# Ln(x) less than root(x)

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1. Dec 21, 2014

I was going through some important points give in my textbook and I saw this:
$\log_e x < \sqrt x$
How did they get this?
I know calculus so you can show this using differentiation, etc.
One possible way is that they took
$f(x)=\sqrt x-\log_e x$
And tried to prove it is always greater than zero.

2. Dec 21, 2014

### ShayanJ

You can write $\sqrt x$ as $e^{\ln \sqrt x}=e^{\frac 1 2 \ln x}$. Now we have $$\lim_{x\to 0} \ln x=-\infty$$ and $$\lim_{y\to -\infty} e^y=0$$. So for the least value of $\sqrt x$, $\ln x < \sqrt x$ and because $\sqrt x$ is strictly increasing and $\frac 1 x < \frac{1}{2\sqrt x}$ for $x>4$, the inequality is always satisfied.

EDIT: Looks like rewriting $\sqrt{x}$ wasn't necessary, but I don't change it because it may give you a feeling for the inequality.

Last edited: Dec 21, 2014
3. Dec 21, 2014

For x<4 why can't there be a point of intersection? You haven't specified what happens for x<4
I understood everything else. Thanks for the reply.

4. Dec 21, 2014

### ShayanJ

When $x \to 0$, we have $\sqrt x=0$ and $\ln x \to -\infty$. In the interval $(0,4]$, $\ln x$ increases faster than $\sqrt x$, but that slight excess can't make up for the huge gap in such a small interval.

5. Dec 22, 2014

### chiro

Check the derivatives - if one derivative is smaller than another and the initial values are different then you have shown that one is always smaller than the other provided this requirement is met.