1. thanks i understand now, the answer is 2+ radical 5 rite?

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

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

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

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

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