, $\mathbf{(-7)^3 \times \Big(\frac{1}{-7}\Big)^{-9} \div (-7)^{10} = \Big(\frac{1}{7}\Big)^{-2}}$.

We know that,

49 = 7²

Simplifying,

$$\begin{aligned}\Rightarrow \frac{49 \times z^{-3}}{7^{-3} \times 10 \times z^{-5}}\\[1em] \Rightarrow \frac{7^2 \times z^{-3}}{7^{-3} \times 10 \times z^{-5}}\\[1em] \Rightarrow \frac{7^{2 - (-3)} \times z^{-3 - (-5)}}{10}\\[1em] \Rightarrow \frac{7^{2 + 3} \times z^{-3 + 5}}{10}\\[1em] \Rightarrow \frac{7^{5} \times z^{2}}{10}\end{aligned}$$

Hence, $\mathbf{\frac{49 \times z^{-3}}{7^{-3} \times 10 \times z^{-5}} = \frac{7^{5} \times z^{2}}{10}}$.

We know that,

9 = 3² and 27 = 3³

Simplifying,

$$\begin{aligned}\Rightarrow \frac{9^3 \times 27 \times t^4}{(3)^2 \times (3)^4 \times t^2}\\[1em] \Rightarrow \frac{(3^2)^3 \times 3^3 \times t^4}{3^2 \times 3^4 \times t^2}\\[1em] \Rightarrow \frac{3^6 \times 3^3 \times t^4}{3^2 \times 3^4 \times t^2}\\[1em] \Rightarrow \frac{3^{6 + 3} \times t^4}{3^{2 + 4} \times t^2}\\[1em] \Rightarrow \frac{3^{9} \times t^4}{3^{6} \times t^2}\\[1em] \Rightarrow 3^{9 - 6} \times t^{4 - 2}\\[1em] \Rightarrow 3^{3} \times t^{2}\\[1em] \Rightarrow 27,t^2\end{aligned}$$

Hence, $\mathbf{\frac{9^3 \times 27 \times t^4}{(3)^2 \times (3)^4 \times t^2} = 27,t^2}$.

Simplifying,

$$\begin{aligned}\Rightarrow \frac{(3^{-2})^2 \times (5^2)^{-3} \times (t^{-3})^2}{(3^{-2})^5 \times (5^3)^{-2} \times (t^{-4})^3}\\[1em] \Rightarrow \frac{3^{-4} \times 5^{-6} \times t^{-6}}{3^{-10} \times 5^{-6} \times t^{-12}}\\[1em] \Rightarrow 3^{-4 - (-10)} \times 5^{-6 - (-6)} \times t^{-6 - (-12)}\\[1em] \Rightarrow 3^{-4 + 10} \times 5^{-6 + 6} \times t^{-6 + 12}\\[1em] \Rightarrow 3^{6} \times 5^{0} \times t^{6}\\[1em] \Rightarrow 3^{6} \times 1 \times t^{6}\\[1em] \Rightarrow 3^{6} \times t^{6}\end{aligned}$$

Hence, $\mathbf{\frac{(3^{-2})^2 \times (5^2)^{-3} \times (t^{-3})^2}{(3^{-2})^5 \times (5^3)^{-2} \times (t^{-4})^3} = 3^{6} \times t^{6}}$.

We know that,

15 = 3 x 5, 500 = 2² x 5³ and 6 = 2 x 3

Simplifying,

$$\begin{aligned}\Rightarrow \frac{2^{-5} \times 15^{-5} \times 500}{5^{-6} \times 6^{-5}}\\[1em] \Rightarrow \frac{2^{-5} \times (3 \times 5)^{-5} \times (2^2 \times 5^3)}{5^{-6} \times (2 \times 3)^{-5}}\\[1em] \Rightarrow \frac{2^{-5} \times 3^{-5} \times 5^{-5} \times 2^2 \times 5^3}{5^{-6} \times 2^{-5} \times 3^{-5}}\\[1em] \Rightarrow \frac{2^{-5 + 2} \times 3^{-5} \times 5^{-5 + 3}}{2^{-5} \times 3^{-5} \times 5^{-6}}\\[1em] \Rightarrow \frac{2^{-3} \times 3^{-5} \times 5^{-2}}{2^{-5} \times 3^{-5} \times 5^{-6}}\\[1em] \Rightarrow 2^{-3 - (-5)} \times 3^{-5 - (-5)} \times 5^{-2 - (-6)}\\[1em] \Rightarrow 2^{-3 + 5} \times 3^{0} \times 5^{-2 + 6}\\[1em] \Rightarrow 2^{2} \times 1 \times 5^{4}\\[1em] \Rightarrow 4 \times 625\\[1em] \Rightarrow 2500\end{aligned}$$

