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If we are given an expression $(-8)^\frac26$, how do we solve it?

If it is $(-8)^\frac13$, we can find the cube root of -8 which is -2. However, if we square it first and find the sixth root, we get +2 (or maybe $\pm2$, which still isn't the same as just -2.

Also realised that when solving logarithmic and exponential questions, I had often made the assumption that if $b^a=b^c$, then $a=c$, either explicitly, or via some formula. I don't know if this is really true now.

So is there some convention by which we solve non-integral powers or is it just undefined to do so or something?

P.s. Talking of non-integral powers, will a quantity like $(-8)^{√3}$, will it be positive or negative. (Basically that's an irrational power now.)

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  • $\begingroup$ With irrational exponents $r$ and negative real numbers $x$, the expression $x^r$ is left undefined in the set of real numbers. Complex numbers are a different story (where Euler's formula can give you an interpretation). $\endgroup$ – John Coleman Aug 9 '16 at 15:22
  • $\begingroup$ One usually uses Euler's formula: $$e^{xi}=\cos(x)+i\sin(x)$$ in particular $$(-8)^{\sqrt3}=e^{\sqrt3\pi i}8^{\sqrt3}$$ $\endgroup$ – Simply Beautiful Art Aug 9 '16 at 15:26
  • $\begingroup$ @ghosts_in_the_code: It depends on what level you are at. For complex numbers see the above comments. If you are taking an Algebra/Precalculus class, usually you assume that the base of the exponential is positive and not equal to 1. Under this assumption, yes, you can conclude that $b^a=b^c$ implies $a=c$, by taking the logarithm base $b$ of both sides. If $b$ is negative, the logarithm base $b$ will give you infinitely many answers! $\endgroup$ – Ioannis Souldatos Aug 9 '16 at 15:32
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In the context of complex numbers, $a^b$ (for $a \ne 0$) is a "multi-valued function", defined as $\exp(b \log(a))$ where $\log(a)$ is any of the branches of the natural logarithm of $a$. This is multivalued because you can add any integer multiple of $2\pi i$ to a value of $\log(a)$ and get another one.

Since $2/6 = 1/3$, $(-8)^{2/6}$ is the same as $(-8)^{1/3}$. This is $\exp((1/3) \log(-8))$, and $\log(-8) = 3 \ln(2) + (2n+1) \pi i$ for integer $n$, so we get $2 \exp((2n+1) \pi i/3)$ which has three different possible values: $2 \exp(\pi i/3) = 1 + \sqrt{3}\; i$, $2 \exp(\pi i) = -2$ and $2 \exp(5 \pi i/3) = 1 - \sqrt{3}\; i$.

What you apparently meant was $((-8)^2)^{1/6}$, which is not the same thing. The "laws of exponents" have to be modified when complex numbers are involved.
$$((-8)^2)^{1/6} = 64^{1/6} = \exp((1/6) \log(64)) = \exp(\ln(2) + n \pi i/3)$$ which has $6$ possible values: the three we had before as well as $2$, $-1+\sqrt{3}\; i$ and $-1-\sqrt{3}\; i$.

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