Exponents (Powers)

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

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

Hence,

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

Hence, option 3 is the correct option.

$$\begin{aligned}\Big(\frac{2}{3}\Big)^3 \times \Big(\frac{3}{2}\Big)^6\\[1em] = \Big(\frac{2 \times 2 \times 2}{3 \times 3 \times 3}\Big) \times \Big(\frac{3 \times 3 \times 3 \times 3 \times 3 \times 3}{2 \times 2 \times 2 \times 2 \times 2 \times 2}\Big)\\[1em] = \Big(\frac{8}{27}\Big) \times \Big(\frac{729}{64}\Big)\\[1em] = \Big(\frac{8 \times 729}{27 \times 64}\Big)\\[1em] = \frac{5832}{1728} \\[1em] = \frac{27}{8}\end{aligned}$$

Hence, option 2 is the correct option.

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

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

And $a^0 = 1$

Hence,

$$\begin{aligned}8^0 + 8^{-1} + 4^{-1}\\[1em] = 1 + \frac{1}{8} + \frac{1}{4} \\[1em] = \frac{1}{1} + \frac{1}{8} + \frac{1}{4}\end{aligned}$$

LCM of 1, 8 and 4 is 2 x 2 x 2 = 8

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

Hence, option 2 is the correct option.

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

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

$$\begin{aligned}(-5)^5 \times (-5)^{-3}\\[1em] = (-5)^5 \times \Big(\frac{-1}{5}\Big)^3\\[1em] = (-5 \times -5 \times -5 \times -5 \times -5) \times \Big(\frac{-1 \times -1 \times -1}{5 \times 5 \times 5}\Big)\\[1em] = -3125 \times \Big(\frac{-1}{125}\Big)\\[1em] = \Big(\frac{-3125 \times -1}{125}\Big)\\[1em] = \Big(\frac{3125}{125}\Big)\\[1em] = 25\end{aligned}$$

Hence, option 4 is the correct option.

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

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

Hence,

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

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

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

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

Hence,

$$\begin{aligned}(3^{-1} \times 4^{-1}) ÷ 6^{-1}\\[1em] = \Big(\frac{1}{3}^1 \times \frac{1}{4}^1\Big) ÷ \Big(\frac{1}{6}\Big)^1\\[1em] = \Big(\frac{1 \times 1}{3 \times 4}\Big) ÷ \Big(\frac{1}{6}\Big)\\[1em] = \Big(\frac{1}{12}\Big) ÷ \Big(\frac{1}{6}\Big)\\[1em] = \Big(\frac{1}{12}\Big) \times \Big(\frac{6}{1}\Big)\\[1em] = \Big(\frac{1 \times 6}{12 \times 1}\Big)\\[1em] = \frac{6}{12}\\[1em] = \frac{1}{2}\end{aligned}$$

Hence, $(3^{-1} \times 4^{-1}) ÷ 6^{-1} = \frac{1}{2}$

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

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

Hence,

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

LCM of 2 and 3 is 2 x 3 = 6

$$\begin{aligned}= \Big(\frac{1 \times 3}{2 \times 3} + \frac{1 \times 2}{3 \times 2}\Big)^3\\[1em] = \Big(\frac{3}{6} + \frac{2}{6}\Big)^3\\[1em] = \Big(\frac{3 + 2}{6}\Big)^3\\[1em] = \Big(\frac{5}{6}\Big)^3\\[1em] = \Big(\frac{5 \times 5 \times 5}{6 \times 6 \times 6}\Big)\\[1em] = \frac{125}{216}\end{aligned}$$

Hence,$(2^{-1} + 3^{-1})^3 = \frac{125}{216}$

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

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

Hence,

$$\begin{aligned}(3^{-1} ÷ 4^{-1})^2\\[1em] = \Big(\frac{1}{3}^1 ÷ \frac{1}{4}^1\Big)^2\\[1em] = \Big(\frac{1}{3} ÷ \frac{1}{4}\Big)^2\\[1em] = \Big(\frac{1}{3} \times \frac{4}{1}\Big)^2\\[1em] = \Big(\frac{1 \times 4}{3 \times 1}\Big)^2\\[1em] = \Big(\frac{4}{3}\Big)^2\\[1em] = \Big(\frac{4 \times 4}{3 \times 3}\Big)\\[1em] = \Big(\frac{16}{9}\Big)\\[1em] = 1\Big(\frac{7}{9}\Big)\end{aligned}$$

Hence, $(3^{-1} ÷ 4^{-1})^2 = 1\Big(\frac{7}{9}\Big)$

$$\begin{aligned}(2^2 + 3^2) \times \Big(\frac{1}{2}\Big)^2\\[1em] = (2 \times 2 + 3 \times 3) \times \Big(\frac{1 \times 1}{2 \times 2}\Big)\\[1em] = (4 + 9) \times \Big(\frac{1}{4}\Big)\\[1em] = (13) \times \Big(\frac{1}{4}\Big)\\[1em] = \Big(\frac{13 \times 1}{4}\Big)\\[1em] = \Big(\frac{13}{4}\Big)\\[1em] = 3\Big(\frac{1}{4}\Big)\end{aligned}$$

Hence, $(2^2 + 3^2) \times \Big(\frac{1}{2}\Big)^2 = 3\Big(\frac{1}{4}\Big)$

$$\begin{aligned}(5^2 - 3^2) \times \Big(\frac{2}{3}\Big)^{-3}\\[1em] = (5 \times 5 - 3 \times 3) \times \Big(\frac{3}{2}\Big)^3\\[1em] = (25 - 9) \times \Big(\frac{3 \times 3 \times 3}{2 \times 2 \times 2}\Big)\\[1em] = (16) \times \Big(\frac{27}{8}\Big)\\[1em] = \Big(\frac{16 \times 27}{8}\Big)\\[1em] = \Big(\frac{432}{8}\Big)\\[1em] = 54\end{aligned}$$

Hence, $(5^2 - 3^2) \times \Big(\frac{2}{3}\Big)^{-3} = 54$

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

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

$$\begin{aligned}\Big[\Big(\frac{1}{4}\Big)^{-3} - \Big(\frac{1}{3}\Big)^{-3}\Big] ÷ \Big(\frac{1}{6}\Big)^{-3}\\[1em] = \Big[\Big(\frac{4}{1}\Big)^3 - \Big(\frac{3}{1}\Big)^3\Big] ÷ \Big(\frac{6}{1}\Big)^3\\[1em] = [(4 \times 4 \times 4) - (3 \times 3 \times 3)] ÷ (6 \times 6 \times 6)\\[1em] = [64 - 27] ÷ (216)\\[1em] = [37] ÷ (216)\\[1em] = [37] \times \Big(\frac{1}{216}\Big)\\[1em] = \Big(\frac{37 \times 1}{216}\Big)\\[1em] = \Big(\frac{37}{216}\Big)\end{aligned}$$

Hence, $\Big[\Big(\frac{1}{4}\Big)^{-3} - \Big(\frac{1}{3}\Big)^{-3}\Big] ÷ \Big(\frac{1}{6}\Big)^{-3} = \frac{37}{216}$.

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

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

$$\begin{aligned}\Big[\Big(-\frac{3}{4}\Big)^{-2}\Big]^2\\[1em] = \Big[\Big(\frac{-4}{3}\Big)^2\Big]^2\\[1em] = \Big[\Big(\frac{-4 \times (-4)}{3 \times 3}\Big)\Big]^2\\[1em] = \Big[\Big(\frac{16}{9}\Big)\Big]^2\\[1em] = \Big(\frac{16 \times 16}{9 \times 9}\Big)\\[1em] = \Big(\frac{256}{81}\Big)\\[1em] = 3\frac{13}{81}\end{aligned}$$

Hence, $\Big[\Big(-\frac{3}{4}\Big)^{-2}\Big]^2 = 3\frac{13}{81}$

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

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

$$\begin{aligned}\Big[\Big(\frac{3}{5}\Big)^{-2}\Big]^{-2}\\[1em] = \Big[\Big(\frac{5}{3}\Big)^2\Big]^{-2}\\[1em] = \Big[\Big(\frac{5 \times 5}{3 \times 3}\Big)\Big]^{-2}\\[1em] = \Big(\frac{25}{9}\Big)^{-2}\\[1em] = \Big(\frac{9}{25}\Big)^2\\[1em] = \frac{9 \times 9}{25 \times 25}\\[1em] = \frac{81}{625}\end{aligned}$$

