<s>[INST] Solve equation $(2+ \sqrt{5})^{\frac{x}{2}}+2 \cdot \left(\frac{7 - 3 \sqrt{5}}{2}\right)^{\frac{x}{4}}=5$ Find the value of $x$ given the equation,
$$(2+ \sqrt{5})^{\frac{x}{2}}+2 \cdot \left(\frac{7 - 3 \sqrt{5}}{2}\right)^{\frac{x}{4}}=5$$
I think they are powers of $ \frac {1} {\Phi} $ or $ \Phi $. In case I think the first would be $ \Phi ^ 3 $ and the second $ (\frac {1} {\Phi}) ^ 4 $. Is it true? 
How to solve the problem?
 [/INST] Recognize 
$$\left(\frac{7 - 3 \sqrt{5}}{2}\right)^{x/4} 
= \left(\frac{3 - \sqrt{5}}{2}\right)^{x/2}
= \left(\frac{\sqrt{5}-1}{2}\right)^{x}$$
$$(2+ \sqrt{5})\left(\frac{3 - \sqrt{5}}{2}\right) = \frac2{\sqrt5-1}$$
and rewrite the given equation as
$$\left(\frac2{\sqrt5-1}\right)^{x/2}
+2 \cdot \left(\frac{\sqrt{5}-1}{2}\right)^{2x}=5\left(\frac{\sqrt{5}-1}{2}\right)^{x}$$
or
$$2a^5-5a^3+1 = 0, \>\>\>\text{where}\>\>\> a = \left(\frac{\sqrt{5}-1}{2}\right)^{x/2}$$
Then, factorize to get,
$$(a^2+a-1)(2a^3-2a^2-a-1)=0$$
The first factor $a^2+a-1=0$ yields,
$$a = \left(\frac{\sqrt{5}-1}{2}\right)^{x/2} = \frac{\sqrt{5}-1}{2}$$
which leads to the first solution $x=2$. On the other hand, $$2a^3-2a^2-a-1=0$$ has one real root given by 
$$a = \left(\frac{\sqrt{5}-1}{2}\right)^{x/2} 
= \frac13 + \frac13\sqrt[3]{10+\frac{15}4\sqrt6}+ \frac13\sqrt[3]{10-\frac{15}4\sqrt6}$$
which leads to the second solution,
$$x= \frac{2\ln\left( 1+ \sqrt[3]{10+\frac{15}4\sqrt6}+ \sqrt[3]{10-\frac{15}4\sqrt6}\right)-2\ln3}{\ln\frac{\sqrt{5}-1}{2} }$$
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