find function f(x) such that the following holds find function f(x) such that the following holds
\begin{align}
e^{\sum_{i=1}^{n} f(x_i)} = \sum_{i=1}^{n} x_i
\end{align}
where $0<x_i<1 \quad \forall I$ and n is an integer n>1.
or, if the above is not possible, how about a function f such that f is a function of x_i and some other element such as sum of all the x_i's or the product of all the xi's.
 A: Let $g(x)=e^{f(x)}$.
Then
$\prod_{k=1}^n g(x_k)
=\sum_{k=1}^n x_k$.
Putting all $x_k = x$,
$g^n(x) = nx$
so
$g(x) = (nx)^{1/n}$.
Then
$\sum_{k=1}^n x_k
=\prod_{k=1}^n g(x_k)
=\prod_{k=1}^n (nx_k)^{1/n}
=n\prod_{k=1}^n (x_k)^{1/n}
$.
so
$\frac1{n}\sum_{k=1}^n x_k
=\prod_{k=1}^n (x_k)^{1/n}
=(\prod_{k=1}^n x_k)^{1/n}
$.
By the arithmetic-geometric mean inequality,
this is true only when
all the $x_k$ are equal.
Therefore, for $n \ge 2$,
there is no such function.
A: As you hadn't requested analytical formula for $f$, then let me suggest following:
Suppose $x_1 = e$ and $x_2=e^2$. Then we have:
$e + e^2 = e(1+e) = e^{1+ln(1+e)}$. So $f(x_1) = 1$ and $f(x_2) = ln(1+e)$. And we can consider function
$$ f(x) =
\left\{
\begin{array}{ll}
    1  & \mbox{if }\ x=e \\
    ln(1+e) & \mbox{if }\ x=e^2
\end{array}
\right. $$
For $n>1$ same. Let's take $x_1 = e$, $x_2=e^2$, ..., $x_n=e^n$.Then we have
$$e+e^2+...+e^n = e(1+e+..+e^{n-1})=\\
e^{1+ln(1+e+..+e^{n-1})} = e^{1+1+...+1+ln(1+..+\frac{1}{e^{n-1}})}$$
So function is defined by: $f(e) = 1, f(e^2)=1, ..., f(e^n) = ln(1+..+\frac{1}{e^{n-1}})$.
If $0<x_i<1$, is important, then same idea works for negative powers.
$$e^{-1}+e^{-2}+e^{-3}= e^{-1}e^{ln(1+e^{-1}+e^{-2})}= e^{-1-1+ln(e+1+e^{-1})}$$
