Show that $$x=ne^{-x}$$ with $n \in \mathbb{N}$ and $n\ge1$ has one solution $x_n$ such that: $$x_n = \log(n)-\log(\log(n)) + o(\log(n))= \log(n)-\log(\log(n)) + o(1)$$
My try:
Let be $$f(x)=x-ne^{-x}$$
$f(0)=-n<0$ and $f'(x)=1+ne^{-x} >0$ for all $x$, so $f(x)$ is increasing and there will be $x_n \in (0, +∞)$ for which $f(x_n)=0$.
Which is the next step? Show that $x_n+\log(x_n)=\log(n)$…?