Inversion of weighted sum of exponentials

Question

I'm interested in the analytic solution to the problem:

$\sum_i^n a_i e^{b_i x}=c$

In the case that I'm studying all the terms $b_i<0$. I'm not sure if it is possible to invert a generic weighted average in the form:

$y=\sum_i^n a_i e^{b_i x}$

Can someone suggest me a computationally fast solution to this problem?

Answer 1

You can rewrite you problem by letting $k :=e^x$:

$$\begin{align} \sum_i a_ie^{b_i x} = \sum_i a_i \left(e^x \right)^{b_i} = \sum_i a_i k^{b_i} \end{align}$$.

Now you want to solve the following equation for $k$:

$$\sum_i a_i k^{b_i} = c$$

According to Abel–Ruffini theorem, there is no way to solve this equation putting $k$ in one side of the equation and all the rest on the other (you can't isolate the $k$) if your polynomial is of degree 5 or higher.

Answer 2

For numerical solution I vote to Eduardo's answer. I am an engineer not a mathematician. But nevertheless wanna give it a try.

For analytical solution, the simplest I can think of fitting b values and a values into curves as functions and then maybe

$\int_i^n(a(y)k^{b(y)},dy)= c(n-i)$ , where $k:=e^x$