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

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?