How to prove that $2$ is the only solution of the equation $x=\sqrt{2}^x$? I tried to prove that $2$ is the only solution to the equation $x=\sqrt{2}^x$ without any results. 
Here's my try : Let $f:[0,+\infty)\rightarrow\mathbb{R}$ such that $f(x)=\sqrt{2}^x-x$. Thus, $f'(x)=\sqrt{2}^x\ln\sqrt2-1$. $f'$ converges to $0$ when $x=\log_{\sqrt2}\frac{1}{\ln\sqrt2}$. The derivate should keep it's sign to prove that $f$ has exactly one solution, I can't understand what is going on. Please, help
 A: It is not the only solution: $x=4$ is another solution. Actually, you can show, in your notations, that 


*

*$f'(x)<0\;$ for $\;0\le x<\dfrac{2(\ln 2-\ln(\ln 2))}{\ln2}\approx 3.06$,

*$f'(x)>0\;$ for $\;x> \dfrac{2(\ln 2-\ln(\ln 2))}{\ln2}$.


So there are only two solutions: $2$ and $4$.
A: Looks like $2$ isn't the only solution. If you take the power $\frac{1}{x}$ to each side you get 
$$x^{\frac{1}{X}} = 2^{\frac{1}{2}}$$
Obviously, $x=2$ works, but also notice that
$$4^{\frac{1}{4}} = (2^2)^{\frac{1}{4}} = 2^{\frac{1}{2}}$$
So $x=4$ is also a solution.
A: The function $f(x)=\frac {1}{x}\ln x$ is strictly increasing for $x\in [1,e]$ and strictly decreasing for $x\geq e$ because $f'(x)=\frac {1-\ln x}{x^2}.$ Also $f(1)=0=\lim_{x\to \infty}f(x).$ So for any $u\in (0,e)$ there is a unique $v>e$ such that $f(u)=f(v).$ That is, $u^{1/u}=v^{1/v}.$ In particular when $u=2$ we have $v=4.$
A: Let $f(x) = e^{\frac x 2\ln 2}$ and rewrite your equation to $f(x)=x$. If $x = 2$ were a unique solution, then the line $y= x$ would be tangent to graph of $f(x)$. But, $f'(x) = \frac{\ln 2}2e^{\frac x 2\ln 2}$, hence $f'(2) = \ln 2\neq 1$, which is slope of $y = x$. Thus, the solution is not unique.
