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Mathematics 14 Online
OpenStudy (anonymous):

Solve for x e^-2x-2xe^-2x=0

OpenStudy (anonymous):

\[e ^{-2x}-2xe ^{-2x}=0\]

OpenStudy (anonymous):

Factoring, you have\[e^{-2x}(1-2x)=0\] What do you know about exponentials?

OpenStudy (anonymous):

Very little

OpenStudy (anonymous):

I'll narrow it down a bit. For any value of x, what will e^(ax) be?

OpenStudy (anonymous):

1

OpenStudy (anonymous):

Not quite... That's only true for x = 0. The point I'm trying to get across is that \[e^{ax}>0 \text{ for any constant a and value of x.}\]

OpenStudy (anonymous):

oh ok

OpenStudy (anonymous):

Do you see how you can use this fact?

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