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
ln x x 1 / e
apparently works: Can someone give me a formal proof of this inequality?
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Answers (1)

Trace Arias
Answered 2022-09-20 Author has 6 answers
Consider the fuction
f ( x ) = log ( x ) x 1 / e
Its derivative
f ( x ) = e x 1 e e x
cancels for x = e e and, for this value f ( x ) = 0; the second derivative test shows that this is a maximum( f ( e e ) = e 1 2 e ). Then
log ( x ) x 1 / e
is always satisfied.

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