How would I rearrange an equation like this where there are 4 unique constants that are non-integers in front of the exponent and in front of e so as to solve for t?
$$x(t) = Ae^{Bt} + Ce^{Dt}$$
If A and C were equal this would be possible, but all four are unique and non-integers. Is it even possible?
I am trying to determine at what time $t$, $x(t)$ will equal some value. Naturally there is a range for which this equation will have a solution but I can't seem to get anywhere with my algebra.
Thanks!
As Paul commented, taking into account all the conditions described, only numeric methods will give the solution.
Where I see a potential problem is that you could have zero, one or two solutions.
Consider $$x(t) = Ae^{Bt} + Ce^{Dt}$$ $$x'(t)= ABe^{Bt} + CDe^{Dt}$$ $$x''(t)= AB^2e^{Bt} + CD^2e^{Dt}$$
The first derivative would potentially cancel for $$t_*=-\frac{\log \left(-\frac{A B}{C D}\right)}{B-D}$$ which does not exist in the real domain if $ABCD >0$. This is the "nice" case and Newton method will work like a charm if you consider for the zero of the better conditioned function $$f(t)=\log(Ae^{Bt} + Ce^{Dt})-\log(x)\tag 1$$ which will almost look like a straight line.
For example, consider the case $A=234$, $B=0.345$, $C=567$, $D=0.123$, $x=123456$. Plot the original function and its transform to see how different is the problem.
Being lazy, let us start iterating with $t_0=0$; Newton method will generate the following iterates $$\left( \begin{array}{cc} n & t_n \\ 0 & 0 \\ 1 & 26.81753360 \\ 2 & 18.11584802 \\ 3 & 18.04389549 \\ 4 & 18.04388025 \end{array} \right)$$
But, if $ABCD <0$ this would be a different story.