$\log(n)$ is what power of $n$? Sorry about asking such an elementary question, but I have been wondering about this exact definition for a while. What power of $n$ is $\log(n)$. I know that it is $n^\epsilon$ for a very small $\epsilon$, but what value is $\epsilon$ exactly?
 A: Suppose we seek such $\varepsilon$, that
$$
n^\varepsilon = \log(n).
$$
Consider $n>1$. 
Then (if $\log(n)$ is natural logarithm, to the base $e$)
$$
n^\varepsilon = e^{\varepsilon\log(n)}, \qquad \log(n) = e^{\log(\log(n))},
$$
then powers of $e$ must be equal:
$$
\varepsilon \log(n) = \log(\log(n)),
$$
$$
\varepsilon = \frac{\log(\log(n))}{\log(n)}.
$$
Examples:
$n=10$:   $\varepsilon = 0.3622156886...$;
$n=10^2$:  $\varepsilon = 0.3316228421...$;
$n=10^3$: $\varepsilon = 0.2797789811...$;
$n=10^6$: $\varepsilon = 0.1900611565...$;
$n=10^9$: $\varepsilon = 0.1462731331...$.
A: Perhaps  T. Bongers has answered the question you meant to ask, but given your mention of the definition of $\log$ I'm not so sure.  To answer your question literally, the function $n \mapsto \log(n)$ is not equal to the function $n \mapsto n^\epsilon$ for any number $\epsilon$ (not even if $\epsilon$ is "very small.)  It is a different kind of function altogether, with very different properties, and it is certainly not defined as a power function.
A: No: If $\epsilon$ is any positive number, then $n^{\epsilon}$ grows faster than $\log{n}$. This can be made precise in the statement
$$\lim_{n \to \infty} \frac{\log{n}}{n^{\epsilon}} = 0$$
for all $\epsilon > 0$. To prove this, just note that by L'Hospital's rule,
$$\lim_{n \to \infty} \frac{\log{n}}{n^{\epsilon}} = \lim_{n \to \infty} \frac{\frac{1}{n}}{\epsilon n^{\epsilon - 1}} = \frac{1}{\epsilon}  \lim_{n \to \infty} \frac{1}{n^{\epsilon}} = 0$$
