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Solve for x e^-2x-2xe^-2x=0
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\[e ^{-2x}-2xe ^{-2x}=0\]
Factoring, you have\[e^{-2x}(1-2x)=0\] What do you know about exponentials?
Very little
I'll narrow it down a bit. For any value of x, what will e^(ax) be?
1
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Not quite... That's only true for x = 0. The point I'm trying to get across is that \[e^{ax}>0 \text{ for any constant a and value of x.}\]
oh ok
Do you see how you can use this fact?
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