Prove $\frac{f(b)}{f(a)}=e^{(b-a)\frac{f'(c)}{f(c)}}$ 
Let $f:[a,b]\rightarrow \mathbb{R}$ be a positive and differentiable function. Prove: $$\frac{f(b)}{f(a)}=e^{(b-a)\frac{f'(c)}{f(c)}}$$ 

I tried to apply the MVT and make some algebraic manipulations, but got stuck.
Any help appreciated.
 A: Let $g(x)=\ln(f(x))$. According to the mean value theorem,
$$ \frac{g(b)-g(a)}{b-a}=g^{\prime}(c)=\frac{f^{\prime}(c)}{f(c)} $$
for some $c\in(a,b)$. This implies that
$$ \ln\Big(\frac{f(b)}{f(a)}\Big)=(b-a)\frac{f^{\prime}(c)}{f(c)} $$
hence
$$ \frac{f(b)}{f(a)}=\exp\Big((b-a)\frac{f^{\prime}(c)}{f(c)}\Big) $$
A: High-level idea: The reflex to use the MVT is a good one. Now, the issue is that you do not want to apply it directly to $f$, but to some other function $g$ related to $f$. In particular, the term $e^{b-a}$ strongly hints that you're going to take an exponential after that; and seeing $\frac{f'(c)}{f(c)}$ should prompt you to think of $(\ln f)' = \frac{f'}{f}$. These two together suggest we set $g=\ln f$, and apply the MVT to $g$: good thing, this is legitimate, as $f$ is assumed positive. So let's try that...

Let $g\colon [a,b]\to \mathbb{R}$ be defined by $g\stackrel{\rm def}{=} \ln f$. $g$ is well-defined, and differentiable, as $f$ is positive and differentiable.
Applying the Mean Value Theorem on $g$, we get that there exists $c\in(a,b)$ such that
$$
g(b)-g(a) = g'(c) (b-a)
$$
and, exponentiating both sides, we obtain
$$
e^{g(b)-g(a)} = e^{g'(c)(b-a)}.
$$
Recalling that $e^{g(b)-g(a)} = \frac{e^{g(b)}}{e^{g(a)}} = \frac{f(b)}{f(a)}$ and $g'(c) = \frac{f'(c)}{f(c)}$ concludes the proof.
