I was bored and started solving the following equation:
$$2^x = x^2$$
I can see two solutions: $x = 2$ and $x = 4$. WolframAlpha tells me there is one more, but I can hardly get the two I mentioned above (at which I arrived with empirical methods). I fear this might be outside the realm of my Maths skills (up to multivariable calculus).
If you are asking (I assume you are) about how to get these two results analytically, I am almost certain that there is no way to get a closed solution to the equation $a^x = x^a$, except for special cases like $a=2$ in your case. In that case, using plots to have a look at how many solutions you can expect is useful:
http://www.wolframalpha.com/input/?i=plot+2%5Ex-x%5E2
You can see that there is another solution for some negative value of $x$, which you probably cannot get without numeric methods.
For the positive solutions, some things can actually be said without resorting to numeric methods.
You can show that for $x<2$, the function $2^x - x^2$ is positive, since it is increasing at $x=0$, decreasing at $x=2$ and is concave on $[0,2]$.
For $x>2$, you can show that because the function is convex, it can have at most two zeroes on $$[2,\infty).
Together, these two points show that the only positive solutions to your equation are $2$ and $4$
In Wolfram, use
solve x^2 - 2^x = 0
To find the negative solution to be about -0.766665
https://www.wolframalpha.com/input/?i=solve+x%5E2+-+2%5Ex+%3D+0
$2^x = x^2$. You can use any latex commands you find useful then. – 5xum Nov 05 '14 at 14:25