Hence, $\mathbf{\frac{2^{-5} \times 15^{-5} \times 500}{5^{-6} \times 6^{-5}} = 2500}$.

Let the required number be $x$. So,

$$\begin{aligned}\Rightarrow \Big(\frac{3}{-2}\Big)^{-3} \div x = \Big(\frac{2}{3}\Big)^2\\[1em] \Rightarrow \Big(\frac{-2}{3}\Big)^{3} \div x = \Big(\frac{2}{3}\Big)^2\\[1em] \Rightarrow \frac{-8}{27} \div x = \frac{4}{9}\\[1em] \Rightarrow \frac{-8}{27} \times \frac{1}{x} = \frac{4}{9}\\[1em] \Rightarrow \frac{1}{x} = \frac{4}{9} \times \frac{27}{-8}\\[1em] \Rightarrow \frac{1}{x} = \frac{4 \times 27}{9 \times (-8)}\\[1em] \Rightarrow \frac{1}{x} = \frac{108}{-72}\\[1em] \Rightarrow \frac{1}{x} = \frac{-3}{2}\\[1em] \Rightarrow x = \frac{-2}{3}\end{aligned}$$

Hence, $\mathbf{\Big(\frac{3}{-2}\Big)^{-3}}$ should be divided by $\mathbf{\frac{-2}{3}}$ to get $\mathbf{\Big(\frac{2}{3}\Big)^2}$.

We know that 9 = 3²

$$\begin{aligned}\Rightarrow 9^m \div 3^{-2} = 9^4\\[1em] \Rightarrow (3^2)^m \div 3^{-2} = (3^2)^4\\[1em] \Rightarrow 3^{2m} \div 3^{-2} = 3^{8}\\[1em] \Rightarrow 3^{2m - (-2)} = 3^{8}\\[1em] \Rightarrow 3^{2m + 2} = 3^{8}\end{aligned}$$

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

$$\begin{aligned}\Rightarrow 2m + 2 = 8\\[1em] \Rightarrow 2m = 8 - 2\\[1em] \Rightarrow 2m = 6\\[1em] \Rightarrow m = \frac{6}{2}\\[1em] \Rightarrow m = 3\end{aligned}$$

Hence, the value of m is 3.

Solving,

$$\begin{aligned}\Rightarrow \Big(\frac{-5}{7}\Big)^{-4} \times \Big(\frac{-5}{7}\Big)^{12} = \Big\lbrace\Big(\frac{-5}{7}\Big)^3\Big\rbrace^x \times \Big(\frac{-5}{7}\Big)^{-1}\\[1em] \Rightarrow \Big(\frac{-5}{7}\Big)^{-4 + 12} = \Big(\frac{-5}{7}\Big)^{3x} \times \Big(\frac{-5}{7}\Big)^{-1}\\[1em] \Rightarrow \Big(\frac{-5}{7}\Big)^{8} = \Big(\frac{-5}{7}\Big)^{3x - 1}\end{aligned}$$

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

$$\begin{aligned}\Rightarrow 8 = 3x - 1\\[1em] \Rightarrow 8 + 1 = 3x\\[1em] \Rightarrow 9 = 3x\\[1em] \Rightarrow x = \frac{9}{3}\\[1em] \Rightarrow x = 3\end{aligned}$$

Hence, the value of x is 3.

