 David Troyer

2022-01-01

$\mathrm{log}\left(n\right)$ is what power of $n$? temnimam2

No: If $\in$ is any positive number, then ${n}^{ϵ}$ grows faster than $\mathrm{log}n$. This can be clarified in the assertion.
$\underset{n\to \mathrm{\infty }}{lim}\frac{\mathrm{log}n}{{n}^{ϵ}}=0$
for all $ϵ>0$. To prove this, just note that by LHospitals rule,
$\underset{n\to \mathrm{\infty }}{lim}\frac{\mathrm{log}n}{{n}^{ϵ}}=\underset{n\to \mathrm{\infty }}{lim}\frac{\frac{1}{n}}{ϵ{n}^{ϵ-1}}=\frac{1}{ϵ}\underset{n\to \mathrm{\infty }}{lim}\frac{1}{{n}^{ϵ}}=0$ einfachmoipf

Perhaps T. Bongers has answered the question you meant to ask, but given your mention of the definition of $\mathrm{log}$ I'm not so sure. To answer your question literally, the function $n\to \mathrm{log}\left(n\right)$ is not equal to the function $n\to {n}^{ϵ}$ for any number $ϵ$ (not even if $ϵ$ 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 $ϵ$, that
${n}^{ϵ}=\mathrm{log}\left(n\right).$
Consider n>1. Then (if $\mathrm{log}\left(n\right)$ is natural logarithm, to the base e)
$\begin{array}{}{n}^{ϵ}={e}^{ϵ\mathrm{log}\left(n\right)}\\ \mathrm{log}\left(n\right)={e}^{\mathrm{log}\left(\mathrm{log}\left(n\right)\right)}\text{then powers of e must be equal:}\\ ϵ\mathrm{log}\left(n\right)=\mathrm{log}\left(\mathrm{log}\left(n\right)\right),\\ ϵ=\frac{\mathrm{log}\left(\mathrm{log}\left(n\right)\right)}{\mathrm{log}\left(n\right)}.\\ \text{Examples:}\\ n=10:ϵ=0.3622156886...;\\ n={10}^{2}:ϵ=0.3316228421...;\\ n={10}^{3}:ϵ=0.2797789811...;\\ n={10}^{6}:ϵ=0.1900611565...;\\ n={10}^{9}:ϵ=0.1462731331....\end{array}$

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