Hence, $\Big[\Big(\frac{3}{5}\Big)^{-2}\Big]^{-2} = \frac{81}{625}$

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

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

$$\begin{aligned}(5^{-1} \times 3^{-1}) ÷ 6^{-1}\\[1em] = \Big(\frac{1}{5}^1 \times \frac{1}{3}^1\Big) ÷ \frac{1}{6}^1\\[1em] = \Big(\frac{1 \times 1}{5 \times 3}\Big) ÷ \frac{1}{6}^1\\[1em] = \Big(\frac{1}{15}\Big) ÷ \frac{1}{6}\\[1em] = \Big(\frac{1}{15}\Big) \times \frac{6}{1}\\[1em] = \Big(\frac{1 \times 6}{15 \times 1}\Big)\\[1em] = \frac{6}{15}\\[1em] = \frac{2}{5}\end{aligned}$$

Hence, $(5^{-1} \times 3^{-1}) ÷ 6^{-1} = \frac{2}{5}$

As we know by Prime factorization of 1125, 1125 = 3 x 3 x 5 x 5 x 5

1125 = 3² x 5³

3^m x 5ⁿ = 3² x 5³

Hence, m = 2 and n = 3.

$$\begin{aligned}9 \times 3^x = (27)^{2x - 3}\\[1em] \Rightarrow 3^2 \times 3^x = (3^3)^{2x - 3}\\[1em] \Rightarrow 3^2 \times 3^x = (3)^{3(2x - 3)}\\[1em] \Rightarrow 3^2 \times 3^x = (3)^{6x - 9}\\[1em] \Rightarrow 3^2 = (3)^{6x - 9} ÷ 3^x\\[1em] \Rightarrow 3^2 = (3)^{6x - 9 - x} \space [∵a^m ÷ a^n = a^{m-n}] \\[1em] \Rightarrow 3^2 = (3)^{5x - 9}\\[1em] \Rightarrow 3^2 = 3^{5x} ÷ 3^9\\[1em] \Rightarrow 3^2 \times 3^9 = 3^{5x}\\[1em] \Rightarrow 3^{2 + 9} = 3^{5x} \space [∵a^m \times a^n = a^{m+n}] \\[1em] \Rightarrow 3^{11} = 3^{5x}\\[1em] \Rightarrow 5x = 11 \space [∵a^m = a^n ⟹m = n] \\[1em] \Rightarrow x = \frac{11}{5} \\[1em] \Rightarrow x = 2\frac{1}{5}\end{aligned}$$

If $9 \times 3^x = (27)^{2x-3}$, then $x = 2\frac{1}{5}$.

$x = 3^m$ and $y = 3^{m + 2}$. Hence,

$$\begin{aligned}\frac{x}{y} = \frac{3^m}{3^{m + 2}}\\[1em] = \frac{3^m}{3^m \times 3^2}\\[1em] = \frac{1}{3^2}\\[1em] = \frac{1}{9}\end{aligned}$$

Hence, option 2 is the correct option.

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

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

$$\begin{aligned}\Big[\frac{x^{-2}}{3y^{-1}}\Big]^{-1}\\[1em] = \Big[\frac{y^1}{3x^2}\Big]^{-1}\\[1em] = \Big[\frac{3x^2}{y}\Big]\end{aligned}$$

Hence, option 1 is the correct option.

According to product property,

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

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

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

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

Hence, option 3 is the correct option.

$$\begin{aligned}\Big[\frac{m}{n}\Big]^{x-1} = \Big[\frac{n}{m}\Big]^{x-5}\\[1em] \Rightarrow \Big[\frac{m}{n}\Big]^{x-1} = \Big[\frac{m}{n}\Big]^{5-x}\end{aligned}$$

According to the property,$a^m = a^n \Rightarrow m = n$

$$\begin{aligned}\Rightarrow (x - 1) = (5 - x)\\[1em] \Rightarrow x + x = 5 + 1\\[1em] \Rightarrow 2x = 6\\[1em] \Rightarrow x = \frac{6}{2}\\[1em] \Rightarrow x = 3\end{aligned}$$

Hence, option 1 is the correct option.

$$\begin{aligned}\Big[\frac{1}{7}\Big]^{-3} \times 7^{-1} \times \Big[\frac{1}{49}\Big]\\[1em] = \Big[\frac{7}{1}\Big]^3 \times \Big[\frac{1}{7}\Big]^1 \times \Big[\frac{1}{49}\Big]\\[1em] = \Big[\frac{7 \times 7 \times 7}{1 \times 1 \times 1}\Big] \times \Big[\frac{1}{7}\Big] \times \Big[\frac{1}{49}\Big]\\[1em] = \Big[\frac{343}{1}\Big] \times \Big[\frac{1}{7}\Big] \times \Big[\frac{1}{49}\Big]\\[1em] = \Big[\frac{343 \times 1 \times 1}{1 \times 7 \times 49}\Big]\\[1em] = \Big[\frac{343}{343}\Big]\\[1em] = 1\end{aligned}$$

Hence, option 4 is the correct option.

$$\begin{aligned}1^8 \times 3^0 \times 5^3 \times 2^2\\[1em] = (1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1) \times 1 \times (5 \times 5 \times 5) \times (2 \times 2)\\[1em] = 1 \times 1 \times 125 \times 4 \\[1em] = 500\end{aligned}$$

Hence, $1^8 \times 3^0 \times 5^3 \times 2^2 = 500$

$$\begin{aligned}(4^7)^2 \times (4^{-3})^4\\[1em] = (4)^{7 \times 2} \times \Big[\frac{1}{4}\Big]^{3 \times 4}\\[1em] = (4)^{14} \times \Big[\frac{1}{4}\Big]^{12}\\[1em] = (4)^{14} \times (4)^{-12}\\[1em] = (4)^{14-12} \\[1em] = 4^2 \\[1em] = 16\end{aligned}$$

Hence, $(4^7)^2 \times (4^{-3})^4 = 16$

According to quotient property,

$a^m ÷ a^n = a^{m - n}$

$$\begin{aligned}(2^{-9} ÷ 2^{-11})^3\\[1em] = (2^{-9 - (-11)})^3\\[1em] = (2^{-9 + 11})^3\\[1em] = (2^{2})^3\\[1em] = (2)^{2\times3}\\[1em] = 2^{6}\\[1em] = 2 \times 2 \times 2 \times 2 \times 2 \times 2\\[1em] = 64\end{aligned}$$

Hence, $(2^{-9} ÷ 2^{-11})^3 = 64$

$$\begin{aligned}\Big[\frac{2}{3}\Big]^{-4} \times \Big[\frac{27}{8}\Big]^{-2}\\[1em] = \Big[\frac{3}{2}\Big]^4 \times \Big[\frac{8}{27}\Big]^2\\[1em] = \Big[\frac{3^4}{2^4}\Big] \times \Big[\frac{2^3}{3^3}\Big]^2\\[1em] = \Big[\frac{3^4}{2^4}\Big] \times \Big[\frac{2^6}{3^6}\Big]\\[1em] = \Big[\frac{2^6}{3^6}\Big] \times \Big[\frac{2^{-4}}{3^{-4}}\Big] \\[1em] = \Big[\frac{2^{6-4}}{3^{6-4}}\Big]\\[1em] = \Big[\frac{2^2}{3^2}\Big]\\[1em] = \Big[\frac{4}{9}\Big]\end{aligned}$$

Hence, $\Big[\frac{2}{3}\Big]^{-4} \times \Big[\frac{27}{8}\Big]^{-2} = \Big[\frac{4}{9}\Big]$

$$\begin{aligned}\Big[\frac{56}{28}\Big]^0 ÷ \Big[\frac{2}{5}\Big]^3 \times \frac{16}{25}\\[1em] = 1 ÷ \Big[\frac{2}{5}\Big]^3 \times \frac{16}{25} \space [∵ a^0 = 1] \\[1em] = 1 \times \Big[\frac{5}{2}\Big]^3 \times \frac{16}{25}\\[1em] = \Big[\frac{5}{2}\Big]^3 \times \Big[\frac{2^4}{5^2}\Big]\\[1em] = \Big[\frac{5^3}{2^3}\Big] \times \Big[\frac{2^4}{5^2}\Big]\\[1em] = 5^3 \times 2^{-3} \times 2^4 \times 5^{-2} \\[1em] = (5)^{3-2} \times (2)^{4-3}\\[1em] = (5)^{1} \times (2)^{1}\\[1em] = 10\end{aligned}$$

$\Big[\frac{56}{28}\Big]^0 ÷ \Big[\frac{2}{5}\Big]^3 \times \frac{16}{25} = 10$

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

$(12)^{-2} \times 3^3 = \frac{3}{16}$

$$\begin{aligned}(-5)^4 \times (-5)^6 ÷ (-5)^9\\[1em] = (-5)^{4+6-9}\\[1em] = (-5)^1\\[1em] = -5\end{aligned}$$