Solving,

$$\begin{aligned}\Rightarrow \Big(\frac{-2}{3}\Big)^{-13} \times \Big(\frac{3}{-2}\Big)^{8} = \Big(\frac{-2}{3}\Big)^{-2x+1}\\[1em] \Rightarrow \Big(\frac{-2}{3}\Big)^{-13} \times \Big(\frac{-2}{3}\Big)^{-8} = \Big(\frac{-2}{3}\Big)^{-2x+1}\\[1em] \Rightarrow \Big(\frac{-2}{3}\Big)^{-13 + (-8)} = \Big(\frac{-2}{3}\Big)^{-2x+1}\\[1em] \Rightarrow \Big(\frac{-2}{3}\Big)^{-21} = \Big(\frac{-2}{3}\Big)^{-2x+1}\end{aligned}$$

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

$$\begin{aligned}\Rightarrow -21 = -2x + 1\\[1em] \Rightarrow -21 - 1 = -2x\\[1em] \Rightarrow -22 = -2x\\[1em] \Rightarrow x = \frac{-22}{-2}\\[1em] \Rightarrow x = 11\end{aligned}$$

Hence, the value of x is 11.

We know that 125 = 5³

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

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

$$\begin{aligned}\Rightarrow 2x - 1 = -3x + 9\\[1em] \Rightarrow 2x + 3x = 9 + 1\\[1em] \Rightarrow 5x = 10\\[1em] \Rightarrow x = \frac{10}{5}\\[1em] \Rightarrow x = 2\end{aligned}$$

Hence, the value of x is 2.

We know that 9 = 3², 27 = 3³ and 81 = 3⁴

$$\begin{aligned}\Rightarrow \frac{9^n \times 3^5 \times (27)^3}{3 \times (81)^4} = 27\\[1em] \Rightarrow \frac{(3^2)^n \times 3^5 \times (3^3)^3}{3 \times (3^4)^4} = 3^3\\[1em] \Rightarrow \frac{3^{2n} \times 3^5 \times 3^{9}}{3^1 \times 3^{16}} = 3^3\\[1em] \Rightarrow \frac{3^{2n + 5 + 9}}{3^{1 + 16}} = 3^3\\[1em] \Rightarrow \frac{3^{2n + 14}}{3^{17}} = 3^3\\[1em] \Rightarrow 3^{2n + 14 - 17} = 3^3\\[1em] \Rightarrow 3^{2n - 3} = 3^3\end{aligned}$$

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

$$\begin{aligned}\Rightarrow 2n - 3 = 3\\[1em] \Rightarrow 2n = 3 + 3\\[1em] \Rightarrow 2n = 6\\[1em] \Rightarrow n = \frac{6}{2}\\[1em] \Rightarrow n = 3\end{aligned}$$

Hence, the value of n is 3.

(i) Solving,

$$\begin{aligned}\Rightarrow \Big(\frac{-4}{5}\Big)^2\\[1em] \Rightarrow \frac{(-4) \times (-4)}{5 \times 5}\\[1em] \Rightarrow \frac{16}{25}\end{aligned}$$

The multiplicative inverse of a non-zero number is its reciprocal.

Thus, multiplicative inverse of $\frac{16}{25} = \frac{25}{16}$.

Hence, multiplicative inverse of $\mathbf{\Big(\frac{-4}{5}\Big)^2} = \mathbf{\frac{25}{16}}$.

(ii) As 8 is an even number and a negative number raised to an even power gives a positive value,

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

Thus, 7⁸ = (-7)⁸.

Solving,

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

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

$$\begin{aligned}\Rightarrow x + 3 = 8\\[1em] \Rightarrow x = 8 - 3\\[1em] \Rightarrow x = 5\end{aligned}$$

Hence, the value of x is 5.

(iii) We know that 729 = 3 x 3 x 3 x 3 x 3 x 3 = 3⁶

$$\begin{aligned}\Rightarrow 3^n = 729\\[1em] \Rightarrow 3^n = 3^6\end{aligned}$$

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

$\Rightarrow n = 6$

Hence, the value of n is 6.

(iv) We know that,

a⁰ = 1

⇒ 7⁰ + 5⁰

⇒ 1 + 1

⇒ 2

Hence, the value of (7⁰ + 5⁰) is 2.

(i) Solving,

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

Since the value of (-3)^-2 is $\frac{1}{9}$ and not $\frac{1}{-9}$, the statement is false.

