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.
1/n
then the rest of the question is aboutn
. Which one? For integer exponents, $a^n \cdot b^n = (a \cdot b)^n$ holds regardless of signs. $\endgroup$