There are some cases where it is not so simple to decide which function grows faster asymptotically.
For example, in the following cases, why (intuitively) $g(n)$ should grow faster than $f(n)$, or vice-versa?
$$
\begin{array}{c|c}
f(n) & g(n)\\
\hline
10 \log(n) & \log(n^2)\\
\frac{n^2}{\log(n)} & n(\log(n))^2\\
n^{0.1} & \log(n)^{10}\\
\log(n)^{\log(n)} & \frac{n}{\log(n)}\\
\sqrt n & \log(n)^3\\
n\,2^n & 3^n\\
\log(n)^{\log(n)} & 2^{(\log_2 n)^2}
\end{array}
$$
It is easy to type some values on a calculator or to plot the functions to see which one grows faster, but during an exam I cannot do this of course.
Best Answer
Ultimately, it suffices to evaluate the limit $$ \lim_{n \to \infty} \frac{g(n)}{f(n)} $$ if it's $\infty$, then $g$ grows faster. If it's $0$, then $f$ grows faster. If it's anything else, then they grow the same (up to some constant), i.e. $f = \Theta(g)$.
As far as intuition goes, it helps to rewrite things, or sometimes to divide/multiply both sides by the same term.
Rewrite $\log(n^2) = 2 \log(n)$. It should now be obvious.
Divide both sides by $\frac{n}{\log(n)}$. We're now comparing $n$ to $\log(n)^3$. $n$ is asymptotically larger since $n^\alpha$ will always beat $\log(n)^\beta$ for any $\alpha,\beta > 0$ ("polynomial beats log"). Similarly, $e^{n^\alpha}$ will always beat $n^\beta$ for any $\alpha,\beta > 0$ ("exponential beats polynomial").
Same rule as last time. The left side is larger.
Rewrite $\log(n)^{\log(n)} = n^{\log(\log(n))}$ (how?). Perhaps it is now clear that the left side beats $n$, let alone $\frac{n}{\log(n)}$.
Polynomial always beats log (see 2.). Left is larger.
Divide both sides by $2^n$. Now, the right side is larger because exponential always beats polynomial (see 2.). Right is larger.
Rewrite the right side as $n^{\log_2(n)}$ (how?). This beats $n^{\log(\log(n))}$ (why)? So, the right side is larger.
Note on 4: we have $$ \log(n)^{\log(n)} = [e^{\log(\log(n))}]^{\log(n)} = e^{\log(n)\log(\log(n))} = [e^{\log(n)}]^{\log(\log(n))} = n^{\log(\log(n))} $$