Instance of Mean Value Theorem What's an elementary proof that there exists $c\in (a, b)$ with
$$e^b-e^a=(b-a)e^c$$
without calculus, just using the standard properties of the exponential function?
*By calculus I mean derivatives and integrals. The intermediate value theorem is fine, the mean value theorem is not.
**Either elementary definition of the exponential function is fine, but I suspect that continuity, monotonicity and $e^{a+b}=e^a e^b$, $(e^{a})^b=e^{ab}$ should be more than enough to do this.
 A: Just take $$c=\ln\left(\frac{e^b-e^a}{b-a}\right).$$
Edit: $a<c\iff e^a<\frac{e^b-e^a}{b-a}\iff b-a+1<e^{b-a}$. Now $x:=b-a$ is positive. Hence it is obvious that $x+1<e^x=1+x+x^2/2+\cdots$.
Edit$^2$: To show that $c<b$ yields to show $1<e^{b-a}(b-a+1)$. Now clearly $1<xe^x+e^x$ since $e^x>1$ for positive $x$.
A: To use the intermediate value theorem, let $f(x)=e^x-\frac{e^b-e^a}{b-a}$. It is sufficient to check that $f(a)<0$ and $f(b)>0$.
Simple rearrangement shows $f(a)=\frac{e^a}{b-a}((b-a+1)-e^{b-a})$. Since $b>a$, this is negative if $e^{b-a}>(b-a)+1$. This is clear from both the series of $e^x$, as b-a>0, and all terms are positive so truncating to 2 terms only reduces the value.
Similarly, we have $f(b)=\frac{e^a}{b-a}((b-a-1)e^{b-a}+1)$. This is positive if  $e^{b-a}>\frac{1}{1-(b-a)}$. Setting $y=b-a$, this is clear from the series for $e^y$ and the series for $\frac{1}{1-y}=1+y+y^2+...$ (sum of a geometric series). 
A: Something close to what you want.
Let $a, b \in \mathbb{R}$ and let $a < b$. If $f: x \mapsto e^{x} - \frac{e^{b}-e^{a}}{b-a}: \mathbb{R} \to \mathbb{R}$, then $f$ is continuous. But $f(x) > 0$ for $x \to \infty$ and $f(x) < 0$ for $x \to -\infty$, so by Bolzano's theorem (a specialized version of the intermediate value theorem for continuous functions) there is some $c \in \mathbb{R}$ such that
$$
e^{c} = \frac{e^{b} - e^{a}}{b-a}.
$$
But I suspect there is little chance to locate such a point within $]a,b[$ without calculus...
A: $e^{x}$ is a convex increasing function and $\frac{d}{dx}e^x=e^x$, hence if $a<b$
$$ e^{a}<\frac{e^b-e^a}{b-a}<e^b \tag{1}$$
holds, then $\frac{e^b-e^a}{b-a}=e^{\xi}$ for some $\xi\in(a,b)$ by continuity and monotonicity.
