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While I was studying my maths book, I came across this equation:

$ xe^{-x}+2e^{-x}=0 $

I tried to solve it in different ways, but each time I break up some rule. My best try was this:

Let's $u=e^{-x}$, thus we have: $ xu+2u=0 $ By taking $u$ as a common factor we get:

$ u(x+2)=0 $

By dividing both side by $(x+2)$ we get:

$ u=0 $

But $u=e^{-x}$, then:

$ e^{-x}=0 \\ ln(e^{-x}) = ln(0) ?? $

$ln(0)$ is obviously wrong, where did I slip?

3 Answers 3

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When you divide by $x+2$ how do you know $x+2\neq 0$? Indeed you don't! Thats why you should get $u=0$ or $x+2=0$. The first equation has no solutions as $u=e^{-x}>0$. The second gives $x=-2$.

You write $e^{-x}=0$. This equation has no solutions. But you can't write $\ln (e^{-x})=\ln (0)$!!!! This is because $\ln (0)$ is not defined!

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    @HenningMakholm I believe you know I meant $\ln(0)$ and not $\ln(0!!!)$!2012-12-30
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If $xe^{-x}+2e^{-x}=0$, then divide across by $e^{-x}$ (which is non-zero) to get $x+2=0$, which has the solution $x=-2$.

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$a\cdot b=0\implies a=0\text{ or }b=0$ so $u=0\text{ or }x+2=0$.

You deduced $u\neq0$ hence $x+2=0$. That is $x=-2$