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I'm probably missing something due to the long day of going through basic materials and practicing a lot, but I'm puzzled.

$e^x(x^2 - 1) = 0$

Now, I could solve this as follows: $e^xx^2 - e^x = 0 \\ e^xx^2 = e^x \\ x^2 = 1 \\ x = \pm1$

Another way could be to divide both sides by $e^x$, giving $x^2 - 1 = 0$, so: $x^2 = 1$ and $x = \pm 1$

But I could've also choosen to divide bot sides by $x^2- 1$, yielding $e^x = 0$, which has no solution...

What am I doing wrong in this last part?

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    $\begingroup$ The key point is that $e^x$ is never $0$, so we can divide by it. The same can't be said for $x^2-1$. $\endgroup$ – lulu Apr 6 '17 at 17:36
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In the last part when you divide by $x^2-1$ you must assume that $x^2-1 \neq 0$ because you can't divide by $0$. But clearly $x^2-1=0$ so this division is not allowed.

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When you divide you need to make sure you're not dividing by zero. In your example dividing by $x^2-1$ you need to ensure that $x\neq \pm 1$

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$e^x = 0$

Taking $\log$ on both sides,

$x = \log 0$

And $\log 0$ is undefined.

So either no solution or $x = \pm 1$.

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