 # Upper bound of natural logarithm I was playing looking for a good upper bound of natural logarithm and I found that ln x<=x^1/e apparently works: Can someone give me a formal proof of this inequality? kennadiceKesezt 2022-09-19 Answered
Upper bound of natural logarithm
I was playing looking for a good upper bound of natural logarithm and I found that
$\mathrm{ln}x\le {x}^{1/e}$
apparently works: Can someone give me a formal proof of this inequality?
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Consider the fuction
$f\left(x\right)=\mathrm{log}\left(x\right)-{x}^{1/e}$
Its derivative
${f}^{\prime }\left(x\right)=\frac{e-{x}^{\frac{1}{e}}}{ex}$
cancels for $x={e}^{e}$ and, for this value $f\left(x\right)=0$; the second derivative test shows that this is a maximum(${f}^{″}\left({e}^{e}\right)=-{e}^{-1-2e}$). Then
$\mathrm{log}\left(x\right)\le {x}^{1/e}$
is always satisfied.

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