How do you simplify #(x^(1/pi)/y^(2/pi))^pi#?

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Answer 1 · 2018-07-21T07:08:14

#(x^(1/pi) / y^(2/pi))^pi = x / y^2#

Note that if #a > 0# and #b, c# are any real numbers then:

#(a^b)^c = a^(bc)#

So (assuming #x, y > 0#) we find:

#(x^(1/pi) / y^(2/pi))^pi = (x^(1/pi))^pi / (y^(2/pi))^pi = (x^(1/pi * pi)) / (y^(2/pi * pi)) = x / y^2#

Answer 2 · 2018-07-21T12:37:39

#x/y^2#

#(x^(1/pi)/y^(2/pi))^pi=x^(pi/pi)/y^((2pi)/pi)=x/y^2#

Answer 3 · 2018-07-21T13:55:03

#\frac{x}{y^2}#

#(\frac{x^{1/\pi}}{y^{2/\pi}})^\pi#

#=\frac{(x^{1/\pi})^\pi}{(y^{2/\pi})^\pi#

#=\frac{x^{\pi\cdot 1/\pi}}{y^{\pi\cdot 2/\pi}#

#=\frac{x^1}{y^2}#

#=\frac{x}{y^2}#