$(-5)^4 \times (-5)^6 ÷ (-5)^9 = -5$

$$\begin{aligned}\Big[-\frac{1}{3}\Big]^4 ÷ \Big[-\frac{1}{3}\Big]^8 \times \Big[-\frac{1}{3}\Big]^5\\[1em] = \Big[-\frac{1}{3}\Big]^{4-8+5}\\[1em] = \Big[-\frac{1}{3}\Big]^1\\[1em] = -\frac{1}{3}\end{aligned}$$

$\Big[-\frac{1}{3}\Big]^4 ÷ \Big[-\frac{1}{3}\Big]^8 \times \Big[-\frac{1}{3}\Big]^5 = -\frac{1}{3}$

$$\begin{aligned}9^0 \times 4^{-1} ÷ 2^{-4}\\[1em] = 1 \times 4^{-1} ÷ 2^{-4} \space [∵a^0 = 1] \\[1em] = 1 \times 2^{-2} ÷ 2^{-4}\\[1em] = 1 \times 2^{-2-(-4)}\\[1em] = 1 \times 2^{-2+4}\\[1em] = 1 \times 2^{2}\\[1em] = 1 \times 4\\[1em] = 4\end{aligned}$$

$9^0 \times 4^{-1} ÷ 2^{-4} = 4$

$$\begin{aligned}(625)^{-\frac{3}{4}}\\[1em] = (5)^{-4 \times \frac{3}{4}}\\[1em] = (5)^{-3}\\[1em] = \Big[\frac{1}{5}\Big]^3\\[1em] = \frac{1}{125}\end{aligned}$$

$(625)^{-\frac{3}{4}} = \frac{1}{125}$

$$\begin{aligned}\Big[\frac{27}{64}\Big]^{-\frac{2}{3}}\\[1em] = \Big[\frac{3^3}{4^3}\Big]^{-\frac{2}{3}}\\[1em] = \Big[\frac{3}{4}\Big]^{-3 \times \frac{2}{3}}\\[1em] = \Big[\frac{3}{4}\Big]^{-2}\\[1em] = \Big[\frac{4}{3}\Big]^2\\[1em] = \Big[\frac{4 \times 4}{3 \times 3}\Big]\\[1em] = \frac{16}{9}\\[1em] = 1\frac{7}{9}\end{aligned}$$

$\big[\frac{27}{64}\Big]^{-\frac{2}{3}} = 1\frac{7}{9}$

$$\begin{aligned}\Big[\frac{1}{32}\Big]^{-\frac{2}{5}}\\[1em] = \Big[\frac{1^5}{2^5}\Big]^{-\frac{2}{5}}\\[1em] = \Big[\frac{1}{2}\Big]^{-5 \times \frac{2}{5}}\\[1em] = \Big[\frac{1}{2}\Big]^{-2}\\[1em] = \Big[\frac{2}{1}\Big]^2\\[1em] = \Big[\frac{2 \times 2}{1 \times 1}\Big]\\[1em] = \frac{4}{1}\\[1em] = 4\end{aligned}$$

$\big[\frac{1}{32}\Big]^{-\frac{2}{5}} = 4$

$$\begin{aligned}(125)^{-\frac{2}{3}} ÷ (8)^{\frac{2}{3}}\\[1em] = (5^3)^{-\frac{2}{3}} ÷ (2^3)^{\frac{2}{3}}\\[1em] = (5)^{-3\times\frac{2}{3}} ÷ (2)^{3\times\frac{2}{3}}\\[1em] = (5)^{-2} ÷ (2)^{2}\\[1em] = (5)^{-2} \times \Big(\frac{1}{2}\Big)^{2}\\[1em] = (5)^{-2} \times \frac{1}{4}\\[1em] = \Big(\frac{1}{5}\Big)^2 \times \frac{1}{4}\\[1em] = \frac{1}{25} \times \frac{1}{4}\\[1em] = \frac{1}{100}\end{aligned}$$

$(125)^{-\frac{2}{3}} ÷ (8)^{\frac{2}{3}} = \frac{1}{100}$

$$\begin{aligned}(243)^{\frac{2}{5}} ÷ (32)^{-\frac{2}{5}}\\[1em] = (3^5)^{\frac{2}{5}} ÷ (2^5)^{-\frac{2}{5}}\\[1em] = (3)^{5\times\frac{2}{5}} ÷ (2)^{-5\times\frac{2}{5}}\\[1em] = (3)^{2} ÷ (2)^{-2}\\[1em] = (3)^{2} \times \Big(\frac{1}{2}\Big)^{-2}\\[1em] = (3)^{2} \times \frac{2}{1}^{2}\\[1em] = (9) \times (4)\\[1em] = 36\end{aligned}$$

$(243)^{\frac{2}{5}} ÷ (32)^{-\frac{2}{5}} = 36$

As we know for any rational number,

$a^0 = 1$

$$\begin{aligned}(-3)^4 - (\sqrt[4]{3})^0 \times (-2)^5 ÷ (64)^{\frac{2}{3}}\\[1em] =(-3 \times -3 \times -3 \times -3) - 1 \times (-2)^5 ÷ (4^3)^{\frac{2}{3}}\\[1em] = 81 - (-2 \times -2 \times -2 \times -2 \times -2) ÷ (4)^{3\times\frac{2}{3}}\\[1em] = 81 - (-32) ÷ (4)^{2}\\[1em] = 81 + 32 ÷ (4 \times 4)\\[1em] = 81 + 32 ÷ 16\\[1em] = 81 + 2\\[1em] = 83\end{aligned}$$

$(-3)^4 - (\sqrt[4]{3})^0 \times (-2)^5 ÷ (64)^{\frac{2}{3}} = 83$

$$\begin{aligned}(27)^{\frac{2}{3}} ÷ \Big(\frac{81}{16}\Big)^{-\frac{1}{4}}\\[1em] = (3^3)^{\frac{2}{3}} ÷ \Big(\frac{3^4}{2^4}\Big)^{-\frac{1}{4}}\\[1em] = (3)^{3\times \frac{2}{3}} ÷ \Big(\frac{3}{2}\Big)^{-4\times\frac{1}{4}}\\[1em] = (3)^2 ÷ \Big(\frac{3}{2}\Big)^{-1}\\[1em] = (3)^2 ÷ \Big(\frac{2}{3}\Big)\\[1em] = (3)^2 \times \Big(\frac{3}{2}\Big)\\[1em] = 9 \times \Big(\frac{3}{2}\Big)\\[1em] = \frac{9 \times 3}{2}\\[1em] = \frac{27}{2}\\[1em] = 13\frac{1}{2}\end{aligned}$$

$(27)^{\frac{2}{3}} ÷ \Big(\frac{81}{16}\Big)^{-\frac{1}{4}} = 13\frac{1}{2}$

$$\begin{aligned}8^{\frac{4}{3}} + 25^{\frac{3}{2}} - \Big(\frac{1}{27}\Big)^{-\frac{2}{3}}\\[1em] = (2^3)^{\frac{4}{3}} + (5^2)^{\frac{3}{2}} - \Big(\frac{1^3}{3^3}\Big)^{-\frac{2}{3}}\\[1em] = (2)^{3\times\frac{4}{3}} + (5)^{2\times\frac{3}{2}} - \Big(\frac{1}{3}\Big)^{-3\times\frac{2}{3}}\\[1em] = (2)^4 + (5)^3 - \Big(\frac{1}{3}\Big)^{-2}\\[1em] = (2)^4 + (5)^3 - \Big(\frac{3}{1}\Big)^2\\[1em] = 16 + 125 - 9\\[1em] = 132\end{aligned}$$

$8^{\frac{4}{3}} + 25^{\frac{3}{2}} - \Big(\frac{1}{27}\Big)^{-\frac{2}{3}} = 132$

$$\begin{aligned}[(64)^{-2}]^{-3} ÷ [\lbrace (-8)^2\rbrace^3]^2\\[1em] = [(2^6)^{-2}]^{-3} ÷ [\lbrace (-2^3)^2\rbrace^3]^2\\[1em] = [(2)^{-6 \times 2}]^{-3} ÷ [\lbrace (-2)^{3 \times 2}\rbrace^3]^2\\[1em] = [(2)^{-12}]^{-3} ÷ [\lbrace (-2)^{6}\rbrace^3]^2\\[1em] = [(2)]^{-12 \times (-3)} ÷ [\lbrace (-2)\rbrace^{6 \times 3}]^2\\[1em] = (2)^{36} ÷ [(-2)^{18}]^2\\[1em] = (2)^{36} ÷ (-2)^{18 \times 2}\\[1em] = (2)^{36} ÷ (-2)^{36}\\[1em] = \frac{2^ {36}}{-2^{36}}\\[1em] = (2)^{36-36}\times\frac{1^ {36}}{-1^{36}}\\[1em] = (2)^0\times(-1)^{36}\\[1em] = 1\end{aligned}$$

