How to find exponential decay equation from $3$ points when asymptote is unknown If I know I'm dealing with an equation of the form $y=ab^x+c$ , and I know 3 points along the curve $(x_0,y_0), (x_1,y_1), (x_2,y_2)$, it seems like I should be able to calculate $a$, $b$, and $c$. After some googling and trial and error, I've figured out how to do it when the points' $x$ values are equidistant, but I can't figure out how to do it when $x_1$ is an arbitrary value between $x_0$ and $x_2$.
In the case where $x_1$ is the midpoint between $x_0$ and $x_2$,
$$b = \left(\frac{y_2-y_1}{y_1-y_0}\right)^{\frac{1}{x_1-x_0}}.$$
(That's the equation that I found to work after a combination of googling and guessing intuitively, but I can't exactly explain why it works.)
After finding $b$, it's easy to solve for $a$ and $c$ by solving the system of equations for two points:
$$\begin{align} y_0 &=ab^{x_0}+c\\
y_1&=ab^{x_1}+c\end{align}.$$
That comes out to:
$$a = \frac{y_1 - y_0}{b^{x_1}-b^{x_0}}$$
$$c = y_0 - a b^{x_0}$$
But I can't figure out how to find $b$ when $x_1$ is some arbitrary value. The equation for $b$ above doesn't work in that case. The furthest I've gotten is to set up a system of equations:
$$y_0=ab^{x_0}+c$$
$$y_1=ab^{x_1}+c$$
$$y_2=ab^{x_2}+c$$
But no matter how I try to manipulate these equations to try to solve for $b$, I always end up with something like:$y_2-y_1=a(b^{x_2}-b^{x_1}),$ and I can't figure out a way to solve algebraically for $b$.
Does anyone know how to find $b$ in this case?

Hint.
Taking the system
$$
\cases{
y_1=ab^{x_1}+c\\
y_2=ab^{x_2}+c\\
y_3=ab^{x_3}+c
}
$$
after eliminating $a,c$ remains the equation
$$
(y_1-y_2)b^{x_3}+(y_2-y_3)b^{x_1}+(y_3-y_1)b^{x_2}=0
$$
This equation is always verified for $b=1$ and $b = 0$ which are trivial solutions for this equation. For values of $x_k$ which are multiples, the problem to solve is a polynomial root's determination. For the general case, an iterative process like Newton-Raphson can used to determine the possible non trivial solutions, that should be real.
Calling $y_1-y_2=\alpha, y_2-y_3=\beta, y_3-y_1 = \gamma, x_1-x_3= u, x_2-x_3=v$ as $\alpha+\beta+\gamma = 0$, we have equivalently to solve for $b$
$$
\alpha+\beta b^u-(\alpha+\beta)b^v = 0
$$
which could also have no other real solutions than the trivial ones.
