How to prove that log(x)1x>1?

It’s very basic but I’m having trouble to find a way to prove this inequality

log(x)<x

when x>1

(log(x) is the natural logarithm)

I can think about the two graphs but I can't find another way to prove it, and, besides that, I don't understand why should it not hold if x<1

Can anyone help me?

Thanks in advance.

Answer

You may just differentiate
f(x):=logxx,x1, giving
f(x)=1x1=1xx<0forx>1 since
f(1)=1<0 and f is strictly decreasing, then
f(x)<0,x>1, that is
logxx<0,x>1.

Attribution
Source : Link , Question Author : Gianolepo , Answer Author : Olivier Oloa

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