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I would like Mathematica to be able to simplify the expression $\left(x^{1/y}\right)^y$ to $x$. However, running

Simplify[(x^(1/y))^y]

does not accomplish the desired simplification. I have tried including assumptions that $y$ is both real and non-zero in order to avoid any potential pathological cases, but neither seem to do the trick. Are there additional assumptions required in order to get this to work?

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  • $\begingroup$ PowerExpand[(x^(1/y))^y] does what you want. Assumes everything is real and positive. Assumptions would do the trick too. $\endgroup$ – Bill Watts May 14 at 23:48
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You probably meant:

Simplify[(x^(1/y))^y]

(x^(1/y))^y

which still doesn't simplify without assumptions. You can provide assumptions:

Simplify[(x^(1/y))^y, Assumptions->x>0 && y>0]

x

or you can use PowerExpand:

PowerExpand[(x^(1/y))^y]

x

Alternatively, you can give PowerExpand the option Assumptions->True to find out under what conditions the above expansion is valid:

PowerExpand[(x^(1/y))^y, Assumptions->True]

E^(2 I π y Floor[1/2 - Im[Log[x]/y]/(2 π)]) x

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  • $\begingroup$ Thanks! This works, but I don't understand why $x$ has to be positive. Naively taking the case $x=-1$, $y=2$ doesn't seem to give any problem, and I can't think of any situation with negative $x$ that would give the problem. $\endgroup$ – Henry Shackleton May 15 at 2:37
  • $\begingroup$ @HenryShackleton Try $x=-1$ and $y=1/2$ Basically, if $y$ is not an integer, than problems arise. $\endgroup$ – Carl Woll May 15 at 2:42
  • $\begingroup$ I see, thank you for the clarification! Including the assumption that $y$ is an integer is also sufficient for the simplification to go through. $\endgroup$ – Henry Shackleton May 15 at 2:50

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