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

$$\displaystyle{\log{{\left({n}\right)}}}$$ is what power of $$\displaystyle{n}$$?

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temnimam2
No: If $$\displaystyle\in$$ is any positive number, then $$\displaystyle{n}^{\epsilon}$$ grows faster than $$\displaystyle{\log{{n}}}$$. This can be made precise in the statement
$$\displaystyle\lim_{{{n}\rightarrow\infty}}{\frac{{{\log{{n}}}}}{{{n}^{\epsilon}}}}={0}$$
for all $$\displaystyle\epsilon{>}{0}$$. To prove this, just note that by L'Hospital's rule,
$$\displaystyle\lim_{{{n}\rightarrow\infty}}{\frac{{{\log{{n}}}}}{{{n}^{\epsilon}}}}=\lim_{{{n}\rightarrow\infty}}{\frac{{{\frac{{{1}}}{{{n}}}}}}{{\epsilon{n}^{{\epsilon-{1}}}}}}={\frac{{{1}}}{{\epsilon}}}\lim_{{{n}\rightarrow\infty}}{\frac{{{1}}}{{{n}^{\epsilon}}}}={0}$$
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einfachmoipf
Perhaps T. Bongers has answered the question you meant to ask, but given your mention of the definition of $$\displaystyle{\log{}}$$ I'm not so sure. To answer your question literally, the function $$\displaystyle{n}\rightarrow{\log{{\left({n}\right)}}}$$ is not equal to the function $$\displaystyle{n}\rightarrow{n}^{\epsilon}$$ for any number $$\displaystyle\epsilon$$ (not even if $$\displaystyle\epsilon$$ 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.
Vasquez

Suppose we seek such $$\epsilon$$, that
$$n^\epsilon=\log(n).$$
Consider n>1. Then (if $$\log(n)$$ is natural logarithm, to the base e)
$$\begin{array}{}n^\epsilon=e^{\epsilon\log(n)} \\\log(n)=e^{\log(\log(n))} \text{then powers of e must be equal:} \\\epsilon\log(n)=\log(\log(n)), \\\epsilon=\frac{\log(\log(n))}{\log(n)}. \\\text{Examples:} \\n=10: \epsilon=0.3622156886...; \\n=10^2: \epsilon=0.3316228421...; \\n=10^3: \epsilon=0.2797789811...; \\n=10^6: \epsilon=0.1900611565...; \\n=10^9: \epsilon=0.1462731331.... \end{array}$$