Question

Fill in the blanks.

${\left(\frac{{15}^{\frac{1}{4}}}{3^{\frac{1}{2}}}\right)}^{-2}=\mathrm{................}$

Answer 1

Simplify the given expression using the law of radical exponent.

$\begin{array}{l}{\left(\frac{{15}^{\frac{1}{4}}}{3^{\frac{1}{2}}}\right)}^{-2}\\ =\left\{\frac{{\left({15}^{\frac{1}{4}}\right)}^{-2}}{{\left(3^{\frac{1}{2}}\right)}^{-2}}\right\}\\ =\left\{\frac{{\left(15\right)}^{\frac{1}{4}\times -2}}{{\left(3\right)}^{\frac{1}{2}\times -2}}\right\}\left[\because {\left(a^m\right)}^n=a^{mn}\right]\\ =\left\{\frac{{\left(15\right)}^{-\frac{1}{2}}}{{\left(3\right)}^{-1}}\right\}\\ =\frac{3}{\sqrt{15}}\left[\because a^{-n}=\frac{1}{a^n}\right]\\ =\frac{3\sqrt{15}}{\sqrt{15}\times \sqrt{15}}\left[\text{rationalise the denominator}\right]\\ =\frac{3\sqrt{15}}{15}\\ =\frac{\sqrt{15}}{5}\end{array}$

Hence, ${\left(\frac{{15}^{\frac{1}{4}}}{3^{\frac{1}{2}}}\right)}^{-2}=\frac{\sqrt{15}}{5}$