Given three unique points $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ on the line $y = a b^{cx}$, how would I solve for the parameters $a$, $b$, and $c$? I'm hoping that there is only one solution.
I've already found $a$ with substitution:
\begin{equation} \begin{split} \frac{ab^{cx_1}}{ab^{cx_2}} &= \frac{y_1}{y_2} \\\\ \frac{b^{cx_1}}{b^{cx_2}} &= \frac{y_1}{y_2} \\\\ b^{(cx_1-cx_2)} &= y_1 / y_2 \\\\ \therefore b &= (y_1/y_2)^{(cx_1-cx_2)^{-1}} \\\\ \end{split} \end{equation}
Substituting that into the original equation:
\begin{equation} \begin{split} a \cdot ( (y_1/y_2)^{(cx_1-cx_2)^{-1}} ) ^{cx_1} &= y_1 \\\\ a \cdot ( y_1/y_2 )^{x_1/(x_1-x_2)} &= y_1 \end{split} \end{equation}
$$ \therefore a = \frac{y_1}{( y_1/y_2 )^{x_1/(x_1-x_2)}}\quad \text{(not very Q.E.D.)} $$
This lets me state that $b^{cx_1} = \frac{y_1}{a}$ etc., but to solve $b$ and $c$ eludes me, and I'm not entirely sure there's one single solution.
The values of $b$ and $c$ here are not independent. Write $b^{cx}=e^{rx}$ by defining
$$r=c\log b$$
where I will write all logs to base $e$. Then you have $y=ae^{rx}$, so there are two independent parameters, not three. This means that in general three points may not all lie on one curve of this form. The equations read
\begin{align*} y_1&=ae^{rx_1}\\ y_2&=ae^{rx_2}\\ y_3&=ae^{rx_3}\\ &\iff\\ \log y_1&=\log a+rx_1\\ \log y_2&=\log a+rx_2\\ \log y_3&=\log a+rx_3\\ \end{align*} This system has a solution if and only if $x_i\ne x_j$ for some $(i,j)$ and $$(x_1-x_2)(\log y_2-\log y_3)=(x_2-x_3)(\log y_1-\log y_2)$$ In this case, the solution is given by
\begin{align*} r&=\frac{\log y_i-\log y_j}{x_i-x_j}\\ a&=y_1e^{-rx_1} \end{align*}
If you want to write these in terms of $b,c$, then as noted in a comment, $b,c$ are not uniquely determined. Given $r$, if you choose any $b>0$ then you can define $c=r/\log b$.