# Another version of mean value theorem

Let $f$ be a real valued differentiable $n+1$-times on $\Bbb R.$

Show that for each $a<b$ such that,

$$\ln\left(\frac{f(b)+f'(b)+\cdots+ f^{(n)}(b)}{f(a)+f'(a)+\cdots+ f^{(n)}(a)}\right)=b-a$$ there exists $c\in (a,b)$ such that $$f^{(n+1)}(c)= f(c)$$

I have the feeling one should apply the classical mean value theorem. But I don't know to what exact function I should apply it.

Can someone provide me with a hint?

Define $g(x)=\ln(f(x)+\cdots+f^{(n)}(x))$. Then the condition is $$g(b)-g(a)=b-a.$$
By mean-value theorem, there exists $c\in(a,b)$ such that $$g'(c)=\frac{g(b)-g(a)}{b-a}=1.$$
But $$g'(c)=\frac{f'(c)+\cdots+f^{(n+1)}(c)+f(c)-f(c)}{f(c)+\cdots+f^{(n)}(c)}=1+\frac{f^{(n+1)}(c)-f(c)}{f(c)+\cdots+f^{(n)}(c)}.$$ So $g'(c)=1\iff f(c)=f^{(n+1)}(c)$.