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?
 A: 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,
$$x= \frac{2\ln\left( 1+ \sqrt[3]{10+\frac{15}4\sqrt6}+ \sqrt[3]{10-\frac{15}4\sqrt6}\right)-2\ln3}{\ln\frac{\sqrt{5}-1}{2} }$$
A: There are two solutions (see below).
Here is how we can prove this fact in a rigorous way. Let :
$$f(x):=(2+ \sqrt{5})^{\frac{x}{2}}+2 \cdot \left(\frac{7 - 3 \sqrt{5}}{2}\right)^{\frac{x}{4}}\tag{1}$$
which has the form 
$$f(x)=a^x+2b^x \ \text{with} \  a>1, b<1.\tag{2}$$
Its derivative being $f'(x)=\ln(a) a^x +2\ln(b) b^x,$
we have $$f'x)<0 \ \iff \ \ln(a) a^x +2\ln(b) b^x<0$$
$$\iff \left(\frac{a}{b}\right)^x<\underbrace{-2\dfrac{\ln b}{\ln a}}_C$$
$$\iff x<\dfrac{\ln C}{\ln\left(\frac{a}{b}\right)}$$
(the RHS, named $C$, being a positive number because $\ln a>0$ whereas $\ln b<0$). 
Therefore, $f$ is first decreasing, till a certain value $x_0$ then increasing. 
Besides, we have, taking (2) into account : 
$$\lim_{x \to -\infty}f(x)=\lim_{x \to +\infty}f(x)=+\infty$$
Therefore, as $f(0)=3$, the initial equation $f(x)=5$ has two solutions. One of them is clearly

$$x=2$$

(indeed $f(2)=(2+ \sqrt{5})+2\sqrt{\frac{7 - 3 \sqrt{5}}{2}}=2+\sqrt{5}+3-\sqrt{5}=5$.)
The other one is negative :

$$x\approx-1.7863876...$$

(see figure below)

