Totally lost

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April 17th 2006, 11:12 AM
macca101

Totally lost

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

$<br /> (x^2)^{20-k}(\frac{1}{x})^k=x<br />$
April 17th 2006, 11:24 AM
TD!

Apply properties of exponents:

$<br /> \begin{gathered}<br /> \left( {x^2 } \right)^{20 - k} \left( {\frac{1}<br /> {x}} \right)^k = x \hfill \\<br /> x^{2\left( {20 - k} \right)} \left( {x^{ - 1} } \right)^k = x \hfill \\<br /> x^{40 - 2k} x^{ - k} = x \hfill \\<br /> x^{40 - 3k} = x \hfill \\<br /> \Rightarrow 40 - 3k = 1 \Leftrightarrow k = 13 \hfill \\ <br /> \end{gathered} <br />$
April 17th 2006, 11:42 AM
macca101

Thanks very much
April 17th 2006, 11:45 AM
TD!

You're welcome, hopefully clear to you now?
April 17th 2006, 11:48 AM
macca101

Quote:

Originally Posted by TD!

You're welcome, hopefully clear to you now?

It always is after the fact :mad:
April 17th 2006, 11:52 AM
TD!

Better late than never ;)

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