stuck at this Rolle's theorem exercise 
Let $f:[a,b] \rightarrow \mathbb{R_{+}^{*}}$ and $f$ derivative in $[a,b]$. Show that: $\exists c\in (a,b)$:
$$\frac{f\left ( b \right )}{f\left ( a \right )}=\frac{{f}'\left ( c \right )}{f\left ( c \right )}e^{b-a}$$

I tried my best and arrived here : 
$${\ln [f(c)]}'-\frac{f(b)}{e^{b}}\frac{1}{\frac{f(a)}{e^{a}}}=0$$
I would love to hear some ideas or hints. Thanks for your attention!

I had an answer using the same hint as suggested by Paramanand Singh in the comments but that doesn't give the required result. He points out that this is probably a typo, so I'll post this suggestion anyway.
You can derive the Mean Value Theorem (MVT) from Rolle's theorem. Apply it to the function $g$ where $g(x) = \ln f(x)$ on $[a,b]$ to get, for a certain $c \in \; ]a,b[$:
$$g(b)-g(a) = (b-a)g'(c) \implies \ln f(b) - \ln f(a) = (b-a)\frac{f'(c)}{f(c)}$$
Now the left-hand side is $\ln\tfrac{f(b)}{f(a)}$, so applying the exponential function to both sides gives:
$$\frac{f(b)}{f(a)} = \exp\left( (b-a)\frac{f'(c)}{f(c)} \right)$$
Which is not what you wrote, but what might be the intented answer.
