Does this kind of mean value have an explicit form? The original problem that gave rise to this question is to solve
$$
\sum_{i=1}^n\frac1{1+e^{-(x-c_i)}}=\frac n2
$$
(it comes from the need to determine difficulty of a polytomous item under the graded model in Item Response Theory).
For $n=2$ the (well, a) solution is $\frac{c_1+c_2}2$, so it should behave as certain generalized mean of the $c_i$. The question is whether there might be some explicit expression for it.
I just tried some obvious transformations, but without success. For example, the equation is equivalent to
$$
\sum_{i=1}^n\tanh\frac{c_i-x}2=0.
$$
Also, one may switch to polynomials: denoting $e^{-x}$ by $y$ and $e^{-c_i}$ by $a_i$ it becomes
$$
\sum_{i=1}^n\frac1{1+y/a_i}=\frac n2
$$
(solution being now the geometric mean of the $a_i$ for $n=2$). Modulo some transformations this can be reformulated as finding solutions of $t\frac{d\log p}{dt}=\frac n2$ in terms of roots of a given polynomial $p(t)$ of degree $n$.
...decided to add yet another reformulation: one has to find an extremum (a root of the derivative) for
$$
\prod_{i=1}^n\left(1-\frac z{b_i}\right)^{b_i^2-1}
$$
 A: Using your $y$ and $a_i$ coordinates, one is seeking a solution to
$$\sum_{i=1}^n\frac{1}{1+y/a_i}=\frac{n}{2}.$$ Multiplying all this out this becomes some equation of the form $P(y)/Q(y)=0$ with $P(y)$ and $Q(y)$ polynomials in $y$ with coefficients involving the $a_i$.
So one seeks solutions to $P(y)=0$. Now if we let $\sigma_j$ be the $j$th elementary symmetric function in the $a_i$ (so $\sigma_1=a_1+a_2+\cdots+a_n$ and $\sigma_n=a_1a_2\cdots a_n$ then it seems to me that (perhaps up to a constant -- these things are only defined up to constants) we have
$$P(y)=ny^n+(n-2)\sigma_1 y^{n-1}+(n-4)\sigma_2 y^{n-2}+\cdots-n\sigma_n.$$
I didn't check this for all $n$ but I did check it for $n=5$.
This is kind-of bad, because it shows that $y$ is a root of a "random" polynomial. For example if one sets $n=5$ and $a_1,a_2,a_3,a_4,a_5=1,2,3,4,5$ then we get a degree 5 polynomial $5y^5 + 45y^4 + 85y^3 - 225y^2 - 822y - 600$
with rational coefficients whose Galois group over the rationals is $S_5$.  
This suggests to me that there will be no easy formula in general (unless you allow your formula to be powerful enough to spit out roots of quintics).
