Question

If x = 8^2/3 × 32^–2/5, then x^–5 = ________.

Answer 1

$$\begin{gathered} \mathrm{Let}x=8^{\frac{2}{3}}\times {32}^{-\frac{2}{5}} \\ \Rightarrow x={\left(2^3\right)}^{\frac{2}{3}}\times {\left(2^5\right)}^{-\frac{2}{5}} \\ \Rightarrow x={\left(2\right)}^{3\times \frac{2}{3}}\times {\left(2\right)}^{5\times \left(-\frac{2}{5}\right)} \\ \Rightarrow x={\left(2\right)}^2\times {\left(2\right)}^{-2} \\ \Rightarrow x=2^2\times \frac{1}{2^2} \\ \Rightarrow x=1 \\ \mathrm{Now}, \\ {x}^{-5}=\frac{1}{x^5} \\ =\frac{1}{1^5} \\ =1 \end{gathered}$$

Hence, if x = 8^2/3 × 32^–2/5, then x^–5 = 1.