How to prove that $Ax = e^x$ has two solutions when $e < A < \infty$ How to prove that $Ax = e^x$ has two solutions when $e < A < \infty$?
This is easy to visualise graphically, but how can it be shown with algebra?

Consider the function $f(x)=e^x-Ax$. Then $f(0)=1$. We have $f'(x)=e^x-A$, $f''(x)=e^x$. As the second derivative is always positive, our function is convex. The derivative has a single zero at $x=\log A$, so $f$ has a minimum at that point. This means that 
$$
f(x)\geq f(\log A)=A-A\log A.
$$
As $\lim_{x\to\infty}f(x)=\infty$, if the minimum is negative, then $f$ will have two roots (and none if the minimum is positive). 
Assuming $A>0$, we have $A\log A>A$ precisely when $\log A>1$, i.e. $A>e$.
In conclusion, we have


*

*Two points where $Ax=e^x$ when $A>e$;

*One point where $Ax=e^x$ when $A=e$;

*No points where $Ax=e^x$ when $A<e$;

