How do you simplify #[(8(x^3) ^2 y^-3)/(6y^2 x^-2)] [(4(x^2 ) ^2 y^-3)/(12x^-3 y^2 )]# using only positive exponents?

1 Answer
Apr 21, 2016

Answer:

# (4x^15)/(9y^10)#

Explanation:

Using the following #color(blue)" rules of exponents " #

#• a^mxxa^n = a^(m+n)#

example : #2^2xx2^3 = 2^(2+3) = 2^5 = 32 #
#color(red)"-------------------------------------------------"#
#•( a^m)/(a^n) = a^(m-n) #
example# (3^4)/(3^2) = 3^(4-2)=3^2=9 #
and # (2^3)/(2^5) = 2^(3-5)=2^(-2) #
#color(red)"--------------------------------------------------"#
#• a^(-m) hArr 1/(a^m) #
example (from above): #(2^3)/(2^5)=2^(-2)=1/(2^2)=1/4#
#color(red)"----------------------------------------------------"#
#• (a^m)^n = a^(mxxn)= a^(mn)#

example# (2^2)^3=2^(2xx3)=2^6=64#
#color(red)"-----------------------------------------------------"#

now applying these to the question.

first bracket# [(8(x^3)^2y^(-3))/(6y^2x^-2)]=8/6xx(x^6)/(x^-2)xx(y^-3)/y^2 #

#=4/3x^(6-(-2))y^(-3-2)=4/3 x^8 y^-5#
#color(blue)"-------------------------------------------------"#

second bracket # [(4(x^2)^2y^-3)/(12x^-3y^2)]=4/12xxx^4/x^-3xxy^-3/y^2#

#=1/3x^7y^-5 #
#color(blue)"------------------------------------------------"#

Now multiplying the 2 simplifications together

# 4/3 x^8y^-5xx1/3 x^7y^-5 =4/9x^(15)y^(-10)=(4x^(15))/(9y^(10)#