Question

Find the multiplicative inverse of ${\left(\frac{81}{16}\right)}^{(-\frac{3}{4})}$.

Answer 1

Find the multiplicative inverse

Let $x={\left(\frac{81}{16}\right)}^{\left(-\frac{3}{4}\right)}$

The multiplicative inverse of a number is its reciprocal.

So, $\frac{1}{x}=\frac{1}{{\left({\displaystyle \frac{81}{16}}\right)}^{\left(-{\displaystyle \frac{3}{4}}\right)}}$

$$\begin{gathered} \Rightarrow x^{-1}={\left(\frac{81}{16}\right)}^{\frac{3}{4}} \\ \Rightarrow x^{-1}=\frac{3^{4\times {\displaystyle \frac{3}{4}}}}{2^{4\times {\displaystyle \frac{3}{4}}}} \\ \Rightarrow x^{-1}=\frac{2^{-3}}{3^{-3}} \\ \Rightarrow x^{-1}={\left(\frac{3}{2}\right)}^3 \\ \Rightarrow x^{-1}=\frac{27}{8} \end{gathered}$$

Therefore, the multiplicative inverse of ${\left(\frac{81}{16}\right)}^{\left(-\frac{3}{4}\right)}$ will be $\frac{27}{8}$.