I need to solve for all values of a the number of solutions to the equation

$\displaystyle e^{x}=ax$

So far, and I'm very, very unsure about this, I've got:

$\displaystyle e^x=ax \Leftrightarrow \frac {e^x}{x}=a$

We set f(x) = e^x / x and derive, ending up with $\displaystyle \frac {xe^x-e^x}{x^2} = \frac {e^x(x-1)}{x^2}$

meaning that the curve changes at x = 1.

We put up a value? table

x | 0 1

f'(x) | ~ - 0 +

f(x) | ~ e

We look at the limits

$\displaystyle \lim_{x\to 0^+}f(x)=\infty$

$\displaystyle \lim_{x\to +\infty}f(x)=\infty$

So we have a curve with its minimum point at (1,e).

The solutions are:

a < e has 0 solutions

a = e has 1 solution

a > e has 2 solutions

Correct?

Edit:

I forgot to add that we need to check for asymptotes. Since f(x) / x approaches infinity as x->infinity there aren't any asymptotes.