Asymptotics of a solution Let $x(n)$ be the solution to the following equation
$$
x=-\frac{\log(x)}{n} \quad \quad \quad \quad (1)
$$
as a function of $n,$ where $n \in \mathbb N.$
How would you find the asymptotic behaviour of the solution, i.e. a function $f$ of $n$ such that there exist constants $A,B$ and $n_0\in\mathbb N$ so that it holds
$$Af(n) \leq x(n) \leq Bf(n)$$
for all $n > n_0$
?

Call $u_n:t\mapsto t\mathrm e^{nt}$, then $x(n)$ solves $u_n(x(n))=1$. For every $a$, introduce
$$
x_a(n)=\frac{\log n}n-a\frac{\log\log n}n.
$$
Simple computations show that, for every fixed $a$, $u_n(x_a(n))\cdot(\log n)^{a-1}\to1$ when $n\to\infty$. Thus, for every $a\gt1$, there exists some finite index $n(a)$ such that $x(n)\geqslant x_a(n)$ for every $n\geqslant n(a)$, and, for every $a\lt1$,  there exists some finite index $n'(a)$ such that $x(n)\leqslant x_a(n)$ for every $n\geqslant n'(a)$. Finally,  when $n\to\infty$,
$$
nx(n)=\log n-\log\log n+o(\log\log n).
$$
The assertion in your post holds with $f(n)=(\log n)/n$,  $n_0=\max\{n(A),n'(B)\}$, $B=1$ and every $A\lt1$.
