Inequality with ln. Solve $ \frac x {\ln x} > e$

Question

How do I solve this? Wolfram just shows me the solution. I need to solve:

$$\begin{align}\ln(\frac{x}{\ln(x)})&>1\\ \implies\quad\frac{x}{\ln(x)} &> e\end{align}$$

Answer 1

By inspection, $\frac e{\ln(e)}=e$ so if you show $\frac x{\ln(x)}$ is increasing you want $x \gt e$

Answer 2

Your equation reduces to

$$x > e\ln(x)$$

By using the graphic method, since you know how to plot those two easy functions, you will obtain, by plotting $x$ and $e\ln(x)$:

Which shows you that you need

$$x > e$$

By derivative method:

$$x - e\ln(x) > 0$$

Deriving

$$1 - \frac{e}{x} > 0$$

$$\frac{e}{x} < 1$$

$$e < x$$

Answer 3

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

$$\displaystyle\frac{x}{\ln x} > e \quad \forall \ x \in \ ]1,+\infty[ \ \text{with} \ x\neq e$$