Question #13eae

1 Answer
sente
May 19, 2016

#((x^3y^4)/(x^(-2)y^(-4)))^-1=1/(x^5y^8)=x^-5y^-8#

Explanation:

We will use the following properties of exponents:

  • #x^a*x^b = x^(a+b)#

  • #(x^a)^b = x^(ab)#

  • #(xy)^a = x^ay^a#

  • #x^0 = 1# for all #x!=0#

With those, for #x, y != 0# we have

#((x^3y^4)/(x^(-2)y^(-4)))^-1 = ((x^3)^(-1)(y^4)^(-1))/((x^(-2))^(-1)(y^(-4))^(-1))#

#=(x^(3*-1)y^(4*-1))/(x^(-2*-1)y^(-4*-1))#

#=(x^-3y^-4)/(x^2y^4)#

#=((x^-3y^-4)(x^3y^4))/((x^2y^4)(x^3y^4))#

#=(x^(-3+3)y^(-4+4))/(x^(2+3)y^(4+4))#

#=(x^0y^0)/(x^5y^8)#

#=(1*1)/(x^5y^8)#

#=1/(x^5y^8)#

Note that we can also multiply this by #(x^-5y^-8)/(x^-5y^-8)# to find it to be equivalent to #x^-5y^-8#

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