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Math Help - Totally lost

  1. #1
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    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 />
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  2. #2
    TD!
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    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 />
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  3. #3
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    Thanks very much
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  4. #4
    TD!
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    You're welcome, hopefully clear to you now?
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  5. #5
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    Quote Originally Posted by TD!
    You're welcome, hopefully clear to you now?
    It always is after the fact
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  6. #6
    TD!
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    Better late than never
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