# Math Help - Totally lost

1. ## Totally lost

By solving a suitable equation in k, find a positive integer k such that...

$
(x^2)^{20-k}(\frac{1}{x})^k=x
$

2. Apply properties of exponents:

$
\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}
$

3. Thanks very much

4. You're welcome, hopefully clear to you now?

5. Originally Posted by TD!
You're welcome, hopefully clear to you now?
It always is after the fact

6. Better late than never