Question

Find the value of ${\left(\frac{16}{81}\right)}^{-3/4}÷{\left(\frac{243}{32}\right)}^{3/5}$ of ${\left(\frac{9}{4}\right)}^{-1/2}$

Answer 1

Step 1: Factorise each of the terms of the given expression.

Change of into multiply sign

$\begin{array}{l}{\left(\frac{16}{81}\right)}^{-3/4}÷{\left(\frac{243}{32}\right)}^{3/5}\times {\left(\frac{9}{4}\right)}^{-1/2}\\ ={\left(\frac{2\times 2\times 2\times 2}{3\times 3\times 3\times 3}\right)}^{-3/4}÷{\left(\frac{3\times 3\times 3\times 3\times 3}{2\times 2\times 2\times 2\times 2}\right)}^{3/5}\times {\left(\frac{3\times 3}{2\times 2}\right)}^{-1/2}\\ ={\left(\frac{2^4}{3^4}\right)}^{-3/4}÷{\left(\frac{3^5}{2^5}\right)}^{3/5}\times {\left(\frac{3^2}{2^2}\right)}^{-1/2}\end{array}$

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

$\begin{array}{l}{\left(\frac{2^4}{3^4}\right)}^{-3/4}÷{\left(\frac{3^5}{2^5}\right)}^{3/5}\times {\left(\frac{3^2}{2^2}\right)}^{-1/2}\\ ={\left\{{\left(\frac{2}{3}\right)}^4\right\}}^{-3/4}÷{\left\{{\left(\frac{3}{2}\right)}^5\right\}}^{3/5}\times {\left\{{\left(\frac{3}{2}\right)}^2\right\}}^{-1/2}\left[\because {\left(\frac{p}{q}\right)}^n=\frac{p^n}{q^n}\right]\\ ={\left(\frac{2}{3}\right)}^{-4\times \frac{3}{4}}÷{\left(\frac{3}{2}\right)}^{5\times \frac{3}{5}}\times {\left(\frac{3}{2}\right)}^{-2\times \frac{1}{2}}\left[\because {\left(p^n\right)}^m=p^{m\times n}\right]\\ ={\left(\frac{2}{3}\right)}^{-3}÷{\left(\frac{3}{2}\right)}^3\times {\left(\frac{3}{2}\right)}^{-1}\\ ={\left(\frac{3}{2}\right)}^3÷{\left(\frac{3}{2}\right)}^3\times \left(\frac{2}{3}\right)\left[\because p^{-n}=\frac{1}{p^n}\right]\\ =1\times \left(\frac{2}{3}\right)\\ =\frac{2}{3}\end{array}$

Hence, the value of the given expression is $\frac{2}{3}$.