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
