Exponential equation with a negative exponent

Question

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.

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

score 1 · Answer 1 · answered Jun 17 '13 at 13:17

1

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

answered Jun 17 '13 at 13:17

M.H

11,498
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score 0 · Answer 2 · answered Jun 17 '13 at 13:16

0

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

answered Jun 17 '13 at 13:16

ExpIsPositive

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score 0 · Answer 3 · answered Jun 17 '13 at 13:17

0

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

answered Jun 17 '13 at 13:17

Abhra Abir Kundu

8,150

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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

Exponential equation with a negative exponent

3 Answers3