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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 :)

Paul
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gola
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  • The second one is the only way since, being lucky, you have the factors. Please, never develop a product of terms when you want the product to be equal to $0$. – Claude Leibovici Sep 16 '14 at 08:50
  • You should rephrase "only this way" as "simply this way", since there are in fact many other ways to solve it. – barak manos Sep 16 '14 at 08:56

3 Answers3

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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.

Matt Rigby
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1

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 $x=\ln2$

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

barak manos
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