$[(64)^{-2}]^{-3} ÷ [\lbrace (-8)^2\rbrace^3]^2 = 1$

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

LCM of 8 and 16 is 2 x 2 x 2 x 2 = 16

$$\begin{aligned}= \Big(\frac{1 \times 2}{8 \times 2} - \frac{1 \times 1}{16 \times 1}\Big)\Big(\frac{1 \times 2}{8 \times 2} + \frac{1 \times 1}{16 \times 1}\Big)\\[1em] = \Big(\frac{2}{16} - \frac{1}{16}\Big)\Big(\frac{2}{16} + \frac{1}{16}\Big)\\[1em] = \Big(\frac{2-1}{16}\Big)\Big(\frac{2+1}{16}\Big)\\[1em] = \Big(\frac{1}{16}\Big)\Big(\frac{3}{16}\Big)\\[1em] = \Big(\frac{3}{256}\Big)\end{aligned}$$

$(2^{-3} - 2^{-4})(2^{-3} + 2^{-4}) = \Big(\frac{3}{256}\Big)$

According to property of exponents, $a^0 = 1$

Then, $(-5)^0 = 1$

According to property of exponents, $a^0 = 1$

$$\begin{aligned}8^0 + 4^0 + 2^0 \\[1em] = 1 + 1 + 1 \\[1em] = 3\end{aligned}$$

Hence, $8^0 + 4^0 + 2^0 = 3$

According to property of exponents, $a^0 = 1$

$$\begin{aligned}(8 + 4 + 2)^0\\[1em] = (14)^0\\[1em] = 1\end{aligned}$$

Hence, $(8 + 4 + 2)^0 = 1$

According to property of exponents, $a^0 = 1$

$$\begin{aligned}4x^0\\[1em] = 4 \times 1\\[1em] = 4\end{aligned}$$

Hence, $4x^0 = 4$

According to property of exponents, $a^0 = 1$

Then, $(4x)^0 = 1$

According to property of exponents,

$a^0 = 1$

$$\begin{aligned}[(10^3)^0]^5\\[1em] = [(10 \times 10 \times 10)^0]^5\\[1em] = [(1000)^0]^5\\[1em] = [1]^5\\[1em] = [1 \times 1 \times 1 \times 1 \times 1]\\[1em] = 1\end{aligned}$$

$[(10^3)^0]^5 = 1$

According to property of exponents,

$a^0 = 1$

$$\begin{aligned}(7x^0)^2\\[1em] = (7 \times 1)^2\\[1em] = (7)^2\\[1em] = (7\times 7)\\[1em] = 49\end{aligned}$$

$(7x^0)^2 = 49$

$$\begin{aligned}9^0 + 9^{-1} - 9^{-2} + 9^{\frac{1}{2}} - 9^{-\frac{1}{2}}\\[1em] = 1 + \frac{1}{9}^1 - \frac{1}{9}^2 + (3^2)^{\frac{1}{2}} -\frac{1}{9}^{\frac{1}{2}}\\[1em] = 1 + \frac{1}{9}^1 - \frac{1}{9}^2 + (3)^{2\times\frac{1}{2}} -\frac{1^2}{3^2}^{\frac{1}{2}}\\[1em] = 1 + \frac{1}{9}^1 - \frac{1}{9}^2 + (3)^1 -\frac{1}{3}^{2\times\frac{1}{2}}\\[1em] = 1 + \frac{1}{9} - \frac{1}{9}^2 + (3) -\frac{1}{3}^1\\[1em] = 1 + \frac{1}{9} - \frac{1}{9\times 9} + (3) -\frac{1}{3}\\[1em] = 4 + \frac{1}{9} - \frac{1}{81} -\frac{1}{3}\\[1em] = \frac{4}{1} + \frac{1}{9} - \frac{1}{81} -\frac{1}{3}\end{aligned}$$

LCM of 1, 9, 81 and 3 is 3 x 3 x 3 x 3 =81

$$\begin{aligned}= \frac{4 \times 81}{1 \times 81} + \frac{1 \times 9}{9 \times 9} - \frac{1 \times 1}{81 \times 1} -\frac{1 \times 27}{3 \times 27}\\[1em] = \frac{324}{81} + \frac{9}{81} - \frac{1}{81} -\frac{27}{81}\\[1em] = \frac{324 + 9 -1 -27}{81}\\[1em] = \frac{305}{81}\\[1em] = 3\frac{62}{81}\end{aligned}$$

$9^0 + 9^{-1} - 9^{-2} + 9^{\frac{1}{2}} - 9^{-\frac{1}{2}} = 3\frac{62}{81}$

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

$\frac{a^5b^2}{a^2b^{-3}}= a^{3}b^{5}$

$$\begin{aligned}15y^8 ÷ 3y^3\\[1em] = \frac{15}{3}y^{8-3}\\[1em] = 5y^5\end{aligned}$$

$15y^8 ÷ 3y^3 = 5y^5$

$$\begin{aligned}x^{10}y^6 ÷ x^3y^{-2}\\[1em] = x^{10-3}y^{6-(-2)}\\[1em] = x^{7}y^{6+2}\\[1em] = x^{7}y^{8}\end{aligned}$$

$x^{10}y^6 ÷ x^3y^{-2} = x^{7}y^{8}$

$$\begin{aligned}5z^{16} ÷ 15z^{-11}\\[1em] = \frac{5}{15}z^{16-(-11)}\\[1em] = \frac{1}{3}z^{16+11}\\[1em] = \frac{1}{3}z^{27}\end{aligned}$$

$5z^{16} ÷ 15z^{-11} = \frac{1}{3}z^{27}$

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

$(36x^2)^{\frac{1}{2}} = 6x$

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

$(125x^{-3})^{\frac{1}{3}} = 5x^{-1}$

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

$(2x^2y^{-3})^{-2} = \frac{1}{4}y^6x^{-4}$

$$\begin{aligned}(27x^{-3}y^6)^{\frac{2}{3}}\\[1em] = \Big(\frac{27y^6}{x^3}\Big)^{\frac{2}{3}}\\[1em] = \Big(\frac{3^3y^{2\times3}}{x^3}\Big)^{\frac{2}{3}}\\[1em] = \Big(\frac{3y^2}{x}\Big)^{3\times\frac{2}{3}}\\[1em] = \Big(\frac{3y^2}{x}\Big)^{\frac{6}{3}}\\[1em] = \Big(\frac{3y^2}{x}\Big)^2\\[1em] = \frac{9y^4}{x^2}\\[1em] = 9y^4x^{-2}\end{aligned}$$

$(27x^{-3}y^6)^{\frac{2}{3}} = 9y^4x^{-2}$

$$\begin{aligned}(-2x^{\frac{2}{3}}y^{\frac{-3}{2}})^6\\[1em] = \Big(-2^6x^{6\times\frac{2}{3}}y^{6\times\frac{-3}{2}}\Big)\\[1em] = \Big(64x^{\frac{12}{3}}y^{\frac{-18}{2}}\Big)\\[1em] = \Big(64x^4y^{-9}\Big)\\[1em] = \frac{64x^4}{y^9}\end{aligned}$$

$(-2x^{\frac{2}{3}}y^{\frac{-3}{2}})^6 = \frac{64x^4}{y^9}$

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

$(x^{a+b})^{a-b}.(x^{b+c})^{b-c}.(x^{c+a})^{c-a} = 1$

$$\begin{aligned}(\sqrt[5]{x^{20}y^{-10}z^5}) ÷ \frac{x^3}{y^3}\\[1em] = (x^{\frac{20}{5}}y^{\frac{-10}{5}}z^{\frac{5}{5}}) ÷ \frac{x^3}{y^3}\\[1em] = (x^4y^{-2}z^1) ÷ \frac{x^3}{y^3}\\[1em] = (x^4y^{-2}z^1) \times \frac{y^3}{x^3}\\[1em] = (x^{4-3}y^{-2+3}z)\\[1em] = (x^1y^1z)\\[1em] = xyz\end{aligned}$$

