TITLE: Inequality with ln. Solve $ \frac x {\ln x} > e$
QUESTION [2 upvotes]: How do I solve this? Wolfram just shows me the solution.
I need to solve:
$$\begin{align}\ln(\frac{x}{\ln(x)})&>1\\ \implies\quad\frac{x}{\ln(x)} &> e\end{align}$$

REPLY [0 votes]: Since you want that $\displaystyle\frac{x}{\ln x} > e$ you have that $\ln x > 0$ so $x>1$. You have to find when $x > e\ln x$ so $x - e \ln x > 0$. Let $f(x) = x - e \ln x$. Now consider $f'(x) = 1 - \displaystyle\frac ex$ You have that $f'(x) < 0$ if $x \in \ \left]0,e\right[$ and $f'(x) > 0$ if $x \in \ \left]e,\infty\right[$ so in $x=e$ you have a minimum. You can easily find that $f(e)=0$ so the result is that
$$\displaystyle\frac{x}{\ln x} > e \quad \forall \ x \in \ ]1,+\infty[ \ \text{with} \ x\neq e$$
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