Given the non-linear equation $y = A + B \mathrm{e}^{C t}$ and 3 sets of points: ($y_1$, $t_1$), ($y_2$, $t_2$), ($y_3$, $t_3$), can the variables $A$, $B$, and $C$ be calculated analytically?
\begin{align*} y_1 = A + B \mathrm{e}^{C t_1} \\ y_2 = A + B \mathrm{e}^{C t_2} \\ y_3 = A + B \mathrm{e}^{C t_3} \end{align*}
\begin{align*} y_1 - y_2 = B \mathrm{e}^{C t_1} - B \mathrm{e}^{C t_2} \\ y_2 - y_3 = B \mathrm{e}^{C t_2} - B \mathrm{e}^{C t_3} \end{align*}
\begin{align*} \frac{y_1 - y_2}{y_2 - y_3} = \frac{\mathrm{e}^{C t_1} - \mathrm{e}^{C t_2}}{\mathrm{e}^{C t_2} - \mathrm{e}^{C t_3}} \end{align*}
This is where I am stuck. Without using approximations, I can't reduce this function to determine the variable $C$. Is this possible?
In the most general case, as written, you cannot extract $C$ from the third equation since it is highly transcendental. The only solution would be a numerical method such as Newton which would work without any problem provided a reasonable guess; a plot of the function (the third equation) as a function of $C$ will allow you to locate the root and start iterating from this value.
However, there are cases where this is feasible. The simplest I found corresponds to $t_2=2t_1$ and $t_3=3t_1$. For such a case, simplifying the rhs, you should arrive to $$C=\frac{\log \left(\frac{{y_2}-{y_3}}{{y_1}-{y_2}}\right)}{{t_1}}$$ Similarly, if $t_2=2t_1$ and $t_3=4t_1$, you should arrive to $$C=\frac{\log \left(\frac{-\sqrt{{y_1}-{y_2}} \sqrt{{y_1}+3 {y_2}-4 {y_3}}-{y_1}+{y_2}}{2 ({y_1}-{y_2})}\right)}{{t_1}}$$