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

fofopausiomiava

fofopausiomiava

Answered question

2022-10-08

How to prove that log ( x ) < x when x > 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 & Explanation

Helena Bentley

Helena Bentley

Beginner2022-10-09Added 10 answers

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

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