Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 10 down vote accepted

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!

share|cite|improve this answer
@Sp. Note that as above never,never make some exponential terms like $e^a$ equal to $0$. It is absolutely wrong. No additional notes needed for your problem +1 :-) – Babak S. Dec 30 '12 at 9:46
$\ln(0!!!!)$ is defined; it's zero. :-) – Henning Makholm Dec 30 '12 at 10:45
@HenningMakholm I believe you know I meant $\ln(0)$ and not $\ln(0!!!)$! – Nameless Dec 30 '12 at 10:48

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 $x=-2$.

share|cite|improve this answer

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

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.