- #1

etotheipi

Gold Member

2019 Award

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## Main Question or Discussion Point

I was just thinking about this earlier and couldn't come up with a good enough resolution. I'm guessing it's a matter of convention more than anything. If we have ##x^{2} = a##, taking the principle root of both sides gives ##\sqrt{x^{2}} = \sqrt{a} \implies |x| = \sqrt{a}##.

Yet evidently if the rule ##{(a^{b})}^{c} = a^{bc}## is taken to be true, then we end up with ##{(x^{2})}^{\frac{1}{2}} = a^{\frac{1}{2}} \implies x = \sqrt{a}##, which disregards the potential negative root.

If we use the definition ##a^{b} = e^{b\ln{a}}##, then negative bases make no sense since the domain of ##\ln{x}## is greater than zero. And, if I write ##a## in complex form, it turns out I can use the normal power rules to get both answers as expected:

##x^{2} = ae^{2n\pi i} \implies x = \sqrt{a} e^{n\pi i} = \pm \sqrt{a}##

Is the failure of this particular index law in the second example then just something we need to be aware of, or is there a resolution of some sort?

Yet evidently if the rule ##{(a^{b})}^{c} = a^{bc}## is taken to be true, then we end up with ##{(x^{2})}^{\frac{1}{2}} = a^{\frac{1}{2}} \implies x = \sqrt{a}##, which disregards the potential negative root.

If we use the definition ##a^{b} = e^{b\ln{a}}##, then negative bases make no sense since the domain of ##\ln{x}## is greater than zero. And, if I write ##a## in complex form, it turns out I can use the normal power rules to get both answers as expected:

##x^{2} = ae^{2n\pi i} \implies x = \sqrt{a} e^{n\pi i} = \pm \sqrt{a}##

Is the failure of this particular index law in the second example then just something we need to be aware of, or is there a resolution of some sort?