Question

Find the value of the following expression:

${\left(\frac{256}{625}\right)}^{-1/2}$

Answer 1

Step 1: Factorise $256\text{and}625$.

$\begin{array}{l}{\left(\frac{256}{625}\right)}^{-1/2}\\ ={\left(\frac{16\times 16}{25\times 25}\right)}^{-1/2}\\ ={\left(\frac{{16}^2}{{25}^2}\right)}^{-1/2}\end{array}$

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

$\begin{array}{l}{\left(\frac{{16}^2}{{25}^2}\right)}^{-1/2}\\ ={\left(\frac{{25}^2}{{16}^2}\right)}^{1/2}\left[\because p^{-n}=\frac{1}{p^n}\right]\\ ={\left\{{\left(\frac{25}{16}\right)}^2\right\}}^{1/2}\left[\because {\left(\frac{p}{q}\right)}^n=\frac{p^n}{q^n}\right]\\ ={\left(\frac{25}{16}\right)}^{2\times \frac{1}{2}}\left[{\because \left(p^m\right)}^n=p^{m\times n}\right]\\ =\frac{25}{16}\end{array}$

Hence, the value of the given expression is $\frac{25}{16}$.