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):=logx−x,x≥1, giving

f′(x)=1x−1=1−xx<0forx>1 since

f(1)=−1<0 and f is strictly decreasing, then

f(x)<0,x>1, that is

logx−x<0,x>1.

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