Hence, the given statement is False.

(ii) The law of exponents $a^m \times a^n = a^{m+n}$ holds for all integers m and n (positive, negative or zero), where a ≠ 0.

Since m and n are negative integers, this law is still valid.

Hence, the given statement is True.

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

Solving L.H.S.,

⇒ 3⁰ × 4⁰

⇒ 1 × 1

⇒ 1

Solving R.H.S.,

⇒ 12⁰ = 1

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

Hence, the given statement is True.

(iv) Solving L.H.S.,

⇒ 3⁷ - 2⁷

⇒ 2187 - 128

⇒ 2059

Solving R.H.S.,

⇒ 1⁷ = 1

Since, L.H.S. ≠ R.H.S.

Hence, the given statement is False.

(v) Solving,

⇒ (-2)⁵

⇒ (-2) x (-2) x (-2) x (-2) x (-2)

⇒ -32

The multiplicative inverse of -32 is $\frac{1}{-32}$.

Hence, the given statement is True.

Solving,

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

Hence, option 1 is the correct option.

Solving,

$\Rightarrow \Big[\Big(\frac{1}{4}\Big)^{-2} + \Big(\frac{1}{3}\Big)^{-2}\Big] \div \Big(\frac{1}{5}\Big)^{-2}$

⇒ [4² + 3²] ÷ 5²

⇒ [16 + 9] ÷ 25

⇒ 25 ÷ 25

⇒ 1

Hence, option 3 is the correct option.

We know that 4 = 2², so 2⁴ = (2²)² = 4²

$$\begin{aligned}\Rightarrow 2^4 \times 4^3 = 4^x\\[1em] \Rightarrow 4^2 \times 4^3 = 4^x\\[1em] \Rightarrow 4^{2 + 3} = 4^x\\[1em] \Rightarrow 4^5 = 4^x\end{aligned}$$

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

$\Rightarrow x = 5$

Hence, option 1 is the correct option.

Solving,

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

Hence, option 2 is the correct option.

Solving,

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

The multiplicative inverse of $\frac{16}{81}$ is $\frac{81}{16}$.

Hence, option 3 is the correct option.

We know that $a^0 = 1$

$\Rightarrow 2^0 + 3^0 + \Big(\frac{1}{4}\Big)^0$

⇒ 1 + 1 + 1

⇒ 3

Hence, option 4 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \Big(\frac{12}{13}\Big)^4 \times \Big(\frac{13}{12}\Big)^{-8} = \Big(\frac{12}{13}\Big)^{2x}\\[1em] \Rightarrow \Big(\frac{12}{13}\Big)^4 \times \Big(\frac{12}{13}\Big)^{8} = \Big(\frac{12}{13}\Big)^{2x}\\[1em] \Rightarrow \Big(\frac{12}{13}\Big)^{4 + 8} = \Big(\frac{12}{13}\Big)^{2x}\\[1em] \Rightarrow \Big(\frac{12}{13}\Big)^{12} = \Big(\frac{12}{13}\Big)^{2x}\end{aligned}$$

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

$$\begin{aligned}\Rightarrow 12 = 2x\\[1em] \Rightarrow x = \frac{12}{2}\\[1em] \Rightarrow x = 6\end{aligned}$$

Hence, option 2 is the correct option.

Solving,

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

So, the value is 16 and Statement I is true.

$a^{-m} = \frac{1}{a^m}$, where a is any non-zero integer and m is a positive integer, is a standard law of exponents.

So, Statement II is true.

Hence, option 3 is the correct option.

For a number to have a multiplicative inverse, its product with the inverse must equal 1. Since 0 multiplied by any number gives 0 (never 1), the multiplicative inverse of 0 does not exist.

So, Statement I is false.

$\Big(\frac{p}{q}\Big)^{-m} = \frac{1}{\Big(\frac{p}{q}\Big)^m} = \Big(\frac{q}{p}\Big)^m = \frac{q^m}{p^m}$, where p and q are non-zero integers and m is a positive integer, is a standard law of exponents.

So, Statement II is true.

Hence, option 2 is the correct option.

