Assuming $a, b \in \{\mathbb{R^-,R^+}\}$ and $n \in \mathbb{R}$. We often see the statement: $$(a \cdot b)^n = a^n \cdot b^n$$
Why we get the statement? How to prove this?
There should pay attention, the $n \in \mathbb{R}$, not $\mathbb{N}$.
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You can prove it for $n\in\Bbb N$ using commutativity of multiplication and induction – J. W. Tanner Jul 01 '19 at 03:45
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you mean $a,b>0$? – J. W. Tanner Jul 01 '19 at 03:54
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3It does not make sense to write that $a,b\in {\mathbb{R}^- ,\mathbb{R}^+ }$. – azif00 Jul 01 '19 at 04:12
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for avoid $0^0$ – 244boy Jul 01 '19 at 04:15
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1With that you would be saying that $a=\mathbb{R}^-$ or $a=\mathbb{R}^+$. Similarly for $b$. Better you could just write $a\neq 0$. – azif00 Jul 01 '19 at 04:17
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2You probably want $a > 0$ and $b > 0$. Raising a negative number to the power of a rational is dicey enough without worrying about real powers. – Theo Bendit Jul 01 '19 at 04:29
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Similar question:https://math.stackexchange.com/questions/12720/why-is-xn-cdot-xm-xnm?rq=1 – NoChance Jul 01 '19 at 05:32
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Again, for $a<0\land b<0$ the claim fails. It works for $(a<0\land b>0)\lor (a>0\land b<0)\lor(a>0\land b>0)\lor n\in\Bbb Z$. – Jul 01 '19 at 05:45
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why for <0∧<0 it claim fails? – 244boy Jul 01 '19 at 07:49
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For any $a>0$, is defined $a^x=e^{x\log a}$ for all $x\in \mathbb{R}$. So $$(ab)^x=e^{x\log (ab)}=e^{x(\log a +\log b)}=e^{x\log a}\cdot e^{x\log b}=a^xb^x$$ The assumption $a, b> 0$ is clearly necessary for this to be true.
azif00
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1Everything depends on your definitions, for example, you can define $$\log x:=\int_1^x \frac{1}{t}dt$$ for all $x>0$. With this, prove that $\log(xy)=\log(x)+\log(y)$. And finally, you define $a^x=e^{x\log a}$ for any $a>0$ (all this with the purpose that things work as they should). – azif00 Jul 01 '19 at 04:08
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1Of course it's not the only way to prove it, but it's still correct, again, it all depends on your definitions. – azif00 Jul 01 '19 at 04:09
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1@244boy Just have a think about defining real powers. We have a natural definition for $a^n$ where $n \in \Bbb{N}$. We can extend it to $n \in \Bbb{Z}$ (where $a \neq 0$) using multiplicative inverses, and further into $n \in \Bbb{Q}$, by using roots. How would we define $a^\pi$? Bear in mind, you can't define it to be what your calculator says, as your calculator also needs a definition to display anything (plus, it only displays an approximation). – Theo Bendit Jul 01 '19 at 04:36
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1@244boy A naive (but viable) approach would be to define $a^\pi$ as a limit of $a^{r_n}$ where $r_n \in \Bbb{Q}$ and $r_n \to \pi$, but there's much to prove here, including that $a^{r_n}$ has a limit and that it doesn't depend on the choice of $r_n$. The more modern approach is to define $e^x$ or $\ln$ first (as some kind integral, limit, power series, etc; there are a lot of options) and define $a^x$ as Azif00 does in their answer. – Theo Bendit Jul 01 '19 at 04:39