Find explicitly the positive solutions of $2^x=x^2$ 
Find explicitly the positive solutions of the equation $2^x=x^2$

I noticed that $x=2$ and $x=4$ are roots of the equation.
How can I prove that they are the only positive ones? Thanks in advance
 A: Let $f(x)=2^x-x^2$. Then by implicit differentiation $$f'(x)=2^x\ln2-2x,\quad f''(x)=2^x\ln^22-2$$ Now solving $f''(x)=0$ gives $$x_0=1-\frac{\ln\ln^22}{\ln2}$$ which is the global minimum of $f'(x)$. 
Since $$f'(0)=\ln2>0,\quad f'(x_0)<0,\quad f'(4)=16\ln2-8>0$$ and $f'$ is continuous in $\mathbb{R}$, by the Intermediate Value Theorem there are exactly two solutions to $f'(x)=0$.
Can you continue?
A: They're not the only solutions, if we also allow negative nummbers:. Since $2^0>0^2$ and $2^{-1}<(-1)^2$ there exists a solution $x$ with $-1<x<0$.
Those three are the only real solutions. Let $f(x)=2^x-x^2$. Calculate the third derivative and show that $f'''(x)>0$ for every $x$. If $f$ had four zeroes the Mean Value Theorem would show that $f'$ had three zeroes, hence $f''$ would have two zeroes, hence $f'''$ would have a zero.
So since there are only three real solutions and one is negative there are only two positive solutions.
A: For positive $x$, we have $2^x=x^2\iff a^x=x$, where $a=\sqrt2$.  Since $0\lt a\not=1$, the exponential curve $y=a^x$ is concave up, hence any line intersects it at most twice. The line $y=x$ does so at $x=2$ and $x=4$.  So those are the only positive solutions of $2^x=x^2$. (As David Ullrich notes, there is also a negative solution; it corresponds to the interesection of the exponential curve $y=a^x$ and the line $y=-x$.)
