How many solutions does ln x = ax have?

Study of $h$. For $x>0$, $h'(x)=\dfrac{1-\ln x}{x^2}$, so $h$ increases on $(0,e]$ and decreases on $[e,+\infty)$, with maximum $h(e)=\frac1e$. Limits: $h(x)\to-\infty$ as $x\to0^+$ and $h(x)\to0$ as $x\to+\infty$. Also $h(x)<0$ on $(0,1)$, $h(1)=0$, and $h(x)>0$ for $x>1$.

Ranges.
- On $(0,e]$, $h$ is continuous and strictly increasing from $-\infty$ to $\frac1e$, so it takes each value in $(-\infty,\frac1e]$ exactly once.
- On $[e,+\infty)$, $h$ is continuous and strictly decreasing from $\frac1e$ to $0$ (not reached), so it takes each value in $(0,\frac1e]$ exactly once.

Counting.
- $a>\frac1e$: no solution.
- $a=\frac1e$: one solution, $x=e$.
- $0<a<\frac1e$: two solutions, one in $(1,e)$ and one in $(e,+\infty)$.
- $a\le0$: one solution, found on $(0,e]$ (for $a=0$ it is $x=1$).

Meaning. The line $y=ax$ is tangent to the curve $y=\ln x$ exactly when $a=\frac1e$, at the point $x=e$.