Solve $(e^{x+1} -2)(e^{2x} -4) = 0$ ... but there is a problem!

Question

I am a little bit confused. There is this problem:

$$(e^{x+1} -2) (e^{2x} -4) = 0$$

I thought, i could just solve it like this $(a - b)(c - d) = 0 \therefore ac -ad -bc + bd = 0$

After few attempts, i found out, that you can solve it simply this way:

$(e^{x+1} -2)=0$, solve this to get X1

$(e^{2x} -4)=0$, solve this to get X2

But is it allowed? I mean can you just separate the bracket and solve it independently of each other? How is this "Rule" called? Is there a name for it?

Thanks for the answer and please excuse my bad english :)

Answer 1

It follows from the fact $\mathbb{R}$ has no "zero divisors"; i.e. if $x,y \in \mathbb{R}$ and $xy = 0$, then at least one of $x$ and $y$ must be zero.

Answer 2

Consider the following equation $(x-2)(x-1)=0$

to say simply this equation can only be reduced to zero only if $x=2$ or $x=1$

similarly apply this to your equation

$e^{x+1}=2$ or $e^{2x}=4$

$\implies$ $x=\ln2-1$ or $x=\frac{\ln4}2$

simplifying we get

$\implies$ $x=\frac {2\ln2}2$ or $x=\ln2$

Answer 3

The rule is called $${X}\cdot{Y}=0\iff(X=0)\vee(Y=0)$$