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

$\displaystyle

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

$

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- Apr 17th 2006, 11:12 AMmacca101Totally lost
By solving a suitable equation in k, find a positive integer k such that...

$\displaystyle

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

$ - Apr 17th 2006, 11:24 AMTD!
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}

$ - Apr 17th 2006, 11:42 AMmacca101
Thanks very much

- Apr 17th 2006, 11:45 AMTD!
You're welcome, hopefully clear to you now?

- Apr 17th 2006, 11:48 AMmacca101Quote:

Originally Posted by**TD!**

- Apr 17th 2006, 11:52 AMTD!
Better late than never ;)