needs solution of the equation ${(2+{3}^{1/2}})^{x/2}$ + ${(2-{3}^{1/2}})^{x/2}$=$2^x$ $$\left(2+{3}^{1/2}\right)^{x/2} + \left(2-{3}^{1/2}\right)^{x/2} = 2^x.$$
Clearly $x = 2$ is a solution. i need others if there is any. Please help. 
 A: square this equation we have
$$(2+\sqrt{3})^x+(2-\sqrt{3})^x+2=4^x$$
so
$$\left(\dfrac{2+\sqrt{3}}{4}\right)^x+\left(\dfrac{2-\sqrt{3}}{4}\right)^x+\dfrac{2}{4^x}=1$$
It is clear 
$$f(x)=\left(\dfrac{2+\sqrt{3}}{4}\right)^x+\left(\dfrac{2-\sqrt{3}}{4}\right)^x+\dfrac{2}{4^x}$$ is decreasing,because use $y=a^x,0<a<1$ is decreasing
and Note
$$f(2)=1$$
so 
$$x=2$$
or consider
$$f(x)=\left(\dfrac{\sqrt{2+\sqrt{3}}}{2}\right)^x+\left(\dfrac{\sqrt{2-\sqrt{3}}}{2}\right)^x$$
A: Well, you're right: 2 is a solution.
Now, consider the function $$y={(2+{3}^{1/2}})^{x/2}+{(2-{3}^{1/2}})^{x/2}-2^x$$ and compute its derivative.
You obtain this function is always decreasing on $\mathbb{R}$ and then 2 is its only zero.
A: $$\dfrac{(2\pm\sqrt3)^{x/2}}{2^x}=\left(\dfrac{\sqrt{2\pm\sqrt3}}2\right)^x$$
Now $\dfrac{\sqrt{2\pm\sqrt3}}2=\dfrac{\sqrt{4\pm2\sqrt3}}{2\sqrt2}=\dfrac{\sqrt3\pm1}{2\sqrt2}=\sin45^\circ\cos30^\circ\pm\cos45^\circ\sin30^\circ=\sin(45\pm30)^\circ$
$\implies(\sin75^\circ)^x+(\sin15^\circ)^x=1$
$\implies(\sin75^\circ)^x+(\cos75^\circ)^x=1$
Clearly, $x=2$ is a solution.
Now $0<\sin75^\circ<\cos75^\circ<1\implies (\sin75^\circ)^x+(\cos75^\circ)^x$ is decreasing function
