How do you simplify #(4x^4y^-3)^-2# and write it using only positive exponents?

2 Answers
Jan 8, 2017

Answer:

See full simplification process below

Explanation:

First, we we use this rule for exponents to start the simplification:

#(x^color(red)(a))^color(blue)(b) = x^(color(red)(a) xx color(blue)(b))#

#(4x^4y^-3)^-2 ->#

#4^-2x^(4 xx -2)y^(-3 xx -2) ->#

#4^-2x^-8y^6#

Now, we will use this rule for exponents to finalize the simplification leaving only terms with positive exponents:

#x^color(red)(a) = 1/x^color(red)(-a)#

#y^6/(4^(- -2)x^(- -8)#

#y^6/(4^2x^8#

#y^6/(16x^8)#

Jan 8, 2017

Answer:

#(y^6)/(16x^8)#

Explanation:

Using the #color(blue)"laws of exponents"#

#color(red)(bar(ul(|color(white)(2/2)color(black)((a^mb^n)^p=a^(mp)b^(np))color(white)(2/2)|)))#
This can be extended to more than 2 products.

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

#=4^(-2)x^(-8)y^6#

#color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(a^-mhArr1/a^m)color(white)(2/2)|)))#

#rArr4^(-2)x^(-8)y^6=1/4^2xx1/x^8xxy^6/1#

#=(1xx1xxy^6)/(16xxx^8xx1)=(y^6)/(16x^8)#