Question

${\left(\frac{2}{3}\right)}^{-5}$ is equal to
(a) ${\left(\frac{-2}{3}\right)}^5$
(b) ${\left(\frac{3}{2}\right)}^5$
(c) $\frac{2x-5}{3}$
(d) $\frac{2}{3\times 5}$

Answer 1

(b)${\left(\frac{3}{2}\right)}^5$

Rearrange (2/3)⁻⁵ to get a positive exponent.
$$\begin{gathered} {\left(\frac{2}{3}\right)}^{-5}=\frac{1}{{\left({\displaystyle \frac{2}{3}}\right)}^5}\left(a^{-n}=\frac{1}{a^n}\right) \\ =\frac{1}{{\displaystyle \frac{2^5}{3^5}}}\left\{{\left(\frac{a}{b}\right)}^n=\frac{a^n}{b^n}\right\} \\ =\frac{3^5}{2^5} \\ ={\left(\frac{3}{2}\right)}^5 \end{gathered}$$