Solve the following
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) $\frac{2 x-5}{3}$

(d) $\frac{2}{3 \times 5}$

Solution:

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

Rearrange (2/3)−5 to get a positive exponent.

$\left(\frac{2}{3}\right)^{-5}=\frac{1}{\left(\frac{2}{3}\right)^{5}} \quad\left(a^{-n}=\frac{1}{a^{n}}\right)$

$=\frac{1}{\frac{2^{5}}{3^{5}}} \quad\left\{\left(\frac{a}{b}\right)^{n}=\frac{a^{n}}{b^{n}}\right\}$

$=\frac{3^{5}}{2^{5}}$

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

 

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