Question

Find the value of the following expression:

${\left({32}^{-2/5}\right)}^{1/2}$

Answer 1

Step 1: Factorise $32$ into prime numbers.

$\begin{array}{l}{\left({32}^{-2/5}\right)}^{1/2}\\ ={\left\{{\left(32\right)}^{-2/5}\right\}}^{1/2}\\ ={\left\{{\left(2\times 2\times 2\times 2\times 2\right)}^{-2/5}\right\}}^{1/2}\\ ={\left\{{\left(2^5\right)}^{-2/5}\right\}}^{1/2}\end{array}$

Step 2: Simplify the above expression using the laws of rational exponents.

$\begin{array}{l}{\left\{{\left(2^5\right)}^{-2/5}\right\}}^{1/2}\\ ={\left\{{\left(2\right)}^{5\times \left(-\frac{2}{5}\right)}\right\}}^{\frac{1}{2}}\left[\because {\left(p^n\right)}^m=p^{m\times n}\right]\\ ={\left\{{\left(2\right)}^{-2}\right\}}^{\frac{1}{2}}\\ ={\left(2\right)}^{-2\times \frac{1}{2}}\\ ={\left(2\right)}^{-1}\\ =\frac{1}{2}\left[\because p^{-n}=\frac{1}{p^n}\right]\end{array}$

Hence, the value of the given expression is $\frac{1}{2}$.