Solving L.H.S.,

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

Since, L.H.S. = R.H.S., Statement I is true.

$a^m - a^n = a^{m-n}$ is not a valid law of exponents. Subtraction of powers does not simplify by subtracting exponents.

So, Statement II is false.

Hence, option 1 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \Big(\frac{5}{4}\Big)^{-3}\\[1em] \Rightarrow \Big(\frac{4}{5}\Big)^{3}\\[1em] \Rightarrow \frac{4 \times 4 \times 4}{5 \times 5 \times 5}\\[1em] \Rightarrow \frac{64}{125}\end{aligned}$$

Hence, $\mathbf{\Big(\frac{5}{4}\Big)^{-3} = \frac{64}{125}}$.

Solving,

$$\begin{aligned}\Rightarrow \Big(\frac{-3}{7}\Big)^{-2}\\[1em] \Rightarrow \Big(\frac{7}{-3}\Big)^{2}\\[1em] \Rightarrow \frac{7 \times 7}{(-3) \times (-3)}\\[1em] \Rightarrow \frac{49}{9}\\[1em] \Rightarrow 5\frac{4}{9}\end{aligned}$$

Hence, $\mathbf{\Big(\frac{-3}{7}\Big)^{-2} = 5\frac{4}{9}}$.

Solving,

$\Rightarrow \Big[\Big(\frac{1}{2}\Big)^{-3} + \Big(\frac{1}{3}\Big)^{-3} + \Big(\frac{1}{4}\Big)^{-3}\Big] \div (3)^2$

⇒ [2³ + 3³ + 4³] ÷ 3²

⇒ [8 + 27 + 64] ÷ 9

⇒ 99 ÷ 9

$\Rightarrow \frac{99}{9}$

⇒ 11

Hence, the value of the given expression is 11.

Solving,

$$\begin{aligned}\Rightarrow \Big[3^3 - \Big(\frac{1}{2}\Big)^{-3}\Big] \times \frac{1}{19}\\[1em] \Rightarrow \Big[3^3 - 2^3\Big] \times \frac{1}{19}\\[1em] \Rightarrow [27 - 8] \times \frac{1}{19}\\[1em] \Rightarrow 19 \times \frac{1}{19}\\[1em] \Rightarrow \frac{19}{19}\\[1em] \Rightarrow 1\end{aligned}$$

Hence, $\mathbf{\Big[3^3 - \Big(\frac{1}{2}\Big)^{-3}\Big] \times \frac{1}{19} = 1}$.

Solving,

$$\begin{aligned}\Rightarrow [2^{-1} + 3^0 + 5^1 + 7^2 + 9^3] \div \Big(\frac{2}{3}\Big)^{-1}\\[1em] \Rightarrow \Big[\frac{1}{2} + 1 + 5 + 49 + 729\Big] \div \Big(\frac{3}{2}\Big)\\[1em] \Rightarrow \Big[\frac{1}{2} + 784\Big] \div \Big(\frac{3}{2}\Big)\\[1em] \Rightarrow \Big[\frac{1 + 1568}{2}\Big] \div \Big(\frac{3}{2}\Big)\\[1em] \Rightarrow \frac{1569}{2} \times \frac{2}{3}\\[1em] \Rightarrow \frac{1569 \times 2}{2 \times 3}\\[1em] \Rightarrow \frac{1569}{3}\\[1em] \Rightarrow 523\end{aligned}$$

Hence, $\mathbf{[2^{-1} + 3^0 + 5^1 + 7^2 + 9^3] \div \Big(\frac{2}{3}\Big)^{-1} = 523}$.

Solving,

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

LCM of 7 and 8 is 56, and LCM of 3 and 4 is 12

$$\begin{aligned}\Rightarrow \Big(\frac{8}{56} - \frac{7}{56}\Big)^{-1} - \Big(\frac{4}{12} - \frac{3}{12}\Big)^{-1}\\[1em] \Rightarrow \Big(\frac{8 - 7}{56}\Big)^{-1} - \Big(\frac{4 - 3}{12}\Big)^{-1}\\[1em] \Rightarrow \Big(\frac{1}{56}\Big)^{-1} - \Big(\frac{1}{12}\Big)^{-1}\\[1em] \Rightarrow 56 - 12\\[1em] \Rightarrow 44\end{aligned}$$

Hence, $(7^{-1} - 8^{-1})^{-1} - (3^{-1} - 4^{-1})^{-1} = 44$.

