Determining parameters of $y=ab^x+c$ given 3 points We can find the parameters for the equation of a parabola through $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$ by solving the system
\begin{cases} ax_1^2 + bx_1 + c = y_1 \\ ax_2^2 + bx_2 + c = y_2 \\ ax_3^2 + bx_3 + c = y_3. \end{cases}
If we are given 3 points and told that a curve of the form $y=ab^x+c$ passes through them, we might set out to find $a$, $b$, and $c$ by solving the system
\begin{cases} ab^{x_1} + c = y_1 \\ ab^{x_2} + c = y_2 \\ ab^{x_3} + c = y_3. \end{cases}
Can this second system be solved, and if so, how?
 A: The equations being $$a \,b^{x_1}+c=y_1\tag 1$$ $$a \,b^{x_2}+c=y_2\tag 2$$ $$a \,b^{x_3}+c=y_3\tag 3$$ So, as André Nicolas commented, writing differences $$a(b^{x_2}-b^{x_1})=y_2-y_1\tag 4$$ $$a(b^{x_3}-b^{x_2})=y_3-y_2\tag 5$$ Making ratios as André Nicolas commented $$\frac{b^{x_2}-b^{x_1}}{b^{x_3}-b^{x_2}}=\frac{y_2-y_1}{y_3-y_2}\tag 6$$ So, you are left with one nonlinear equation in $b$. When you have $b$, $(4)$ or $(5)$ will give $a$ and then $(1)$, $(2)$ or $(3)$ will give $c$.
Except for very specific cases (such as $x_2=2x_1$, $x_3=3x_1$ for example  would reduce  equation $(6)$ to a polynomial in $b^{x_1}$; equally spaced values for the $x$'s will do  nice job - see at the end of this answer). But, in the most general case, solving equation $(6)$ (which is nonlinear) will require numerical methods (Newton would probably be the simplest).
For illustration purposes, let us consider three data points $(1.5,9.2)$, $(3.1,16.7)$, $(4.7,32.9)$. So, equation $(6)$ write $$\frac{b^{3.1}-b^{1.5}}{b^{4.7}-b^{3.1}}=\frac{75}{162}$$ The plot of the function shows a root close to $b=1.5$; Newton method would converge to $b=1.61821$; now, using this result, $a=3.14089$ and then $c=2.73448$. 
In the particular case where the $x$ values would be equally spaced $(x_2=x_1+\Delta)$, $(x_3=x_2+\Delta)$ equation $(6)$ would greatly simplify leading to $$b^{-\Delta}=\frac{y_2-y_1}{y_3-y_2}$$ and $b$ is immediately obtained (the remaining staying identical).
Edit
Eliminating $a$ and $c$ from equations $(1)$ and $(2)$ and replacing in equation $(3)$ leads to a nicer form for the equation to solve for $b$. It is
$$(y_2-y_3)b^{x_1}+(y_3-y_1)b^{x_2}+(y_1-y_2)b^{x_3}=0$$
