How to prove lnx<x? How can we prove that the inequality ln x<x It is trivial in the cases 0<x<=1. I couldn't do anything for x>1.

jhenezhubby01ff

jhenezhubby01ff

Answered question

2022-09-06

How to prove ln x < x?
How can we prove that the inequality
ln   x < x
It is trivial in the cases 0 < x 1. I couldn't do anything for x > 1

Answer & Explanation

pereishen9g

pereishen9g

Beginner2022-09-07Added 7 answers

Your inequality is equivalent to x < e x for any x. (Substitute x = log t.) It is obvious for x 0. Moreover, if x > 0 then the Taylor expansion of e x gives
x < 1 + x 1 + x + x 2 2 + = e x .
Denisse Fitzpatrick

Denisse Fitzpatrick

Beginner2022-09-08Added 3 answers

This is one way to see it among others.
The function x ln ( 1 + x ) is a concave function (it's twice differentiable and its second derivative is strictly negative). Thus it's below all its tangents.
The tangent at the point ( 0 , 0 ) is the line y = x. Hence
x > 0 , ln ( 1 + x ) x
We deduce from this that
x > 0 , ln x < x

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?