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I've heard before that some index laws don't apply in some situations, like when the base is negative.

For example, I entered $y = (x^{1.1})^{1.2} - x^{1.1\times1.2}$ into Wolfram Alpha and the graph it displayed was not $y=0$, contrary to my expectations.

From what I understand, $(x^a)^b = x^{ab}$ only when $x > 0$ and $a, b \in \mathbb{N}$. Is that correct?

Are there similar restrictions on other index laws? Specifically,

$x^a \times x^b = x^{a+b}$

$x^a \div x^b = x^{a-b}$ ($x \neq 0 $)

$x^\frac{a}{b} =\sqrt[b]{x^a}$ ($b \neq 0 $)

PKBeam
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  • "only when" is the wrong phrase to use here. There exist values of $x,a,b$ such that $(x^{a})^b=x^{ab}$ that don't follow that, for example it happens to be true that $\sqrt{i}^2=\sqrt{i^2}=i$. It has more to do with whether or not you cross a branch cut. It is true however that $(x^a)^b=x^{ab}$ is guaranteed to be true when $x>0$ and $a,b\in\Bbb N$. – JMoravitz Jan 28 '18 at 07:45
  • As for similar restrictions on the others, your first two are equivalent and should work regardless if I'm remembering correctly. As for the third... that is a messy topic. If $a$ and $b$ are both coprime integers with $b$ positive and we are wishing to define this as a function (i.e. only one output), then $x^{\frac{a}{b}}$ is commonly defined as $(\sqrt[b]{x})^a$, which is potentially unequal to $\sqrt[b]{x^a}$. In the case that $a$ and $b$ are not coprime, i.e. $\frac{a}{b}$ is not in simplest form, then $\frac{a}{b}$ should be rewritten in simplest form and those values be used instead – JMoravitz Jan 28 '18 at 07:51
  • When either or both are allowed to be irrational., it might be best to scrap our earlier work and go instead with the definition of $x^a = \textbf{exp}(a\ln x)$. See for example this post. As for possible errors in other index laws, see this section of the wiki page. – JMoravitz Jan 28 '18 at 07:56

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