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

Question

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?

Answer 1

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!

Answer 2

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 $x=-2$.

Answer 3

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

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