Concise Solution for Algebra Problem Consider the following expression 
$$\sum_{i=1}^n \left(\frac{p\alpha_ie^{\alpha_i\cdot x}}{1-p+pe^{\alpha_i\cdot x}}\right) = c$$
where $\alpha_i, c \in \mathbb{R}$ and $p \in (0,1).$
How do I solve for $x$ from the equation?
 A: Here's what I tried and it was too long to be a comment.
By integrating both sides:
$$\sum_{i=1}^n\ln(1-p+p\;e^{\alpha_i x})=cx+c_1$$
and since this must hold for $x=0$, then $c_1=0$.
Therefore
$$\prod_{i=1}^n\left(1+p(e^{\alpha_i x}-1)\right)=e^{cx}$$
Set $y:=e^x$ to convert the above equation to a nonlinear (and seemingly unsolvable in general) equation:
$$\prod_{i=1}^n\left(1+p(y^{\alpha_i}-1)\right)=y^{c}$$
A: We have $$\sum_{i=1}^n \left(\frac{p\alpha_ie^{\alpha_i\cdot x}}{1-p+pe^{\alpha_i\cdot x}}\right) = \sum_{i=1}^{n}\frac{d}{dx} \ln(1-p+p e^{\alpha_i x}) = \frac{d}{dx} \ln\left(\prod_{i=1}^{n} (1-p+pe^{\alpha_i x})\right).$$
Distributing the product (substituting $q=1-p$ might simplify notation) and differentiating gives you a rational function in the variables $Y_i := e^{\alpha_{i}x}$, returning to the original equation this means that the question can be posed as finding the zeros of a multivariate polynomial in $Y_i$ (with $Y_i = e^{\alpha_i x})$. At this point it seems unlikely to me that there is a closed form for the solution $x$.
