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From the first sight, this equation:

$\exp(-2at)=-\exp(-2bt)$

has no solution.

However, Worfram Mathematica clams, it exists. I am wondering, what is the most common to solve it: perhaps, Taylor expansion? Minus in from of the second exponent forbids using the log-mathod. Thank you very much in advance.

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It has no real solution. But it says $e^{2bt-2at}=-1$, which is not hard to solve if you are OK with complex numbers. –  Gerry Myerson Jun 17 '13 at 13:16
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3 Answers

Hint: $$e^{2bt-2at}=-1=e^{i\pi}$$

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Since $\exp$ is a positive function, $-\exp$ will be negative, hence $\exp(\text{whatever}_1)=-\exp (\text{whatever}_2)$ has no real solutions.

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$\displaystyle e^{2at}=-e^{2bt}\Rightarrow {e^{2(a-b)t}}=-1\Rightarrow {e^{2(a-b)t}}=e^{i\pi}$

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As a general rule, strings of symbolic implications are not considered very good answers. Fortunately, it is usually the case that a well chosen sentence can be added to help convey things to the reader. Maybe you have one in mind? –  rschwieb Jun 17 '13 at 13:43
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