Solve $\sum_{i=1}^n a_i\exp(-b_ix) = 1$

Question

Problem. Solve $\displaystyle \sum_{i=1}^n a_i\exp(-b_ix) = 1$ where $n$ is a positive integer and $a_i, b_i (i=1,2,\dots,n)$ are positive constants.

I am not sure this can be solved analytically. The LHS is strictly decreasing and continuous from $+\infty$ to $-\infty$ so the solution exists and is unique.

Answer 1

Assuming that the $b_i$'s are rational numbers of irreducible denominators $d_i$, if we set

$$t:=\exp\left(-\frac x{\text{lcm}(d_i)}\right),$$

we get the polynomial equation

$$\sum_{i=0}^n a_i t^{n_i}-1=0$$ where $n_i=b_i\text{lcm}(d_i)$.

Hence this proves that the equation does not have an analytical solution in the general case.

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If you consider the logarithm of the LHS (and look for its zeroes), it has two oblique asymptotes. For large positive $x$, it reduces to $\log a_{\min}-b_{\min}x$ and for large negative $x$, to $\log a_{\max}-b_{\max}x$ (the $\min$ and $\max$ indexes refer to $b$). This gives you an approximation of the root at the crossing point.