jhenezhubby01ff

2022-09-06

How to prove $\mathrm{ln}x?
How can we prove that the inequality

It is trivial in the cases $0. I couldn't do anything for $x>1$

### Answer & Explanation

pereishen9g

Your inequality is equivalent to $x<{e}^{x}$ for any $x$. (Substitute $x=\mathrm{log}t$.) It is obvious for $x\le 0$. Moreover, if $x>0$ then the Taylor expansion of ${e}^{x}$ gives
$x<1+x\le 1+x+\frac{{x}^{2}}{2}+\cdots ={e}^{x}.$

Denisse Fitzpatrick

This is one way to see it among others.
The function $x↦\mathrm{ln}\left(1+x\right)$ 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 $\left(0,0\right)$ is the line $y=x$. Hence
$\mathrm{\forall }x>0,\phantom{\rule{thinmathspace}{0ex}}\mathrm{ln}\left(1+x\right)\le x$
We deduce from this that
$\mathrm{\forall }x>0,\phantom{\rule{thinmathspace}{0ex}}\mathrm{ln}x

Do you have a similar question?

Recalculate according to your conditions!