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In middle school, I was taught that $(x^a)^b = x^{ab}$. Of course, we were only looking at $a, b \in \mathbb{N}$.

So today I was playing around with squares, and I tried to do this: $$(x^\sqrt{2})^{\sqrt{2}} = x^{\sqrt{2}\cdot\sqrt{2}} = x^2$$ But then doing some tests in Wolfram alpha told me that I'm wrong for $x<0$. I know that we're dealing with complex numbers here, but I don't see why this would break that law.

My question is: Why can't I use this law in this situation? Is there an intuitive explanation for this?

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    So are you suggesting that $x<0 \implies (x^\sqrt{2})^{\sqrt{2}} \neq x^2$? – 2017-01-06
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    @barakmanos yes, according to Wolfram alpha: https://www.wolframalpha.com/input/?i=plot+((x)%5Esqrt(2))%5Esqrt(2)+from+-5+to+5 – 2017-01-06
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    Do you know how to evaluate $(-1)^{\sqrt{2}}$? (It depends on choice of branch cut for the complex logarithm...) – 2017-01-06
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    The key here is that you can only manipulate quantities that are well-defined (or that you know how to define). For instance, $(-1)^1 = -1$, but writing $\sqrt{-1}^2 = -1$ presupposes $\sqrt{-1}$ is defined. – 2017-01-06
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    Well, I've put [$(x^{\sqrt(2)})^{\sqrt(2)} = x^2$](https://www.wolframalpha.com/input/?i=((x)%5Esqrt(2))%5Esqrt(2)+%3D+x%5E2) in WolframAlpha, and the answer is *true*, so... – 2017-01-06
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    @barakmanos Above "True" W|A says "assuming x>0." – 2017-01-06
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    @arctictern: Oh... Right... – 2017-01-06
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    A very good explanation can be found in http://math.stackexchange.com/questions/608023/what-is-1-frac23?rq=1 – 2017-01-06

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For $a>0$ and $b\in \Bbb N$, $a^0 = 1$ and $a^b = a \times \cdots \times a$ ($b$ times) for $b \ge 1$. For $b \in\Bbb Z$, $a^b = \frac1{a^{-b}}$. For $b\in \Bbb R \setminus \Bbb Z$:

$$a^b = \exp(b \ln a)$$

No need to freak out, though, because all the rules you like still hold. Note that there are other possible ways to define $a^b$ when $b$ is non-integer real number, but this is the simplest one of them.

However, for $a < 0$,

$$a^b = \exp(b \log a)$$

where $\log$ is (some branch of) the complex logarithm function, which is defined for $z \in \Bbb C^*$ as:

$$\log z = \ln|z| + i\arg(z)$$

where some branch of $\arg$ is pre-chosen. And $\exp$ is the complex exponential function.

Now consider that for $a<0$, one has:

$$a^{bc} = \exp(bc \log a), \text{ and} \\ (a^b)^c = \exp(c\log(a^b))$$

These two would be equal only if $\log a^b = b \log a$, but unfortunately that's not true. For instance, assuming that we are dealing with the principal branch of logarithm, $\log((-1)^2) = \log 1 = 0$, while $\log(-1) = \ln|-1| + i\arg(-1) = i\pi$, and $0 \neq 2i\pi$.