Solve $2^x=x^2$

Your equation has two obvious solutions which are $x=2$ and $x=4$. The last solution is not rational ($x \approx -0.766665$) and cannot be obtained using simple functions. You cannot get the last root using logarithms.

Consider the function$$f(x):=(\ln 2)x-2\ln x$$ then $f^\prime (x)=\ln 2-2/x$. Then it easily follows that $f^\prime (x)>0$ when $x>4$ and $f^\prime (x)< 0$ when $x<2$. That is $f$ is increasing when $x>4$ and it is decreasing when $x<2$. Also $4$ and $2$ are zeros of $f$. Hence it follows that these are the only zero for $x>0$.

For, $x<0$ put $x=-y$ and consider the function $$g(y)=-(\ln 2)y-2\ln y$$ Then $g^\prime (y)=-\ln 2-2/y<0$ for all $y>0$ i.e. the function is strictly decreasing and hence it has exactly one root for $x<0$.

There is a special function, $W_0(x)$ that is the inverse of $f(x)=xe^x$ when the latter is restricted to $x\in [-1,\infty)$. Using this, expressions of the form $Y=Xe^X$ can be solved as $X=W_0(Y)$. You want to find the solution(s) to the equation $2^x=x^2$. Rewrite $2^x$ as $e^{\ln(2)x}$ and raise each side to the power of $\frac{1}{2}$. We then arrive at $$x=e^{\frac{\ln(2)}{2}x}$$Multiple both sides by $\frac{-\ln(2)}{2}e^{\frac{-\ln(2)}{2}}x$ to arrive at $$\frac{-\ln(2)}{2} x e^{\frac{-\ln(2)}{2}x}=\frac{-\ln(2)}{2}$$Apply $W_0$ to both sides to get $$\frac{-\ln(2)}{2}x=W_0\left(\frac{-\ln(2)}{2}\right)$$ Multiply through to find $$x=\frac{-2}{\ln(2)} W_0\left(\frac{-\ln(2)}{2}\right)$$Which is equal to 2.