# Resolving index laws for fractional exponents

• B
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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?

## Answers and Replies

Mark44
Mentor
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.
For rational exponents, the rule above applies only when the base is positive.
For example ##((-4)^2)^{1/2} = 4##, but ##((-4)^{1/2})^2## is not a real number.
etotheipi said:
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?

• etotheipi
Stephen Tashi
Science Advisor
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.

An underlying problem is that, in texts treating only the real numbers, ##a^r## is not defined for rational numbers ##r##. Instead, ##a^r## is defined in a way that depends on how the rational number ##r## is denoted.

For example, consider ##r = 5/3 = 10/6 ##. Typical secondary school texts in the USA give the definition:

For integers ##m,n## if ##a^{\frac{1}{n}}## is a real number then ##a^{\frac{m}{n}}## is defined to be ##(a^{\frac{1}{n}})^m##.

Using this definition, students may be expected to compute ##(-1)^{\frac{5}{3}}## but they are expected to say that ##(-1)^{\frac{10}{6}}## does not exist - or perhaps the text avoids confusing the students by never asking for such a computation.

• etotheipi