$(\sqrt[5]{x^{20}y^{-10}z^5}) ÷ \frac{x^3}{y^3} = xyz$

$$\begin{aligned}\Big[\frac{256a^{16}}{81b^4}\Big]^{\frac{-3}{4}}\\[1em] = \Big[\frac{2^{8}a^{16}}{3^4b^4}\Big]^{\frac{-3}{4}}\\[1em] = \Big[\frac{2^{2}a^{4}}{3b}\Big]^{4\times\frac{-3}{4}}\\[1em] = \Big[\frac{2^{2}a^{4}}{3b}\Big]^{-3}\\[1em] = \Big[\frac{3b}{2^{2}a^{4}}\Big]^{3}\\[1em] = \Big[\frac{3^3b^3}{2^{6}a^{12}}\Big]\\[1em] = \Big[\frac{27b^3}{64a^{12}}\Big]\\[1em] = \Big[\frac{27}{64}\Big]a^{-12}b^3\end{aligned}$$

$\Big[\frac{256a^{16}}{81b^4}\Big]^{\frac{-3}{4}} = \Big[\frac{27}{64}\Big]a^{-12}b^3$

$$\begin{aligned}(a^{-2}b)^{-2}.(ab)^{-3}\\[1em] = (a^{-2\times(-2)}b^{-2}).(a^{-3}b^{-3})\\[1em] = (a^{4}b^{-2}).(a^{-3}b^{-3})\\[1em] = (a^{4+(-3)}b^{-2+(-3)})\\[1em] = (a^{4-3}b^{-2-3})\\[1em] = (a^{1}b^{-5})\end{aligned}$$

$(a^{-2}b)^{-2}.(ab)^{-3} = (a^{1}b^{-5}) = \frac{a}{b^5}$

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

$(x^ny^{-m})^4\times(x^3y^{-2})^{-n} = (\frac{x^{n}y^{2n}}{y^{4m}})$

$$\begin{aligned}\Big[\frac{125a^{-3}}{y^6}\Big]^{\frac{-1}{3}}\\[1em] = \Big[\frac{5^{3}a^{-3}}{y^6}\Big]^{\frac{-1}{3}}\\[1em] = \Big[\frac{5a^{-1}}{y^2}\Big]^{3\times\frac{-1}{3}}\\[1em] = \Big[\frac{5a^{-1}}{y^2}\Big]^{-1}\\[1em] = \Big[\frac{y^2}{5a^{-1}}\Big]^{1}\\[1em] = \Big[\frac{ay^2}{5}\Big]\end{aligned}$$

$\Big[\frac{125a^{-3}}{y^6}\Big]^{\frac{-1}{3}} = \frac{ay^2}{5}$

$$\begin{aligned}\Big[\frac{32x^{-5}}{243y^{-5}}\Big]^{\frac{-1}{5}}\\[1em] = \Big[\frac{2^{5}x^{-5}}{3^5y^{-5}}\Big]^{\frac{-1}{5}}\\[1em] = \Big[\frac{2x^{-1}}{3y^{-1}}\Big]^{5\times\frac{-1}{5}}\\[1em] = \Big[\frac{2x^{-1}}{3y^{-1}}\Big]^{-1}\\[1em] = \Big[\frac{2y}{3x}\Big]^{-1}\\[1em] = \Big[\frac{3x}{2y}\Big]\end{aligned}$$

$\Big[\frac{32x^{-5}}{243y^{-5}}\Big]^{\frac{-1}{5}}= \Big[\frac{3x}{2y}\Big]$

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

$(a^{-2}b)^{\frac{1}{2}} \times (ab^{-3})^{\frac{1}{3}} = \frac{1}{a^{\frac{2}{3}}b^{\frac{1}{2}}}$

$$\begin{aligned}(xy)^{m-n}.(yz)^{n-l}.(zx)^{l-m}\\[1em] = (x^{m-n}y^{m-n}).(y^{n-l}z^{n-l}).(z^{l-m}x^{l-m})\\[1em] = x^{(m-n)+(l-m)}y^{(m-n)+(n-l)}z^{(n-l)+(l-m)}\\[1em] = x^{-n+l}y^{m-l}z^{n-m}\\[1em] = \frac{x^ly^mz^n}{x^ny^lz^m}\end{aligned}$$

$(xy)^{m-n}.(yz)^{n-l}.(zx)^{l-m} = x^{-n+l}y^{m-l}z^{n-m}$

To prove: $\Big[\frac{x^a}{x^{-b}}\Big]^{a-b}.\Big[\frac{x^b}{x^{-c}}\Big]^{b-c}.\Big[\frac{x^c}{x^{-a}}\Big]^{c-a} = 1$

Taking LHS:

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

∴ LHS = RHS

$\Big[\frac{x^a}{x^{-b}}\Big]^{a-b}.\Big[\frac{x^b}{x^{-c}}\Big]^{b-c}.\Big[\frac{x^c}{x^{-a}}\Big]^{c-a} = 1$

$$\begin{aligned}\frac{x^{5+n}\times(x^2)^{3n+1}}{x^{7n-2}}\\[1em] = \frac{x^{5+n}\times(x)^{2(3n+1)}}{x^{7n-2}}\\[1em] = \frac{x^{5+n}\times(x)^{6n+2}}{x^{7n-2}}\\[1em] = \frac{x^{(5+n)+(6n+2)}}{x^{7n-2}}\\[1em] = \frac{x^{5+n+6n+2}}{x^{7n-2}}\\[1em] = x^{(7+7n)-(7n-2)}\\[1em] = x^{7+7n-7n+2}\\[1em] = x^{9}\end{aligned}$$

$\frac{x^{5+n}\times(x^2)^{3n+1}}{x^{7n-2}} = x^{9}$

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

$\frac{a^{2n+1}\times a^{(2n+1)(2n-1)}}{a^{n(4n-1)}\times (a^2)^{2n+3}} = \frac{1}{a^{n+6}}$

To Prove: $(m + n)^{-1}(m^{-1} + n^{-1}) = (mn)^{-1}$

Taking LHS:

$$\begin{aligned}(m + n)^{-1}(m^{-1} + n^{-1})\\[1em] = \Big(\frac{1}{m + n}\Big)\Big(\frac{1}{m} + \frac{1}{n}\Big)\\[1em] = \Big(\frac{1}{m + n}\Big)\Big(\frac{n}{mn} + \frac{m}{nm}\Big)\\[1em] = \Big(\frac{1}{m + n}\Big)\Big(\frac{n + m}{mn}\Big)\\[1em] = \frac{1 \times (n + m)}{(m + n) \times mn}\\[1em] = \frac{1}{mn}\\[1em] = (mn)^{-1} \\[1em] = \text{RHS}\end{aligned}$$

∴ LHS = RHS

Hence, $(m + n)^{-1}(m^{-1} + n^{-1}) = (mn)^{-1}$

To prove:$\Big[\frac{x^a}{x^b}\Big]^{\frac{1}{ab}}\Big[\frac{x^b}{x^c}\Big]^{\frac{1}{bc}}\Big[\frac{x^c}{x^a}\Big]^{\frac{1}{ca}} = 1$

Taking LHS:

$$\begin{aligned}\Big[\frac{x^a}{x^b}\Big]^{\frac{1}{ab}}\Big[\frac{x^b}{x^c}\Big]^{\frac{1}{bc}}\Big[\frac{x^c}{x^a}\Big]^{\frac{1}{ca}}\\[1em] = \Big[x^{a-b}\Big]^{\frac{1}{ab}}\Big[x^{b-c}\Big]^{\frac{1}{bc}}\Big[x^{c-a}\Big]^{\frac{1}{ca}}\\[1em] = \Big[x\Big]^{\frac{a-b}{ab}}\Big[x\Big]^{\frac{b-c}{bc}}\Big[x\Big]^{\frac{c-a}{ca}}\\[1em] = \Big[x\Big]^{\frac{a}{ab}-\frac{b}{ab}}\Big[x\Big]^{\frac{b}{bc}-\frac{c}{bc}}\Big[x\Big]^{\frac{c}{ca}-\frac{a}{ca}}\\[1em] = \Big[x\Big]^{\frac{1}{b}-\frac{1}{a}}\Big[x\Big]^{\frac{1}{c}-\frac{1}{b}}\Big[x\Big]^{\frac{1}{a}-\frac{1}{c}}\\[1em] = \Big[x\Big]^{\frac{1}{b}-\frac{1}{a}+\frac{1}{c}-\frac{1}{b}+\frac{1}{a}-\frac{1}{c}}\\[1em] = \Big[x\Big]^{\frac{1}{b}-\frac{1}{b}-\frac{1}{a}+\frac{1}{a}+\frac{1}{c}-\frac{1}{c}}\\[1em] = x^0\\[1em] = 1 \\[1em] = \text{RHS}\end{aligned}$$

