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Simplify. `root(4)(32x^11y^15)/root(4)(2x^3y^2)`   ` ` 

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kristenmarieb... | Student, Grade 10 | Valedictorian

Posted April 17, 2013 at 3:47 AM via web

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Simplify.

`root(4)(32x^11y^15)/root(4)(2x^3y^2)`

 

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pramodpandey | College Teacher | Valedictorian

Posted April 17, 2013 at 4:09 AM (Answer #1)

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 We know

`x^m/y^m=(x/y)^m`

Thus apply this rule

`root(4)(32x^11y^15)/root(4)(2x^3y^2)=root(4)((32x^11y^15)/(2x^3y^2))`

`=(16x^(11-3)y^(15-2))^(1/4)`

`=(2^4x^8y^13)^(1/4)`

`=(2^4)^(1/4)(x^8)^(1/4)(y^13)^(1/4)`

`=2x^2y^3(y)^(1/4)`

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oldnick | Valedictorian

Posted April 17, 2013 at 2:42 PM (Answer #2)

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`root(4)(32x^11y^15)/root(4)(2x^3y^2)=` `root(4)((32x^11y^15)/(2x^3y^2))` `=root(4)(16x^8y^13)` `=root(4)(2^4(x^2)^4 (y^3)^4 y)` `=2x^2y^3root(4)(y)`

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