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Solve x for 0=xe^(2x)
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The only time f(x)=0 is when x=0, because there is no value of x that will make e^2x=0.
I was thinking of using ln, but then i remember ln0 not right
If A * B = 0 either A is 0 or B is 0.
\[e ^{2x}>0 (always)\] so only solution is x=0
@NotTim You are right. ln is not defined for x=0.
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Somewhat unrelated, but the derivative is 2x(x)e^(2x)+e^2x, correct?
Sorry, that should be ln is not defined for\[x \le 0\]
2xe^(2x)+e^2x = e^(2x)(2x + 1)
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