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Exponents - Solved Examples


Example 1. Evaluate the following exponents

(a) $6^4$
Solution $6^4$ = 6 * 6 * 6 * 6 = 1296


(b) $(-7)^{-2}$
Solution $(-7)^{-2} = \frac{1}{(-7)^2} = \frac{1}{49}$


(c) $ (\frac{-3}{4})^{-3}$
Solution $(\frac{-3}{4})^{-3} = (\frac{4}{-3})^{3} = (\frac{-4}{3})^{3} = (\frac{-4}{3}) × (\frac{-4}{3}) × (\frac{-4}{3}) = (\frac{-64}{27})$




Example 2. Evaluate:

(a) $6^4 × 6^2$
Solution $6^4 × 6^2 = 6^{4 + 2} = 6^6 = 46656$


(b) $2^4 × 2^{-3}$
Solution $2^4 × 2^{-3} = 2^{4 - 3} = 2^1 = 2$


(c) $(\frac{3}{4})^4 × (\frac{5}{4})^{-2}$
Solution
$\begin{align*} \left (\frac{3}{4} \right )^4 × \left (\frac{5}{4} \right )^{-2} & = \left (\frac{3}{4} \right )^4 × \left (\frac{4}{5} \right )^{2}
\\ & = \frac{3^4}{4^4} × \frac{4^2}{5^2}
\\ & = \frac{3^4}{4^2 × 5^2}
\\ & = \frac{81}{16 × 25}
\\ & = \frac{81}{400}
\end{align*} $


(d) $(5^2)^3$
Solution $(5^2)^3 = 5^{2 * 3} = 5^6 = 15625$


(e) $\left ( \left [\frac{-5}{4} \right ]^2 \right )^{-3}$
Solution
$\begin{align*} \left ( \left [\frac{-5}{4} \right ]^2 \right )^{-3} & = \left [\frac{-5}{4} \right ]^{2 × ({-3})} \\ & = \left [\frac{-5}{4} \right ]^{-6} \\ & = \left [\frac{4}{-5} \right ]^{6} \\ & = \left [\frac{-4}{5} \right ]^{6} \\ & = \left [\frac{-4^6}{5^6} \right ] \\ & = \frac{-4096}{15625} \\ \end{align*} $


(f) $ ( 3^{-1} + 6^{-1} ) ÷ \left (\frac{3}{4} \right )^{-1}$
Solution
$\begin{align*} ( 3^{-1} + 6^{-1} ) ÷ \left (\frac{3}{4} \right )^{-1} & = \left ( \frac{1}{3} + \frac{1}{6} \right) ÷ \left (\frac{4}{3} \right )^{1} \\ & = \left ( \frac{2 + 1}{6} \right) ÷ \left (\frac{4}{3} \right ) \\ & = \left ( \frac{3}{6} \right) ÷ \left (\frac{4}{3} \right ) \\ & = \left ( \frac{1}{2} \right) ÷ \left (\frac{4}{3} \right ) \\ & = \left ( \frac{1}{2} × \frac{3}{4} \right) \\ & = \frac{3}{8} \\ \end{align*} $


(g) If $ \left (\frac{3}{14} \right)^{-4} × \left (\frac{3}{14} \right) ^ {3x} = \left (\frac{3}{14} \right)^{5}$, then $x$ = ?
Solution
$\begin{align*} \left (\frac{3}{14} \right)^{-4} × \left (\frac{3}{14} \right) ^ {3x} = \left (\frac{3}{14} \right)^{5} & ⇒ -4 + 3x = 5 \\ & ⇒ 3x = 9 \\ & ⇒ x = 3 \\ \end{align*} $





Example 3.

(a) By what number should $ \left (\frac{1}{2} \right)^{-1}$ be multiplied so that the product is $\left (\frac{-5}{4} \right )^{-1}$ ?
Solution
Let the required number be $x$. Then,
$\begin{align*} \left (\frac{1}{2} \right)^{-1} × {x} = \left (\frac{-5}{4} \right )^{-1} & ⇒ x = \frac {\left (\frac{-5}{4} \right )^{-1}} {\left (\frac{1}{2} \right)^{-1}} \\ & ⇒ x = \frac {\left (\frac{1}{2} \right)} {\left (\frac{-5}{4} \right )} \\ & ⇒ x = \left (\frac{1}{2} \right) × \left (\frac{-4}{5} \right ) \\ & ⇒ x = \frac{-4}{10} \\ & ⇒ x = \frac{-2}{5} \end{align*} $


(b) By what number should $ \left (\frac{-2}{3} \right)^{-3}$ be divided so that the quotient is $\left (\frac{4}{9} \right )^{-2}$ ?
Solution
Let the required number be $x$. Then,
$\begin{align*} \frac {\left (\frac{-2}{3} \right)^{-3}} {x} = \left (\frac{4}{9} \right )^{-2} & ⇒ \frac {\left (\frac{-2}{3} \right)^{-3}} {\left (\frac{4}{9} \right )^{-2}} = x \\ & ⇒ x = \frac {\left (\frac{4}{9} \right )^{2}} {\left (\frac{-2}{3} \right)^{3}} \\ & ⇒ x = \left (\frac{4}{9} \right )^{2} ÷ \left (\frac{-2}{3} \right)^{3} \\ & ⇒ x = \left (\frac{4}{9} \right )^{2} × \left (\frac{3}{-2} \right)^{3} \\ & ⇒ x = \left (\frac{4}{9} \right )^{2} × \left (\frac{-3}{2} \right)^{3} \\ & ⇒ x = \left (\frac{4 × 4 × -3 × -3 × -3}{9 × 9 × 2 × 2 × 2} \right ) \\ & ⇒ x = \left (\frac{4 × -3 }{9 × 2 } \right ) \\ & ⇒ x = \frac{-2}{3} \\ \end{align*} $