How to compare log and exponential 
$x^{1/3}$


$　 (\log_{2}{x})^3$

How to compare them?
 A: Cubing, consider
$$f(x)=x-\frac{\log ^9(x)}{\log ^9(2)}$$
$$f'(x)=1-\frac{9 \log ^8(x)}{x \log ^9(2)}$$
$$f''(x)=\frac{9 (\log (x)-8) \log ^7(x)}{x^2 \log ^9(2)}$$
The first derivative cancels at two  points
$$x_{1,2}=\exp\Big[-8 W(\pm k) \Big]\qquad \qquad k=\frac{\log ^{\frac{9}{8}}(2)}{8 \sqrt[4]{3}}$$ where $W(.)$ is Lambert function (numerically, $x_1=0.622419$ and $x_2=1.71274$).
$$f(x_1) \sim 0.655209 \qquad \qquad  f''(x_1)=+28.7147$$
$$f(x_2) \sim 1.610341 \qquad \qquad  f''(x_2)=-8.09651$$
So, $x_1$ corresponds to a minimum and $x_2$ to a maximum.
Now, by inspection, $f(2)=1$. So, using Newton method, the iterates will be
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & 2.00000 \\
 1 & 2.18208 \\
 2 & 2.13183 \\
 3 & 2.12495 \\
 4 & 2.12483
\end{array}
\right)$$
Then $f(x) >0$ for any $x < 2.12483$
A: Have you tried this? If not, take a look. Yes, there is no method to exactly compare them, unless you consider this special function $W_z(k)$, as a known function.
https://www.wolframalpha.com/input/?i=x%5E%281%2F3%29-log2%28x%29%5E3
If you refer to the behavior of log/power to infinity (i.e. for x sufficiently large), always the power $x^n, n>0$ goes faster to $\infty$ than $(log_mx)^p$, where $p,m>1$.
