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Math Help - Logs and Radicals

  1. #16
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    thanks i understand now, the answer is 2+ radical 5 rite?
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  2. #17
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    \displaystyle log_{\frac{1}{2+\sqrt{5}}}(-2+\sqrt{5})=x-4\Rightarrow \left(\frac{1}{2+\sqrt{5}}\right)^{x-4}=\sqrt{5}-2

    The answer is an integer.

    To bring the x-4 down, you must take the log of both sides.

    Hint: use \displaystyle log_{\frac{1}{2+\sqrt{5}}

    Spoiler:
    <br />
log_{\frac{1}{2+\sqrt{5}}}\left[\left(\frac{1}{2+\sqrt{5}}\right)^{x-4}\right]=log_{\frac{1}{2+\sqrt{5}}}(\sqrt{5}-2) \Rightarrow (x-4)log_{\frac{1}{2+\sqrt{5}}}\left(\frac{1}{2+\sqrt  {5}}\right)=1
    Last edited by dwsmith; December 16th 2010 at 06:35 PM. Reason: forgot an
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