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?
 A: 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 $A<e$;

A: This is a proof that uses the intermedate value theorem (I don't know an algebraic proof of this fact):
Let $g(x)=e^x-Ax$. $g(0)=1$ and $g(1)=e-A<0$. Using the intermediate value theorem, we deduce that there is a root $r_1$ of $g(x)$ in the interval $[0,1]$
Since $g$ goes increases without bound after sufficiently large x, thus there exists M>1 such that $g(M)>0$. By the intermediate value theorem, we know that there exists another root $r_2$ of $g$ in the interval $]1,M[$
A: Here is one with calculus.
If $f(x)=e^x-Ax$ then:


*

*$f(0)=1>0.$

*$f(1)=e-A<0.$

*$f(e)=e^e-Ae>e^e-e^2>0.$


Now use intermediate value theorem.
Since $f$ is differentiable in $\mathbb{R}$ with $f'(x)=e^x-A$ and $f'(x)<0 \iff x<\ln A , \ \ f'(x)>0 \iff x>\ln A \Rightarrow f(x)=0$  has at most two solutions.
