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 $$ \prod_{i=1}^n\left(1-\frac z{b_i}\right)^{b_i^2-1} $$