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.

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 $x = 1$.
