Solve equation $(2+ \sqrt{5})^{\frac{x}{2}}+2 \cdot \left(\frac{7 - 3 \sqrt{5}}{2}\right)^{\frac{x}{4}}=5$

Question

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?

Answer 1

Recognize

$$\left(\frac{7 - 3 \sqrt{5}}{2}\right)^{1/4} = \frac{\sqrt{5}-1}{2},\>\>\> (2+ \sqrt{5})= \left(\frac2{\sqrt5-1}\right)^3$$

and rewrite the given equation $(2+ \sqrt{5})^{\frac{x}{2}}+2 \cdot \left(\frac{7 - 3 \sqrt{5}}{2}\right)^{\frac{x}{4}}=5$ as

$$2a^5-5a^3+1 = 0$$ with $a = (\frac{\sqrt{5}-1}{2})^{\frac x2}$. Then, factorize

$$(a^2+a-1)(2a^3-2a^2-a-1)=0$$

The first equation $a^2+a-1=0$ leads to $a = \frac{\sqrt{5}-1}{2}$ and the solution $x=2$, while the second equation $2a^3-2a^2-a-1=0$ 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

$$x= \frac{2\ln\left( 1+ \sqrt[3]{10+\frac{15}4\sqrt6}+ \sqrt[3]{10-\frac{15}4\sqrt6}\right)-2\ln3}{\ln\frac{\sqrt{5}-1}{2} }$$

Answer 2

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)