Is $\ln(x)$ ever greater than $x$? Is $\forall x \in \mathbb{R}, \ln(x) \lt x$ a true statement?
Just wondering for some convergence related thing
 A: 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 $(log_{10} x)^4 < x$.
A: 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
$$
\ln(x) < x.
$$
A: 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 $x > 0$.
