Prove that there is $c \in (0,1)$ such that $e^{f'(c)}f(0)^{f(c)}=f(1)^{f(c)}$ Let $f:R \rightarrow (0, +\infty)$ be continuous and differentiable. 
Prove that there is $c \in (0,1)$ such that:
$$e^{f'(c)}f(0)^{f(c)}=f(1)^{f(c)}$$
Direct application of Lagrange's theorem doesn't tell me much. I was not successful with choosing some new function, too.
 A: The other answers give good hints, but I'll try to maybe explain how you would come up with this yourself. First, you could try to algebraically manipulate this to see if we can find something we can work with. Let's take the natural log of both sides so we'll hopefully get rid of the exponents. We get that $$\ln(e^{f'(c)}f(0)^{f(c)})=\ln(f(1)^{f(c)}).$$ We can use the usual log rules to simplify this to $$f'(c) + f(c)\ln(f(0)) = f(c)\ln(f(1)).$$ If we subtract the second term on the right and factor we end up with $$f'(c) = f(c)\cdot(\ln(f(1))-\ln(f(0))).$$ Lastly we divide by $f(c)$ to get $$\frac{f'(c)}{f(c)}=\ln(f(1))-\ln(f(0)).$$ Now, if you let $g(x)=\ln(f(x))$ and use the chain rule we see that $g'(x) = \frac{f'(x)}{f(x)}$. So the entire expression becomes $$g'(c)=\frac{g(1)-g(0)}{1-0}.$$ This suggests you should look at the function $\ln(f(x))$ and apply Lagrange's Theorem. You could then work backwards from what I did here to arrive at the answer.
A: Hint :
$$e^{f'(c)}f(0)^{f(c)} = f(1)^{f(c)} \Leftrightarrow f'(c) = f(c) 
\ln \frac{f(1)}{f(0)} \Leftrightarrow \frac{f'(c)}{f(c)} = \ln \frac{f(1)}{f(0)} $$
$$\implies$$
$$f(c) = k_1\exp \left[ \ln \left(\frac{f(1)}{f(0)}\right) c\right]$$
Note : Any logarithmic operation can be applied since $f: \mathbb R \to \mathbb R_+$.
Alternativelly : If you are into Banach Theory, the following may seem of use :
$$e^{f'(c)}f(0)^{f(c)} = f(1)^{f(c)} \Leftrightarrow f'(c) = f(c)\ln \frac{f(1)}{f(0)} $$
$$\implies$$
$$D\left(f\ln \frac{f(1)}{f(0)}\right)(c) = f(c)\ln \frac{f(1)}{f(0)}$$
The differential operator $D$ is linear and $C^1(0,1)$ is a cool Banach space. It also is 
$$f(c) \ln \frac{f(1)}{f(0)} \in C^1(0,1)$$
A: Hint: Define $g(x):=\ln (f(x))$ and use Mean value theorem for this function restricted to $[0,1]$.
