By solving a suitable equation in k, find a positive integer k such that...
$\displaystyle
(x^2)^{20-k}(\frac{1}{x})^k=x
$
Apply properties of exponents:
$\displaystyle
\begin{gathered}
\left( {x^2 } \right)^{20 - k} \left( {\frac{1}
{x}} \right)^k = x \hfill \\
x^{2\left( {20 - k} \right)} \left( {x^{ - 1} } \right)^k = x \hfill \\
x^{40 - 2k} x^{ - k} = x \hfill \\
x^{40 - 3k} = x \hfill \\
\Rightarrow 40 - 3k = 1 \Leftrightarrow k = 13 \hfill \\
\end{gathered}
$