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As shown by Wolfram Alpha, $(x^2)^{0.5}$ is equal to |x|, but if you tried to simplify it to $x^{2\times {0.5}}$, it would just be $x^1$, or $x$.

Is there some unwritten rule about that distribution law that means you can't do it with fractional exponents?

Edit: What confuses me the most is how Wolfram Alpha also believes $(x^2)^{0.5}$ = x while actually showing a graph of |x|

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  • $\begingroup$ Yes. For rational values of $q$, the value $x^q$ is only defined for positive values of $x$. $\endgroup$ – 5xum Mar 5 '14 at 11:49
  • $\begingroup$ @5xum don't you mean fractional values of q? $(-5)^2$ is very well defined, where neither -5 is positive nor 2 is irrational. $\endgroup$ – Guy Mar 5 '14 at 11:58
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    $\begingroup$ Oh no, not again ... $\endgroup$ – Michael Hoppe Mar 5 '14 at 12:22
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With iterated exponentiation the order of the operations are read from the top down. So in your example one has $$(-5)^{2^{0.5}}=(-5)^{\left (2^{0.5} \right )}=(-5)^{\sqrt{2}}$$ which is undefined over the reals. Also note that in general $$x^{y^{z}} \neq x^{yz}$$

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  • $\begingroup$ I was confused about the whole weird-dollar-signs-everywhere thing until I realised MathJax is blocked thanks to a https/http mismatch. This answer explains a lot, thanks. $\endgroup$ – iirelu Mar 5 '14 at 12:03
  • $\begingroup$ Actually now that you've edited it I don't understand anything. The ((-5)^2)^0.5 going to (25)^0.5 made more sense to me. $\endgroup$ – iirelu Mar 5 '14 at 12:05
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To be precise, $$(x^y)^z \ne x^{yz} \text{for negative $x$.}$$

This is because if $x \not \in \mathbb R^+$ then $x^y$ is not well defined for $y \not \in \mathbb Z$. In this particular case, $(-5^2)^{0.5}$ becomes, $25^{0.5}$. This is generally defined to be $\sqrt{25} = 5$. Your question is the same as asking why the sqaure root function has one value for one $x$, even though the function $f(x)=x^2$ is many one. This is because, contrary to what many people think, the square root is not an inverse of the square function, simply because the square function is not invertible, being many-one.

For a weak analogy, $\tan^{-1}(\tan(x))$ is not necessarily equal to $x$.

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Whenever we have grouping symbols we work from the inside out , following PEMDAS , WIA included. If you want the exponents to cancel do it this way , $$ (x^{0.5})^2$$

Your numerical example becomes $$((-5)^{0.5})^2 = ( \sqrt{-5})^2 = (i \sqrt{5})^2 = i^2 5 = -5 $$

And in general for real numbers x , $$( \sqrt{x})^2 = x$$ $$\sqrt{x^2} = |x| $$

Put it in WIA to see what it gives you.

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