∴ LHS = RHS

Hence,$\Big[\frac{x^a}{x^b}\Big]^{\frac{1}{ab}}\Big[\frac{x^b}{x^c}\Big]^{\frac{1}{bc}}\Big[\frac{x^c}{x^a}\Big]^{\frac{1}{ca}} = 1$

To prove: $\frac{1}{1+x^{a-b}}+\frac{1}{1+x^{b-a}}=1$

Taking LHS:

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

∴ LHS = RHS

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

$$\begin{aligned}12^{-5} \times 12^{2n+1} = 12^{13} ÷ 12^7\\[1em] \Rightarrow 12^{(-5)+(2n+1)} = 12^{13 - 7}\\[1em] \Rightarrow 12^{-5+2n+1} = 12^{6}\\[1em] \Rightarrow 12^{-4+2n} = 12^{6}\\[1em] \Rightarrow -4+2n = 6\\[1em] \Rightarrow 2n = 6+4\\[1em] \Rightarrow 2n = 10\\[1em] \Rightarrow n = \frac{10}{2}\\[1em] \Rightarrow n = 5\\[1em]\end{aligned}$$

If $12^{-5} \times 12^{2n+1} = 12^{13} ÷ 12^7$ then n = 5.

$$\begin{aligned}\frac{a^{2n-3}\times(a^2)^{n+1}}{(a^4)^{-3}} = (a^3)^3 ÷ (a^6)^{-3}\\[1em] \Rightarrow \frac{a^{2n-3}\times(a)^{2(n+1)}}{(a)^{4\times(-3)}} = (a)^{3\times3} ÷ (a)^{6\times(-3)}\\[1em] \Rightarrow \frac{a^{2n-3}\times(a)^{2n+2}}{(a)^{-12}} = a^9 ÷ a^{-18}\\[1em] \Rightarrow \frac{a^{(2n-3)+(2n+2)}}{(a)^{-12}} = a^{9-(-18)}\\[1em] \Rightarrow \frac{a^{2n-3+2n+2}}{(a)^{-12}} = a^{9+18}\\[1em] \Rightarrow \frac{a^{4n-1}}{(a)^{-12}} = a^{27}\\[1em] \Rightarrow a^{(4n-1)-(-12)} = a^{27}\\[1em] \Rightarrow a^{4n-1+12} = a^{27}\\[1em] \Rightarrow a^{4n+11} = a^{27}\\[1em] \Rightarrow 4n+11 = 27\\[1em] \Rightarrow 4n = 27-11\\[1em] \Rightarrow 4n = 16\\[1em] \Rightarrow n = \frac{16}{4}\\[1em] \Rightarrow n = 4\end{aligned}$$

If $\frac{a^{2n-3}\times(a^2)^{n+1}}{(a^4)^{-3}} = (a^3)^3 ÷ (a^6)^{-3}$, then n = 4

$$\begin{aligned}\frac{a^{2n+3}.a^{(2n+1)(n+2)}}{(a^3)^{2n+1}.a^{n(2n+1)}}\\[1em] = \frac{a^{2n+3}.a^{2n^2+4n+n+2}}{a^{3(2n+1)}.a^{2n^2+n}}\\[1em] = \frac{a^{2n+3}.a^{2n^2+5n+2}}{a^{6n+3}.a^{2n^2+n}}\\[1em] = \frac{a^{(2n+3)+(2n^2+5n+2)}}{a^{(6n+3)+(2n^2+n)}}\\[1em] = \frac{a^{2n+3+2n^2+5n+2}}{a^{6n+3+2n^2+n}}\\[1em] = \frac{a^{2n^2+7n+5}}{a^{2n^2+7n+3}}\\[1em] = a^{(2n^2+7n+5)-(2n^2+7n+3)}\\[1em] = a^{2n^2+7n+5-2n^2-7n-3}\\[1em] = a^{2}\\[1em]\end{aligned}$$

$\frac{a^{2n+3}.a^{(2n+1)(n+2)}}{(a^3)^{2n+1}.a^{n(2n+1)}} = a^2$

$$\begin{aligned}\frac{x^{2n+7}.(x^2)^{3n+2}}{x^{4(2n+3)}}\\[1em] = \frac{x^{2n+7}.x^{2(3n+2)}}{x^{4(2n+3)}}\\[1em] = \frac{x^{2n+7}.x^{6n+4}}{x^{8n+12}}\\[1em] = \frac{x^{(2n+7)+(6n+4)}}{x^{8n+12}}\\[1em] = \frac{x^{2n+7+6n+4}}{x^{8n+12}}\\[1em] = \frac{x^{8n+11}}{x^{8n+12}}\\[1em] = x^{(8n+11)-(8n+12)}\\[1em] = x^{8n+11-8n-12}\\[1em] = x^{-1}\\[1em] = \frac{1}{x}\end{aligned}$$

$\frac{x^{2n+7}.(x^2)^{3n+2}}{x^{4(2n+3)}} = \frac{1}{x}$

$$\begin{aligned}(2^{-3} + 3^{-2})\times 7^0\\[1em] = \Big(\frac{1}{2^{3}} + \frac{1}{3^{2}}\Big)\times 1\\[1em] = \Big(\frac{1}{8} + \frac{1}{9}\Big)\end{aligned}$$

LCM of 8 and 9 is 2 x 2 x 2 x 3 x 3 = 72

$$\begin{aligned}= \Big(\frac{1 \times 9}{8 \times 9} + \frac{1 \times 8}{9 \times 8}\Big)\\[1em] = \Big(\frac{9}{72} + \frac{8}{72}\Big)\\[1em] = \Big(\frac{9 + 8}{72}\Big)\\[1em] = \Big(\frac{17}{72}\Big)\end{aligned}$$

$(2^{-3} + 3^{-2})\times 7^0 = \Big(\frac{17}{72}\Big)$

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

LCM of 1 and 2 is 2

$$\begin{aligned}= \Big(\frac{1 \times 2}{1 \times 2} + \frac{1 \times 1}{2 \times 1}\Big)\times 9\\[1em] = \Big(\frac{2}{2} + \frac{1}{2}\Big)\times 9\\[1em] = \Big(\frac{2+1}{2}\Big)\times 9\\[1em] = \Big(\frac{3}{2}\Big)\times 9\\[1em] = \Big(\frac{3 \times 9}{2}\Big)\\[1em] = \frac{27}{2}\\[1em] = 13\frac{1}{2}\end{aligned}$$

$(8^0 + 2^{-1})\times 3^2 = 13\frac{1}{2}$

$$\begin{aligned}\Big[\Big(\frac{1}{6}\Big)^{-1} - \Big(\frac{1}{5}\Big)^{-1}\Big]^{-2}\\[1em] = [6^1 - 5^1]^{-2}\\[1em] = [6 - 5]^{-2}\\[1em] = [1]^{-2}\\[1em] = \frac{1}{1}^2\\[1em] = 1^2\\[1em] = 1\end{aligned}$$

$\Big[\Big(\frac{1}{6}\Big)^{-1} - \Big(\frac{1}{5}\Big)^{-1}\Big]^{-2} = 1$

$$\begin{aligned}\Big[\Big\lbrace\Big(-\frac{1}{3}\Big)^{-2}\Big\rbrace^2\Big]^{-1}\\[1em] = [\lbrace (-3)^2\rbrace^2]^{-1}\\[1em] = [\lbrace 9\rbrace^2]^{-1}\\[1em] = [81]^{-1}\\[1em] = \frac{1}{81}\end{aligned}$$

$\Big[\Big\lbrace\Big(-\frac{1}{3}\Big)^{-2}\Big\rbrace^2\Big]^{-1} = \frac{1}{81}$

$$\begin{aligned}\frac{5^{n+2}-5^{n+1}}{5^{n+1}}\\[1em] = \frac{5^{n+2}}{5^{n+1}} -\frac{5^{n+1}}{5^{n+1}}\\[1em] = 5^{(n+2)-(n+1)} -5^{(n+1)-(n+1)}\\[1em] = 5^{n+2-n-1} -5^{n+1-n-1}\\[1em] = 5^{1} -5^{0}\\[1em] = 5 -1\\[1em] = 4\end{aligned}$$

