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Within the broader scope of simplifying my answer to a calculus problem, I have what I think are some really basic questions on the powers of exponents.

So, obviously $\sqrt{6} = \sqrt{2} \cdot \sqrt{3}$. It seems that such a rule of raising some number (6 in this case) to the $\frac{1}{n}$ ($n \in [0, \infty)$ for our purposes) is equivalent to the product of its factors raised to the $\frac{1}{n}$. Based on some trial runs in Wolframalpha that I've run, however, this doesn't seem to always be the case.

For example: Case 1: $(3)^n \cdot (2)^n = (6)^n$ (this worked out)

Case 2: $(-3)^n \cdot (-2)^n = (6)^n$ (doesn't seem to work out)

Case 3: $(-2)^n \cdot (3)^n = (-6)^n$ (works out in both directions, no matter which term I attach the negative sign to)

I haven't quite been able to attach a rule to this one or find one, even though this seems like a relatively foundational concept. Now, oddly enough, I have an example of Case 3 that doesn't seem to work out. For example, I have \begin{align*} \Bigg(- \frac{x^2}{a^2}\Bigg)^n \end{align*} for $a > 0$, $n \geq 0$, and any value of $x \in \mathbb{R}$. Following Case 3, I should be able to write \begin{align*} \Bigg(- \frac{x^2}{a^2}\Bigg)^n = \Bigg(-1 \cdot \frac{x^2}{a^2}\Bigg)^n = (-1)^n \Bigg(\frac{x^2}{a^2}\Bigg)^n \end{align*} However, Wolframalpha, the only tool I know of that would let me verify something like this, requires that $x$, $a$, and $n$ be positive for this to be true.

My questions, then, are: (a) is this step legal and (b) what are the rules of thumb regarding this?

Thanks in advance, and my apologies if this is rather trivial.

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    $\begingroup$ You start with 1/n then the rest of the question is about n. Which one? For integer exponents, $a^n \cdot b^n = (a \cdot b)^n$ holds regardless of signs. $\endgroup$
    – dxiv
    Dec 31, 2017 at 21:08
  • $\begingroup$ Sorry about that. I'll revise it to make it clearer. But, yes, $n$ would be an integer, so I suppose that settles it. $\endgroup$
    – user465188
    Dec 31, 2017 at 21:09
  • $\begingroup$ As an aside, when referring to the exponent being a fraction, you do run into issues when referring to principal roots when the bases are negative. For example $-1=i\cdot i=(-1)^{\frac{1}{2}}\cdot (-1)^{\frac{1}{2}}\neq ((-1)\cdot (-1))^{\frac{1}{2}}=1^{\frac{1}{2}}=1$ $\endgroup$
    – JMoravitz
    Dec 31, 2017 at 21:12

1 Answer 1

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Case 2: $(-3)^n \cdot (-2)^n = (6)^n$ (doesn't seem to work out)

Yes, it does work out: $\,(-3)^n \cdot (-2)^n = \big((-3)\cdot(-2)\big)^n = 6^n\,$.

For integer $\,n\,$, the identity $\,a^n \cdot b^n = (a \cdot b)^n\,$ holds true for all real $\,a, b \in \mathbb{R}\,$ regardless of signs. This follows directly from the commutativity and associativity of real multiplication:

$$ a^n \cdot b^n = (a \cdot a \cdot \ldots \cdot a) \cdot (b \cdot b\cdot \ldots \cdot b)=(a\cdot b) \cdot (a \cdot b) \cdot \ldots \cdot (a\cdot b) = (a \cdot b)^n $$

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  • $\begingroup$ This is very helpful. Thank you! If only I were able to find such a nice identity before.. $\endgroup$
    – user465188
    Dec 31, 2017 at 21:13

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