Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

From the first sight, this equation:


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.

share|cite|improve this question
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

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

share|cite|improve this answer

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

share|cite|improve this answer

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.