$\frac{5^{n+2}-5^{n+1}}{5^{n+1}} = 4$

$$\begin{aligned}\frac{1}{(125)^{x-7}}=5^{2x-1}\\[1em] \Rightarrow \frac{1}{(5^3)^{x-7}}=5^{2x-1}\\[1em] \Rightarrow \frac{1}{(5)^{3(x-7)}}=5^{2x-1}\\[1em] \Rightarrow \frac{1}{5^{3x-21}}=5^{2x-1}\\[1em] \Rightarrow \Big(\frac{1}{5}\Big)^{3x-21}=5^{2x-1}\\[1em] \Rightarrow 5^{-(3x-21)}=5^{2x-1}\\[1em] \Rightarrow 5^{-3x+21}=5^{2x-1}\\[1em] \Rightarrow -3x + 21 = 2x - 1\\[1em] \Rightarrow 1 + 21 = 2x + 3x\\[1em] \Rightarrow 22 = 5x\\[1em] \Rightarrow x = \frac{22}{5}\end{aligned}$$

If $\frac{1}{(125)^{x-7}}=5^{2x-1}$ then $x = \frac{22}{5}$

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

If $\Big(\frac{2}{3}\Big)^3 \times \Big(\frac{2}{3}\Big)^{-4} = \Big(\frac{2}{3}\Big)^{2x+1}$, then $x = -1$.

$$\begin{aligned}4^n ÷ 4^{-3} = 4^5\\[1em] \Rightarrow 4^{n-(-3)} = 4^5\\[1em] \Rightarrow n - (-3) = 5\\[1em] \Rightarrow n + 3 = 5\\[1em] \Rightarrow n = 5 - 3\\[1em] \Rightarrow n = 2\end{aligned}$$

If $4^n ÷ 4^{-3} = 4^5$, then $n = 2$.

$$\begin{aligned}\frac{81\times3^{n+1}-9\times3^n}{81\times3^{n+2}-9\times3^{n+1}}\\[1em] = \frac{3^4\times3^{n+1}-3^2\times3^n}{3^4\times3^{n+2}-3^2\times3^{n+1}}\\[1em] = \frac{3^{4+(n+1)}-3^{2+n}}{3^{4+(n+2)}-3^{2+(n+1)}}\\[1em] = \frac{3^{n+5}-3^{2+n}}{3^{n+6}-3^{n+3}}\\[1em] = \frac{(3^{n+2}.3^3)-3^{2+n}}{(3^{n+3}.3^3)-3^{n+3}}\\[1em] = \frac{(3^{n+2})}{(3^{n+3})}\frac{3^3-1}{3^{3}-1}\\[1em] = (3^{(n+2)-(n+3)})\frac{\cancel {3^3-1}}{\cancel {3^3-1}}\\[1em] = (3^{n+2-n-3})\\[1em] = (3^{-1})\\[1em] = \frac{1}{3}\end{aligned}$$

$\frac{81\times3^{n+1}-9\times3^n}{81\times3^{n+2}-9\times3^{n+1}} = \frac{1}{3}$

Finding prime factors of 1250,

$$\begin{aligned}2^{n-7} \times 5^{n-4} = 1250\\[1em] \Rightarrow 2^{n-7} \times 5^{n-4} = 2^1 \times 5^4\end{aligned}$$

On comparing the exponent of 2 or 5, we get

$$\begin{aligned}n - 7 = 1 \\[1em] \Rightarrow n = 1 + 7\\[1em] \Rightarrow n = 8\end{aligned}$$

OR

$$\begin{aligned}n - 4 = 4\\[1em] \Rightarrow n = 4 + 4\\[1em] \Rightarrow n = 8\end{aligned}$$

If $2^{n-7} \times 5^{n-4} = 1250$, then $n = 8$.

$$\begin{aligned}(8^0 + 5^0)(8^0 - 5^0)\\[1em] = (1 + 1)(1 - 1)\\[1em] = 2 \times 0\\[1em] = 0\end{aligned}$$

The multiplicative inverse of 0 = $\frac{1}{0}$

And as we know $\frac{1}{0}$ is not defined.

Hence, option 4 is the correct option.

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

Hence, option 2 is the correct option.

$$\begin{aligned}3^4 \times 9^3 = 9^n\\[1em] \Rightarrow 3^4 \times (3^2)^3 = (3^2)^n\\[1em] \Rightarrow 3^4 \times 3^6 = 3^{2n}\\[1em] \Rightarrow 3^{4+6} = 3^{2n}\\[1em] \Rightarrow 3^{10} = 3^{2n}\\[1em] \Rightarrow 10 = 2n\\[1em] \Rightarrow n = \frac{10}{2}\\[1em] \Rightarrow n = 5\end{aligned}$$

Hence, option 1 is the correct option.

$$\begin{aligned}\Big(\frac{5}{6}\Big)^5 \times \Big(\frac{6}{5}\Big)^{-4} = \Big(\frac{5}{6}\Big)^{3x}\\[1em] \Rightarrow \Big(\frac{5}{6}\Big)^5 \times \Big(\frac{5}{6}\Big)^4 = \Big(\frac{5}{6}\Big)^{3x}\\[1em] \Rightarrow \Big(\frac{5}{6}\Big)^{5+4} = \Big(\frac{5}{6}\Big)^{3x}\\[1em] \Rightarrow \Big(\frac{5}{6}\Big)^9 = \Big(\frac{5}{6}\Big)^{3x}\\[1em] \Rightarrow 9 = 3x\\[1em] \Rightarrow x = \frac{9}{3}\\[1em] \Rightarrow x = 3\end{aligned}$$

Hence, option 4 is the correct option.

$$\begin{aligned}\Big(\frac{2}{5}\Big)^{-8} ÷ \Big(\frac{2}{5}\Big)^5\\[1em] \Rightarrow \Big(\frac{2}{5}\Big)^{-8 - 5}\\[1em] \Rightarrow \Big(\frac{2}{5}\Big)^{-13}\end{aligned}$$

Hence, option 2 is the correct option.

We know that,

For any number a ≠ 0, a⁰ = 1.

a⁰ = 1

Statement 1 : (x⁰ + y⁰)(x + y)⁰ = 1

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

⇒ (1 + 1) x 1

⇒ 2 x 1

⇒ 2.

R.H.S. = 1

Since, L.H.S. ≠ R.H.S.

So, statement 1 is false.

Statement 2 : (1 + 1)(1 - 1) = 2 x 0 = 0

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

⇒ (1 + 1)(1 - 1)

⇒ 2 x 0

⇒ 0.

R.H.S. = 0

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

So, statement 2 is true.

Hence, option 4 is the correct option.

According to assertion : (-100)³ = -10,00,000

Solving L.H.S.,

⇒ (-100)³

⇒ (-100) x (-100) x (-100)

⇒ -10,00,000

As, L.H.S. = R.H.S.

So, assertion (A) is true.

We know that,

(-p)^q = p^q; if q is even because the negative sign is raised to an even power and thus becomes positive.

So, reason (R) is true but reason does not explains assertion.

Hence, option 2 is the correct option.

Using the property:

⇒ a⁰ = 1 for any a ≠ 0

Thus, every non-zero number raised to the power zero (0) is always equal to 1.

So, reason (R) is true.

According to assertion : (7⁰ + 2⁰)(7⁰ - 2⁰) = 0

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

⇒ (7⁰ + 2⁰)(7⁰ - 2⁰)

⇒ (1 + 1)(1 - 1)

⇒ 2 x 0

⇒ 0.

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

So, assertion (A) is true and reason clearly explains assertion.

Hence, option 1 is the correct option.

According to assertion : $\Big(\frac{1}{5}\Big)^{-5} \times \Big(\frac{1}{2}\Big)^{-5} = (10)^{-5}$

Solving L.H.S.,

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

Since, L.H.S. ≠ R.H.S.

So, assertion (A) is false.

We know that,

$p^{-q} = \frac{1}{p^q} \text{ and } \frac{1}{p^{-q}} = p^q$, for p ≠ 0

So, reason (R) is true.

Hence, option 4 is the correct option.

According to assertion : (p - q)^-1(p^-1 - q^-1) = -(pq)^-1

Solving L.H.S.,

$$\begin{aligned}\Rightarrow (p - q)^{-1}(p^{-1} - q^{-1})\\[1em] \Rightarrow \Big(\frac{1}{p - q}\Big)\Big(\frac{1}{p} - \frac{1}{q}\Big)\\[1em] \Rightarrow \Big(\frac{1}{p - q}\Big)\Big(\frac{q - p}{pq}\Big)\\[1em] \Rightarrow -\Big(\frac{1}{p - q}\Big)\Big(\frac{p - q}{pq} \Big)\\[1em] \Rightarrow -\frac{(p - q)}{pq(p - q)}\\[1em] \Rightarrow -\frac{1}{pq}\\[1em] \Rightarrow -(pq)^{-1}\end{aligned}$$

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

So, assertion (A) is true.

