How do you simplify # [(4x^-4 y^2)/ (5x^6 y^-3)]^-3# leaving only positive exponents?

1 Answer
May 16, 2016

Answer:

#(125x^30)/(64y^15)#

Explanation:

Given,

#((4x^-4y^2)/(5x^6y^-3))^color(blue)(-3)#

According to the exponent law , #(xy)^n=x^ny^n#, the #color(blue)(-3)# multiplies by each exponent in the numerator and denominator. In addition, the numbers are raised to #color(blue)(-3)#. Thus,

#=(4^(color(blue)(-3))x^((-4xxcolor(blue)(-3)))y^((2xxcolor(blue)(-3))))/(5^(color(blue)(-3))x^((6xxcolor(blue)(-3)))y^((-3xxcolor(blue)(-3))))#

Simplifying,

#=(4^-3x^12y^-6)/(5^-3x^-18y^9)#

According to the exponent law , #x^-n/1=1/x^n#. Similarly, #1/x^-n=x^n/1#. Thus,

#=(5^3x^(12)*5x^18)/(4^3y^9y^6)#

Simplifying,

#=(125x^(12)x^18)/(64y^9y^6)#

Using the exponent law , #x^mx^n=x^(m+n)#, the exponents in the numerator of base #x# can be added together. Similarly, the exponents in the denominator of base #y# can be added together.

#=(125x^(12+18))/(64y^(9+6))#

#=color(green)(|bar(ul(color(white)(a/a)color(black)((125x^30)/(64y^15))color(white)(a/a)|)))#