# How to simplify $(x^{1/y})^y$?

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?

• PowerExpand[(x^(1/y))^y] does what you want. Assumes everything is real and positive. Assumptions would do the trick too. – Bill Watts May 14 '19 at 23:48

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

• 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. – Henry Shackleton May 15 '19 at 2:37
• @HenryShackleton Try $x=-1$ and $y=1/2$ Basically, if $y$ is not an integer, than problems arise. – Carl Woll May 15 '19 at 2:42
• I see, thank you for the clarification! Including the assumption that $y$ is an integer is also sufficient for the simplification to go through. – Henry Shackleton May 15 '19 at 2:50