Multiplying a^-1 and a¹, we get :

⇒ a^-1 x a¹

⇒ $\frac{1}{a}$ x a

⇒ 1.

Thus, a^-1 and a¹ are multiplicative inverses of each other.

So, reason is true but it does not explains assertion.

Hence, option 2 is the correct option.

$$\begin{aligned}\Big(-\frac{3}{5}\Big)^3\\[1em] = \Big(-\frac{3 \times 3 \times 3}{5 \times 5 \times 5}\Big)\\[1em] = -\frac{27}{125}\end{aligned}$$

Hence, $\Big(-\frac{3}{5}\Big)^3 = -\frac{27}{125}$.

$$\begin{aligned}\Big(\frac{2}{7}\Big)^{-2}\\[1em] = \Big(\frac{7}{2}\Big)^2\\[1em] = \Big(\frac{7 \times 7}{2 \times 2}\Big)\\[1em] = \frac{49}{4}\\[1em] = 24\frac{1}{4}\end{aligned}$$

Hence, $\Big(\frac{2}{7}\Big)^{-2} = 24\frac{1}{4}$.

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

Hence,$\Big[-\frac{2}{5}\Big]^4 \times \Big[-\frac{5}{2}\Big]^2 = \Big[\frac{4}{25}\Big]$.

$$\begin{aligned}\Big[-\frac{3}{7}\Big]^{-5} \times \Big[\frac{7}{3}\Big]^2\\[1em] = \Big[-\frac{7}{3}\Big]^5 \times \Big[\frac{7}{3}\Big]^2\\[1em] = \Big[-\frac{7^5}{3^5}\Big] \times \Big[\frac{7^2}{3^2}\Big]\\[1em] = \Big[-\frac{7^{5+2}}{3^{5+2}}\Big]\\[1em] = \Big[-\frac{7^{7}}{3^{7}}\Big]\end{aligned}$$

Hence,$\Big[-\frac{3}{7}\Big]^{-5} \times \Big[\frac{7}{3}\Big]^2 = \Big[-\frac{7^{7}}{3^{7}}\Big]$.

$$\begin{aligned}\Big\lbrace\Big(-\frac{1}{2}\Big)^{-2}\Big\rbrace^{-3}\\[1em] = \Big\lbrace\Big(-\frac{2}{1}\Big)^2\Big\rbrace^{-3}\\[1em] = \Big\lbrace\Big(-\frac{2 \times 2}{1 \times 1}\Big)\Big\rbrace^{-3}\\[1em] = \Big\lbrace\frac{4}{1}\Big\rbrace^{-3}\\[1em] = \Big\lbrace\frac{1}{4}\Big\rbrace^3\\[1em] = \Big\lbrace\frac{1 \times 1 \times 1 }{4 \times 4 \times 4}\Big\rbrace\\[1em] = \frac{1}{64}\end{aligned}$$

$\Big\lbrace\Big(-\frac{1}{2}\Big)^{-2}\Big\rbrace^{-3} = \frac{1}{64}$

$$\begin{aligned}\Big[\Big\lbrace\Big(-\frac{1}{5}\Big)^2\Big\rbrace^{-2}\Big]^{-1}\\[1em] = \Big[\Big\lbrace\Big(\frac{-1 \times -1}{5 \times 5}\Big)\Big\rbrace^{-2}\Big]^{-1}\\[1em] = \Big[\Big\lbrace\frac{1}{25}\Big\rbrace^{-2}\Big]^{-1}\\[1em] = \Big[\Big\lbrace\frac{25}{1}\Big\rbrace^2\Big]^{-1}\\[1em] = \Big[\Big\lbrace\frac{25 \times 25}{1 \times 1}\Big\rbrace\Big]^{-1}\\[1em] = \Big[\frac{625}{1}\Big]^{-1}\\[1em] = \frac{1}{625}\end{aligned}$$

$\Big[\Big\lbrace\Big(-\frac{1}{5}\Big)^2\Big\rbrace^{-2}\Big]^{-1} = \frac{1}{625}$

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

LCM of 7 and 8 is 2 x 2 x 2 x 7 = 56

And

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

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

LCM of 56 and 1 is 56

$$\begin{aligned}= \Big(\frac{1}{56}\Big) - \Big(\frac{12 \times 56}{1 \times 56}\Big)\\[1em] = \Big(\frac{1}{56}\Big) - \Big(\frac{672}{56}\Big)\\[1em] = \Big(\frac{1 - 672}{56}\Big)\\[1em] = -\frac{671}{56}\\[1em] = -11\frac{55}{56}\end{aligned}$$

$(7^{-1} - 8^{-1}) - (3^{-1} - 4^{-1})^{-1} = -11\frac{55}{56}$

$$\begin{aligned}5^{-7} ÷ 5^{-10} \times 5^{-5}\\[1em] = 5^{(-7) - (-10) + (-5)}\\[1em] = 5^{-7 + 10 - 5}\\[1em] = 5^{3 - 5}\\[1em] = 5^{-2}\\[1em] = \frac{1}{5}^2\\[1em] = \frac{1}{25}\end{aligned}$$

$5^{-7} ÷ 5^{-10} \times 5^{-5} = \frac{1}{25}$

Let the number be $x$. So,

$$\begin{aligned}(-5)^{-1} ÷ x = (-25)^{-1}\\[1em] \Rightarrow \frac{-1}{5} ÷ x = \frac{-1}{25}\\[1em] \Rightarrow \frac{-1}{5} \times \frac{1}{x} = \frac{-1}{25}\\[1em] \Rightarrow -\frac{1}{5x} = -\frac{1}{25}\\[1em] \Rightarrow -5x = 25 \\[1em] \Rightarrow x = \frac{25}{5} \\[1em] \Rightarrow x = 5\end{aligned}$$

Hence, (-5)^-1 should be divided by 5 to give the quotient as (-25)^-1.

$$\begin{aligned}8^{11} ÷ 8^5 = 8^{-3} \times 8^{2n-1}\\[1em] \Rightarrow 8^{11 - 5} = 8^{-3 + (2n-1)}\\[1em] \Rightarrow 8^{6} = 8^{-3 + 2n - 1}\\[1em] \Rightarrow 8^{6} = 8^{-4 + 2n}\\[1em] \Rightarrow 6 = -4 + 2n\\[1em] \Rightarrow 6 + 4 = 2n\\[1em] \Rightarrow 10 = 2n\\[1em] \Rightarrow n = \frac{10}{2} \\[1em] \Rightarrow n = 5\end{aligned}$$

If $8^{11} ÷ 8^5 = 8^{-3} \times 8^{2n-1}$, then $n = 5$.

$$\begin{aligned}9^{n+2} = 240 + 9^n\\[1em] \Rightarrow 9^{n+2} - 9^{n} = 240\\[1em] \Rightarrow 9^{n}[9^2 - 1] = 240\\[1em] \Rightarrow 9^n[81 - 1] = 240\\[1em] \Rightarrow 9^n[80] = 240\\[1em] \Rightarrow 9^n = \frac{240}{80}\\[1em] \Rightarrow 9^n = 3\\[1em] \Rightarrow 3^{2n} = 3^1 \\[1em] \Rightarrow 2n = 1 \\[1em] \Rightarrow n = \frac{1}{2}\end{aligned}$$

If $9^{n+2} = 240 + 9^n$, then $n = \frac{1}{2}$.

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

If $3^{2x-1} = (27)^{x-3}$, then $x = 8$.

$$\begin{aligned}\frac{25^x \times 5^5 \times (125)^3}{5 \times (625)^4} = 125\\[1em] \Rightarrow \frac{5^{2x} \times 5^5 \times (5^3)^3}{5 \times (5^4)^4} = 5^3\\[1em] \Rightarrow \frac{5^{2x} \times 5^5 \times (5)^9}{5 \times (5)^{16}} = 5^3\\[1em] \Rightarrow \frac{5^{2x+5+9}}{5^{1+16}} = 5^3\\[1em] \Rightarrow \frac{5^{2x+14}}{5^{17}} = 5^3\\[1em] \Rightarrow 5^{2x+14-17} = 5^3\\[1em] \Rightarrow 5^{2x-3} = 5^3\\[1em] \Rightarrow 2x - 3 = 3\\[1em] \Rightarrow 2x = 3 + 3\\[1em] \Rightarrow 2x = 6\\[1em] \Rightarrow x = \frac{6}{2}\\[1em] \Rightarrow x = 3\end{aligned}$$

If $\frac{25^x \times 5^5 \times (125)^3}{5 \times (625)^4} = 125$, then $x = 3$.