Solving,

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

Hence, the multiplicative inverse of $\mathbf{\Big(\frac{5}{3}\Big)^{-2}}$ is $\mathbf{\frac{25}{9}}$.

We know that 15 = 3 x 5, 625 = 5⁴ and 10 = 2 x 5

$$\begin{aligned}\Rightarrow \frac{2^{-4} \times 15^{-3} \times 625}{5^2 \times 10^{-4}}\\[1em] \Rightarrow \frac{2^{-4} \times (3 \times 5)^{-3} \times 5^4}{5^2 \times (2 \times 5)^{-4}}\\[1em] \Rightarrow \frac{2^{-4} \times 3^{-3} \times 5^{-3} \times 5^4}{5^2 \times 2^{-4} \times 5^{-4}}\\[1em] \Rightarrow \frac{2^{-4} \times 3^{-3} \times 5^{-3 + 4}}{2^{-4} \times 5^{2 - 4}}\\[1em] \Rightarrow \frac{2^{-4} \times 3^{-3} \times 5^{1}}{2^{-4} \times 5^{-2}}\\[1em] \Rightarrow 2^{-4 - (-4)} \times 3^{-3} \times 5^{1 - (-2)}\\[1em] \Rightarrow 2^{0} \times 3^{-3} \times 5^{3}\\[1em] \Rightarrow \frac{5^3}{3^3}\\[1em] \Rightarrow \Big(\frac{5}{3}\Big)^{3}\end{aligned}$$

Hence, $\mathbf{\frac{2^{-4} \times 15^{-3} \times 625}{5^2 \times 10^{-4}} = \Big(\frac{5}{3}\Big)^{3}}$.

Solving,

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

Hence, $\mathbf{\Big(\frac{9}{4}\Big)^{-1} \times \Big(\frac{2}{3}\Big)^2 \div \Big(\frac{3}{2}\Big)^{-2} = \frac{4}{9}}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{5^{n+2} - 6 \times 5^{n+1}}{13 \times 5^n - 2 \times 5^{n+1}}\\[1em] \Rightarrow \frac{5^{n} \times 5^2 - 6 \times 5^{n} \times 5}{13 \times 5^n - 2 \times 5^{n} \times 5}\\[1em] \Rightarrow \frac{5^{n}(5^2 - 6 \times 5)}{5^n(13 - 2 \times 5)}\\[1em] \Rightarrow \frac{5^{n}(25 - 30)}{5^n(13 - 10)}\\[1em] \Rightarrow \frac{25 - 30}{13 - 10}\\[1em] \Rightarrow \frac{-5}{3}\end{aligned}$$

Hence, $\mathbf{\frac{5^{n+2} - 6 \times 5^{n+1}}{13 \times 5^n - 2 \times 5^{n+1}} = \frac{-5}{3}}$.

To prove: $(a + b)^{-1}(a^{-1} + b^{-1}) = (ab)^{-1}$

Taking L.H.S.,

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

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

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

We know that $\frac{-8}{343} = \frac{(-2)^3}{7^3} = \Big(\frac{-2}{7}\Big)^3$

$$\begin{aligned}\Rightarrow \Big\lbrace\Big(\frac{-2}{7}\Big)^2\Big\rbrace^x \times \Big(\frac{-7}{2}\Big)^{-1} = \frac{-8}{343}\\[1em] \Rightarrow \Big(\frac{-2}{7}\Big)^{2x} \times \Big(\frac{-2}{7}\Big)^{1} = \Big(\frac{-2}{7}\Big)^{3}\\[1em] \Rightarrow \Big(\frac{-2}{7}\Big)^{2x + 1} = \Big(\frac{-2}{7}\Big)^{3}\end{aligned}$$

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

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

Hence, the value of x is 1.

