Solving a "logistic-like" function with two unknowns, two data points I am stuck in trying to solve the following:
Given two points $(x_{1}; y_{1})$ and $(x_{2}; y_{2})$, to determine the parameters $a$ and $b$ in the equation:
$$y=\frac{e^{a+bx} - e^{a}}{1+e^{a+bx}}.$$
In other words, I have two unknowns and two set of points, so there's for sure a solution, but how can the following system of equation be rewritten explicitly on $a$ and $b$?
$$\left\{\begin{matrix}
y_{1}&=\frac{e^{a+bx_{1}} - e^{a}}{1+e^{a+bx_{1}}}\\ 
y_{2}&=\frac{e^{a+bx_{2}} - e^{a}}{1+e^{a+bx_{2}}}
\end{matrix}\right.$$
Should I implement an equation solver instead? In which case, do you have any suggestion on the method?  
Thanks very much, I hope it's not a too stupid question :)
 A: I don't know how to solve exactly (or even if it's possible) but maybe this will help: Let $A = e^a$ and $B = e^b$. It suffices to find $A$ and $B$ and then take logarithms. Your two equations can be written as
$$y_1 = {A(B^{x_1} - 1) \over 1 + AB^{x_1}}$$
$$y_2 = {A(B^{x_2} - 1) \over 1 + AB^{x_2}}$$
You can solve each of these equations for $A$ via some algebra:
$$A = {y_1 \over B^{x_1}(1 - y_1) - 1}$$
$$A = {y_2 \over B^{x_2}(1 - y_2) - 1}$$
So the reciprocals of the right-hand sides are the same (the original equation ensures that $y_1$ and $y_2$ are nonzero):
$${B^{x_1}(1 - y_1) - 1 \over y_1} = { B^{x_2}(1 - y_2) - 1 \over y_2}$$
In other words,
$${1 - y_1 \over y_1}B^{x_1} - {1 - y_2 \over y_2}B^{x_2} = {1 \over y_1} - {1 \over y_2}$$
This doesn't have a general solution you can write out (for example if $x_1$ and $x_2$ are large positive integers it's a polynomial equation of large degree). But it shouldn't be too hard to deal with numerically. Once you know what $B$ is (or have a good enough approximation) you can plug back in above to get $A$.
