Does there exist an $x$ such that $3^x = x^2$? I tried solving for $x$ by using $x \log(3) = \log(x^2)
$$\log(3) = \frac{\log(x^2)}{x}$$
I'm stuck on this part. how do I isolate $x$ by itself?
Any help would be appreciated.
 A: Let $f:\mathbb{R}\rightarrow \mathbb{R}$, $f(x)=3^x-x^2$.
Note that $f$ is continuous and $f(0)>0$ and $f(-1)<0$, then exists $\alpha\in (-1,0)$ such that $f(\alpha)=0$, i.e., $\exists \alpha \in(-1,0)$ such that $3^{\alpha}-\alpha^2=0$.
However, only a numerical method solves this problem.
A: The unique real solution is $$-2 \dfrac{W(\ln(3)/2)}{\ln(3)}$$
where $W$ is the Lambert W function.  There are also complex solutions, corresponding to the different branches of $-2 \dfrac{W(\pm\ln(3)/2)}{\ln(3)}$
A: As Robert Israel wrote, the two roots of the equation are $$x_{\pm}=-2 \dfrac{W(\pm\ln(3)/2)}{\ln(3)}$$ The solution $x_-$ can be discarded if $x$ is supposed to be real because, for this case, the argument of Lambert function is smaller that $-\frac{1}{e}$.
As showed by Michael T, there is a solution such that $-1<x<1$ (in fact subbing in $x=0$, we find that $3^0-0=1$ is positive and then the root is somewhere between $x=-1$ and $x=0$). By inspection, you coul find an evn shorter range for the solution.
As you were told, the solution cannot be expressed using basic functions and, if you do not access Lambert function, only numerical methods could be used. Newton iterative scheme is probably the simplest and, starting from a reasonable guess, this will be updated according to $$x+\frac{x^2-3^x}{3^x \log (3)-2 x}$$ So, being very lazy and starting the iterative process at $x=0$, the successive iterates will then be $-0.910239$, $-0.703169$, $-0.686137$, $-0.686027$ which is the solution. 
More generally, the solution of an equation such as $$a b^x=x^c$$ has a solution which write $$x=-\frac{c W\left(-\frac{a^{\frac{1}{c}} \log (b)}{c}\right)}{\log (b)}$$
A: There is no simple algebraic way to find a numerical value to solve this expression, you would need the Lambert W-function. However, you can affirm that a root exists.
Using the (true) assumption that $3^x - x^2$ is continuous, we can show that a root exists by showing that $3^x - x^2$ is positive for a value of $x$ and negative for another. This follows from the intermediate value theorem.
Subbing in $x = -1$, we find that $3^{-1} - (-1)^2 = \frac 13  - 1 = - \frac 23$ is negative.
Subbing in $x = 1$, we find that $3^1 - (1)^2 = 2$ is positive.
Thus, a root exists and it is somewhere between $x = -1$ and $x = 1$.
