How do you simplify #(sqrt(x) / (y^-2)) ^4 #?

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Answer 1 · 2017-04-04T02:36:47

#(sqrtx/y^-2)^4=x^2y^8#

To start we can deal with the negative exponent using the rule #a^-b=1/a^b#. Then:

#(sqrtx/y^-2)^4=(sqrtx/(1/y^2))^4=(y^2sqrtx)^4#

We should rewrite #sqrtx# with a fractional exponent:

#(y^2sqrtx)^4=(y^2x^(1/2))^4#

Split up the exponent:

#(y^2x^(1/2))^4=(y^2)^4(x^(1/2))^4#

Use the rule #(a^b)^c=a^(bc)#:

#(y^2)^4(x^(1/2))^4=y^(2(4))x^(1/2(4))=x^2y^8#