\log(n) is what power of n?

David Troyer 2022-01-01 Answered
\(\displaystyle{\log{{\left({n}\right)}}}\) is what power of \(\displaystyle{n}\)?

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temnimam2
Answered 2022-01-02 Author has 3456 answers
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
Answered 2022-01-03 Author has 2660 answers
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.
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Vasquez
Answered 2022-01-09 Author has 8850 answers

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

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