Solving $2^x=x^3$ Algebraically 
How can I solve $2^x=x^3$ algebraically?

I could take $\log_2(\cdot)$ on both sides, but I'd still be stuck.

This equation follows the general equation 
$$a^x = x^{b}$$
and can be solved, for $a \neq 1$, as follows:
\begin{align}
a^{x} &= x^{b} \\
e^{x \, \ln a} &= x^{b} \\
x^{b} \, e^{- x \, \ln a} &= 1 \\
x \, e^{- x \, \ln a/b} &= 1 \\
- \frac{x \, \ln a}{b} \, e^{- x \, \ln a/b} &= - \frac{\ln a}{b} \\
u \, e^{u} &= - \frac{\ln a}{b} \, \hspace{5mm} u = - \frac{x \, \ln a}{b} \\
u = - \frac{x \ln a}{b} &= W_{0}\left(- \frac{\ln a}{b}\right) \\
x &= - \frac{b}{\ln a} \, W_{0}\left(- \frac{\ln a}{b}\right),
\end{align}
where $W_{0}(x)$ is the primary branch of the Lambert W-function which is defined as the solution to the equation $W(x) \, e^{W(x)} = x$. Note that in the case $a=1$ the solution is $x = e^{2 \pi i/b}$. 
For the problem proposed it is determined that:
$$2^{x} = x^{3} \, \text{ leads to } \, x = - \frac{3}{\ln 2} \, W\left(- \frac{\ln 2}{3}\right).$$ 
