Re: exponential equation with constant

quiasmox@xxxxxxxxx wrote:
[snip]
> This solution depends on the coincidence that one exponent is exactly
> twice the other.
> This brings me to the question -- is there a way, other than computer
> simulation, to solve a*exp(-ct) = 1 - b*exp(-dt) for t?

Yes.

Since this type of problem arises fairly frequently, I wrote
"Solving A e^(ax) + B e^(bx) = 1" (sci.math, July 2004),

<http://mathforum.org/kb/messag­e.jspa?messageID=3392781>

which expresses the solution as a series.

David W. Cantrell
.

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