How do I find the constants in a sum of exponentials? I have a function $$I(t)=A\cdot e^{-at}+B\cdot e^{-bt}$$ and am curious to discover if there is any way to solve for $a,b$ uniquely, given that $A,B,a,b \in \mathbb{R}^+$.
 A: Admitting that you have two data points $(t_1,t_2)$ (I shall assume that $t_2>t_1$) and that you know $(A,B)$, you want to solve for $(a,b)$ equations
$$I(t_1)=I_1=A \, e^{-a t_1}+B \, e^{-b t_1} \tag 1$$
$$I(t_2)=I_2=A \, e^{-a t_2}+B \, e^{-b t_2} \tag 2$$
Using $(1)$, you can get 
$$a=-\frac{1}{{t_1}}\log \left(\frac{I_1-B\,e^{-b {t_1}}}{A}\right)\tag3$$ Plug it in $(2)$ to get
$$I_2=A\left(\frac{I_1-B\, e^{-b {t_1}}}{A}\right)^{\frac{t_2}{t_1}}+B \, e^{-b t_2} \tag 4$$ where only $b$ appears.
In the most general case, you will need a numerical method to solve it.
For sure, if $t_2=2t_1$, it will reduce to a quadratic equation in $e^{-b {t_1}}$; if $t_2=3t_1$, to a cubic; ; if $t_2=4t_1$, to a quartic (which already would be a nightmare !).
For illustration purposes, let us try using
$$t_1=10 \qquad I_1=14 \qquad t_2=13 \qquad I_2=6 \qquad A=123 \qquad B=456$$ So, you need to find the zero of function
$$f(b)=2 \left(\frac{2}{123}\right)^{3/10} \left(7-228 \,e^{-10 b}\right)^{13/10}+456\, e^{-13
   b}-6$$ for $7-228 e^{-10 b} >0$ that is to say $b>\frac{1}{10} \log \left(\frac{228}{7}\right)\approx 0.34 $.
Computing the derivative, you will notice that it cancels when $b=\frac{1}{10} \log \left(\frac{579}{14}\right)\approx 0.37$ and the second derivative test shows that this is a minimum. Do the solution is larger than $0.37$. By inspection or graphing, you would see that the solution is close to $0.5$. 
So, using Newton method starting at $b_0=0.5$, the following iterates will be generated
$$\left(
\begin{array}{cc}
 n & b_n \\
 0 & 0.5000000000 \\
 1 & 0.5027571929 \\
 2 & 0.5027798838 \\
 3 & 0.5027798854
\end{array}
\right)$$ which is the solution for ten significant figures.
Using $(3)$, we obtain $a=0.2413223106$.
