How many solutions does eˣ = ax have?

Case $a=0$. $e^x=0$ has no solution.

Case $a<0$. Let $h(x)=e^x-ax$. Then $h'(x)=e^x-a>0$, so $h$ is strictly increasing, with $h(x)\to-\infty$ at $-\infty$ (since $-ax\to-\infty$ and $e^x\to0$) and $\to+\infty$ at $+\infty$. Exactly one solution.

Case $a>0$. Then $x=0$ is not a solution, and $x>0$ necessarily (as $e^x>0$ and $ax$ must be positive). Solutions are those of $g(x)=a$ with $g(x)=\dfrac{e^x}x$:
$$g'(x)=\frac{(x-1)e^x}{x^2}.$$
$g$ decreases from $+\infty$ (at $0^+$) to the minimum $g(1)=e$, then increases to $+\infty$. Hence:
- $0<a<e$: no solution,
- $a=e$: one solution, $x=1$,
- $a>e$: two solutions (one in $(0,1)$, one in $(1,\infty)$).

Geometric meaning. The line $y=ax$ through the origin meets the curve $y=e^x$ iff its slope is at least $e$ (or negative): for $a=e$ it is tangent to the exponential at the point $(1,e)$. Consequently $e^x\ge ex$ for all $x$.