(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

solve e^(2 x)-e^x-12 = 0 for x

2. Relevant equations

quadratic

3. The attempt at a solution

We had this question on a test, and after solving it (i'd rather not write out the work), I came to two solutions.

Ln(4)

Ln(-3)

I plugged both answers back into the equation, and they both worked.

My teacher says that Ln(-3) is NOT an answer, because it is an "extraneous solution".

I did some messing around on Wolframalpha, and they seem to agree that ln(-3) is, in fact, a solution.

Can someone please clarify if both answers are in fact solutions, or if one is an extraneous solution. Or if you think he meant something else, please tell me. please keep in mind that i'm still in precalculus, so please explain it in as simple terms as possible.

Thanks.

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# Solve e^(2 x)-e^x-12 = 0 for x

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