Is $\ln(x)$ ever greater than $x$?

Question

Is $\forall x \in \mathbb{R}, \ln(x) \lt x$ a true statement?

Just wondering for some convergence related thing

Answer 1

Given

$$ f(x) = x - \ln(x), $$

then

$$ f'(x) = 1 - \frac{1}{x}. $$

Extreme value for $f'(x) = 0$, so

$$ x = 1 $$

Note that

$$ f''(x) = \frac{1}{x^2}, $$

as $f''(1) = 1$ the extreme value is a minimum, whence $f(x) \ge f(1)$, thus

$$ f(x) \ge 1. $$

Therefore $x - \ln(x) \ge 1 > 0$, whence

$$ \ln(x) < x. $$

Answer 2

First note that by definition of $e^x$ we have $$x< \underbrace{\frac{x^0}{0!}}_{=1}+\underbrace{\frac{x^1}{1!}}_{=x}+\sum_{k=2}^\infty \frac{x^k}{k!} = e^x$$ for every $x >0$. Now since $(\ln(x))'= \frac{1}{x}> 0$ for every $x> 0$ it is clear that $x \mapsto \ln(x)$ is strictly increasing, combining these facts shows $$\ln(x) < \ln(e^x) = x,$$ for every $x > 0$.

Answer 3

We can get a stronger result using the method shown by @johannesvalks. Let

$$ f(x) = x - e \ln x. $$

Then

$$ f'(x) = 1 - \frac{e}{x}. $$

The function $f(x)$ has a local maximum or minimum only when $f'(x) = 0$, namely at

$$ x = e. $$

We can show this is a minimum either by taking the second derivative or by examining $f(x)$ at some other positive value of $x$. Therefore, for all $x > 0$,

$$ f(x) = x - e \ln x \ge f(e) = 0. $$

That is,

$$ x \ge e \ln x. $$

You can use this fact to prove other things such as your statement in a comment that $(log_{10} x)^4 < x$.