Has $e^x = ax^2$ a general solution for all $x$?

Question

I was fiddling around with some math and stumbled upon $\exp(x) = a x^2$, finding myself unable to find a solution. Does it even have a general solution $a$ for all $x$? Some googling brought me to the Lambert W function, which I've never encountered before.

Answer 1

Lambert W solution like this: $$ e^x=ax^2 \\ e^{x/2} = \sqrt{a} \;x \\ \frac{1}{\sqrt{a}}=x\;e^{-x/2} \\ -\frac{1}{2\sqrt{a}}=-\frac{x}{2}\;e^{-x/2} \\ W\left(-\frac{1}{2\sqrt{a}}\right) = -\frac{x}{2} \\ -2W\left(-\frac{1}{2\sqrt{a}}\right)=x $$ ... and, as Robert notes, also solutions for the other square-root.

Answer 2

Which do you want to solve for, $a$ or $x$? Solving for $a$ is easy: $a = x^{-2} \exp(x)$ (assuming of course $x \ne 0$). Solving for $x$ is where you need Lambert: $$ x = - 2 W(\pm 1/(2 \sqrt{a}))$$ where $W$ is a branch of the Lambert W function. If $a$ is real and you're looking for real solutions, there are none if $a \le 0$, one ($-2 W(+1/\sqrt{a})$) if $0 < a < e^2/4$, two if $a = e^2/4$, three if $a > e^2/4$.