solving $(1-e^x)(x)=0$ I am trying to understand the logic behind solving algebraic equations. As I understand, you can only multiply or divide both sides by $x$ if $x\not=0$, so If you do end up dividing or multiplying by a $x$, you should always consider the case where $x$ is $0$.
In the following equation, $$(1-e^x)(x)=0$$ if I assume $(1-e^x)\not=0$ and divide both sides by it, I get $x=0$, which makes the assumption and thus the solution invalid. The same thing happens if I initially assume $x\not=0$. It's pretty obvious that $x=0$ is a solution but I can't figure out the more rigorous method to get there. Do I just plug in 0 and see if it works for equations like these?
 A: $ab = 0 \implies a = 0 \text{  or  } b =0$
$1 - e^x = 0 \implies x = 0$
$x = 0 \implies x = 0$
The solution set is the union of the solution sets of the respective generated equations. Since $\{0\} \cup \{0\} = \{0\}$, the desired solution set is $\{0\}$. The only exception here is when a solution causes the initial expression to become undefined, in which case it is removed from the solution set.
An example of when this would occur would be for the function $x(1-\cos(x))^{-1} = 0$, for which
$x = 0 \implies x \in \{0\}$
$(1-\cos(x))^{-1} \implies x \in \emptyset$
The union of these two solution sets is $\{0\}$, however $x = 0$ is clearly undefined in the original expression, thus the final solution set becomes $\emptyset$
A: You're on the right track, you just need to follow through on your reasoning: If
$$(1-e^x)x=0,$$
and you assume that $1-e^x\neq0$, then $x=0$. But then also $1-e^x=0$, a contradiction. So your assumption must be false, and you can conclude that $1-e^x=0$, and hence that $x=0$.
You could also assume that $x\neq0$, and then $1-e^x=0$. But then also $x=0$, a contradiction. So your assumption must be false, and you can conclude that $x=0$.
