So, I have the same initial value $x_{2015}$, let's say at $t_{2015}$ and multiple ending values $x_{2030}$ at say $t_{2030}$. For example:
$x_{2015} = 2500$ $x_{2030} = 1300$
$x_{2015} = 2500$ $x_{2030} =890$
$x_{2015} = 2500$ $x_{2030} = 1240$
$x_{2015} = 2500$ $x_{2030} = 1100$
I want to use the same decay rate $r$, $x_{2015}$ and $x_{2030}$ as input into the exponential decay function $y = a(1-r)^{t-t_0}$ at every 5 years time step to calculate the values from 2015 to 2100. So going from 2015 to 2020, 2025, at $t=2030$, I want the function to reproduce the $x_{2015}$ and $x_{2030}$ values used as input into the function.
So far, I have manipulated the exponential decay function like this: $$x_t = x_{2015} + \frac{(x_{2030} - x_{2015})}{(1-r)}(1-r^\frac{t - t_{2015}}{t_{2030} - t_{2015}})$$
So at $t = t_{2015}$ I get $x_{2015}$ and at $t = t_{2030}$, the $(1-r)$ cancels out and I get $x_{2030}$. So in the end whichever values I use for $x_{2030}$, $x_{2015}$ and the decay rate $r$, it will reproduce the 2015 and 2030 values and along with the other values at each $t$.
The problem is, I am not sure if this makes enough sense, is there another way I can do this or is this the only way?. I have looked at this function $y_t = Be^{-mt}$, but I can't find a way to even manipulate this to give me what the other equation is giving me above.
Given values at two times you can compute the decay rate from the ratio of those two values and the time between them. As you have different final values, each one will give a different decay rate. You have two choices: accept the differences between the measurements and compute a different decay rate for each of the pieces of data or believe that the four end values correspond to noise on a true value with a true decay rate. You could use the mean or the geometric mean of the end values to compute a decay rate.
The correct equation for exponential decay is $Be^{-mt}$. If we measure $t$ in units of five years you can evaluate $B,m$ for any of your data. If we set the zero of time as $2015$ then $B=2500$. For the first pair of data then $$e^{-3m}=\frac {1300}{2500}\\m=\frac {\ln \frac {1300}{2500}}{-3}\approx 0.21798$$ and you can plug in $t=1,2$ to get the values for $2020,2025$