# Logs and Radicals

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• Dec 16th 2010, 05:51 PM
HELPHELPHELP
thanks i understand now, the answer is 2+ radical 5 rite?
• Dec 16th 2010, 05:57 PM
dwsmith
$\displaystyle log_{\frac{1}{2+\sqrt{5}}}(-2+\sqrt{5})=x-4\Rightarrow \left(\frac{1}{2+\sqrt{5}}\right)^{x-4}=\sqrt{5}-2$

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:
$
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)$
$\Rightarrow (x-4)log_{\frac{1}{2+\sqrt{5}}}\left(\frac{1}{2+\sqrt {5}}\right)=1$
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