Exponential Equilibrium Equation

[ArbyFisher]

Solve for real-valued x, e^-aX + e^-bX = 1, where a and b are arbitrary known constants > 0.

For example, e^-48.12/50 + e^-48.12/100 ~ 1.00

In this case X = 48.12 (to two decimals), a = 1/50 and b = 1/100.
For any specific values of a and b, a computational solution can easily be determined, but a general algebraic solution is desired.

The problem can be variously reformulated, i.e., Y = e^X, Y^c + Y^d = 1, c and d < 0

... alas to no apparent avail. The equation sometimes seems like it should be a queuing theory probability or perhaps some geometric shape or I don't know anymore (obviously) ... thanks in any case.

 

[adriank]

If you reduce it to something like Y^c + Y^d = 1, if c and d are rational then you should be able to write it as a polynomial equation in Y. Unfortunately, if the degree of that polynomial is at least 5, you're not in general going to be able to write a solution explicitly in terms of radicals.

In your specific example, let's say you have e^-x/50 + e^-x/100 = 1. Substituting y = e^-x/100 (so x = -100 ln y) you have y² + y = 1, which has solution y = (-1 ± √5) / 2, but y must be positive so y = (√5 - 1) / 2, so x = -100 ln ((√5 - 1) / 2) = 48.1212.

 

[ArbyFisher]

Exactly. If the numbers are "very nice", as in the example, then the factoring works charmingly. But, if the numbers are only rationally nice, then the solution begins to devolve into combinatorial factoring dementia, and factoring then doesn't appear to be a general solution. Finally, of course, if the numbers are irrational, then ... Thank you much for your reply.