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
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)