How would you solve $6xe^{2x}+3e^{2x}=0$ for $x$
I tried:
$\ln(e^{2x})=\ln(1/6x+3)$
$2x=\ln(1)-\ln(6x+3)$
$2x=-\ln(6x+3)$
but then I am stuck there.
What am I missing?
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2You can't solve that $-$ it's not an equation. Do you mean $6xe^{2x}+3e^{2x}=0$? – TonyK Jul 21 '14 at 11:53
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1Do you mean $6xe^{2x} = 3e^{2x}$? – Michael Albanese Jul 21 '14 at 11:53
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sorry I meant 6xe^(2x)+3e^(2x)=0 – ADGB Jul 21 '14 at 11:54
1 Answers
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$$3e^{2x}(2x+1)=0 \Rightarrow x=-\frac{1}{2}, \text{ as } e^{2x} \neq 0, \forall x$$
evinda
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can you please explain a bit more the answer Evinda, I am not familiar with the reasoning behind the answer: "as e2x≠0,∀x" – ADGB Jul 21 '14 at 11:57
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1From the equality $3e^{2x}(2x+1)=0$ we get: $e^{2x}=0$ or $2x+1=0$..But, $e^{2x}=0$ cannot be true for any $x$,because we know that $e^x>0 , \forall x$.Therefore,it must be $2x+1=0 \Rightarrow x=-\frac{1}{2}$ – evinda Jul 21 '14 at 11:59
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