We know that 8 = 2³, 16 = 2⁴ and 4 = 2²

$$\begin{aligned}\Rightarrow \frac{2^{-n} \times 8^{2n+1} \times 16^{2n}}{4^{3n}} = \frac{1}{16}\\[1em] \Rightarrow \frac{2^{-n} \times (2^3)^{2n+1} \times (2^4)^{2n}}{(2^2)^{3n}} = \frac{1}{2^4}\\[1em] \Rightarrow \frac{2^{-n} \times 2^{6n+3} \times 2^{8n}}{2^{6n}} = 2^{-4}\\[1em] \Rightarrow \frac{2^{-n + 6n + 3 + 8n}}{2^{6n}} = 2^{-4}\\[1em] \Rightarrow \frac{2^{13n + 3}}{2^{6n}} = 2^{-4}\\[1em] \Rightarrow 2^{13n + 3 - 6n} = 2^{-4}\\[1em] \Rightarrow 2^{7n + 3} = 2^{-4}\end{aligned}$$

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

$$\begin{aligned}\Rightarrow 7n + 3 = -4\\[1em] \Rightarrow 7n = -4 - 3\\[1em] \Rightarrow 7n = -7\\[1em] \Rightarrow n = \frac{-7}{7}\\[1em] \Rightarrow n = -1\end{aligned}$$

Hence, the value of n is -1.

Let the required number be $x$. So,

$$\begin{aligned}\Rightarrow \Big[\Big(\frac{-5}{2}\Big)^3\Big]^{-3} \times x = \Big(\frac{-2}{5}\Big)^5\\[1em] \Rightarrow \Big(\frac{-5}{2}\Big)^{-9} \times x = \Big(\frac{-2}{5}\Big)^5\\[1em] \Rightarrow \Big(\frac{-2}{5}\Big)^{9} \times x = \Big(\frac{-2}{5}\Big)^5\\[1em] \Rightarrow x = \Big(\frac{-2}{5}\Big)^5 \div \Big(\frac{-2}{5}\Big)^9\\[1em] \Rightarrow x = \Big(\frac{-2}{5}\Big)^{5 - 9}\\[1em] \Rightarrow x = \Big(\frac{-2}{5}\Big)^{-4}\\[1em] \Rightarrow x = \Big(\frac{-5}{2}\Big)^{4}\\[1em] \Rightarrow x = \frac{(-5)^4}{2^4}\\[1em] \Rightarrow x = \frac{625}{16}\end{aligned}$$

Hence, $\mathbf{\Big[\Big(\frac{-5}{2}\Big)^3\Big]^{-3}}$ should be multiplied by $\mathbf{\frac{625}{16}}$ to get $\mathbf{\Big(\frac{-2}{5}\Big)^5}$.

Let the required number be $x$. So,

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

Hence, $\mathbf{\Big(\frac{-3}{2}\Big)^{-4}}$ should be divided by $\mathbf{\frac{-2}{3}}$ to get $\mathbf{\Big(\frac{-2}{3}\Big)^3}$.

We know that 9 = 3² and 4 = 2²

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

So, $\frac{p}{q} = \frac{2}{3}$

Now,

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

Hence, $\mathbf{\Big(\frac{p}{q}\Big)^{-2} = \frac{9}{4}}$.

We know that $a^0 = 1$

⇒ (5⁰ + 3⁰)(5⁰ - 3⁰)

⇒ (1 + 1)(1 - 1)

⇒ 2 x 0

⇒ 0

Since the multiplicative inverse of a number is obtained by taking its reciprocal, and the reciprocal of 0 is not defined, the multiplicative inverse of 0 does not exist.

Hence, the multiplicative inverse of (5⁰ + 3⁰)(5⁰ - 3⁰) does not exist.

Solving,

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

Hence, $\mathbf{\Big(\frac{x^p}{x^q}\Big)^{p+q} \times \Big(\frac{x^q}{x^r}\Big)^{q+r} \times \Big(\frac{x^r}{x^p}\Big)^{r+p} = 1}$.

Consider the second term.

Since $x^{p-q} = \frac{x^p}{x^q}$,

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

Given,

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

Hence, $\mathbf{\frac{x^p}{x^p + x^q} + \frac{1}{x^{p-q} + 1} = 1}$.