11. For what value of the real number $a$ does the equation $a x=\ln x$ have no solution, one solution, or two solutions?
When $a \leqslant 0$, in the same coordinate system, draw the graphs of the functions $y=a x$ and $y=\ln x$. It is easy to see that the two graphs have exactly one intersection point, so the equation has exactly one solution.
When $a>0$, let $f(x)=a x-\ln x$. Then
$$
f^{\prime}(x)=a-\frac{1}{x}=\frac{a x-1}{x}(x>0) \text {. }
$$

Since when $0\frac{1}{a}$, $f^{\prime}(x)>0$, so $f(x)$ is a strictly decreasing function on the interval $\left(0, \frac{1}{a}\right]$ and a strictly increasing function on the interval $\left[\frac{1}{a},+\infty\right)$.
Thus, the minimum value of $f(x)$ is $f\left(\frac{1}{a}\right)=1+\ln a$. Therefore, when $1+\ln a>0$, i.e., $a>\frac{1}{\mathrm{e}}$, $f(x)>0$, and the equation $f(x)=0$ has no solution;

When $1+\ln a=0$, i.e., $a=\frac{1}{\mathrm{e}}$, the equation $f(x)=0$ has exactly one solution;

When $1+\ln a<0$, i.e., $0<a<\frac{1}{\mathrm{e}}$, the equation $f(x)=0$ has two solutions.

In summary:
When $a>\frac{1}{\mathrm{e}}$, the original equation has no solution;
When $a \leqslant 0$ or $a=\frac{1}{\mathrm{e}}$, the original equation has one solution;
When $0<a<\frac{1}{\mathrm{e}}$, the original equation has two solutions.