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e2x + ex = 0

e2x = -ex

e2x / ex = -1

ex = -1

x = ln(-1)

Actually, that's just a restatement of the original problem, but that's

as far as you can go with it, because there's no such thing as the log

(or the ln) of a negative number. Your equation has no solution.

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Q: How do you solve for x where e2x plus ex equals 0?
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