How to prove that $Ax = e^x$ has two solutions when $e < A < \infty$

Question

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?

Answer 1

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

1. Two points where $Ax=e^x$ when $A>e$;
2. One point where $Ax=e^x$ when $A=e$;
3. No points where $Ax=e^x$ when $A<e$;

Answer 2

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 $]1,M[$

Answer 3

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.