$\log(n)$ is what power of $n$?

Question

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?

Answer 1

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,

$$\lim_{n \to \infty} \frac{\log{n}}{n^{\epsilon}} = \lim_{n \to \infty} \frac{\frac{1}{n}}{\epsilon n^{\epsilon - 1}} = \frac{1}{\epsilon} \lim_{n \to \infty} \frac{1}{n^{\epsilon}} = 0$$

Answer 2

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.

Answer 3

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:
$n=10$: $\varepsilon = 0.3622156886...$;
$n=10^2$: $\varepsilon = 0.3316228421...$;
$n=10^3$: $\varepsilon = 0.2797789811...$;

$n=10^6$: $\varepsilon = 0.1900611565...$;

$n=10^9$: $\varepsilon = 0.1462731331...$.