How do you simplify #(-2r^(-4))^(2)(-3r^(2)2^(8))^(-1) #?

Question

1404 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-07T12:37:46

#-1/(172 r^10)#

There are two laws of indices which we can use to start.

#(xyz)^color(blue)(m) = x^color(blue)(m)y^color(blue)(m)z^color(blue)(m)" and " x^color(red)(-m) = 1/x^color(red)

#(-2r^(-4))^color(blue)(2)(-3r^(2)2^(8))^color(red)(-1) #

#=(-2)^2r^(-8) xx 1/(-3r^(2)2^(8))^color(red)(1) #

#=4/color(teal)(r^8) xx 1/(-3color(teal)(r^2) 2^8)" " larr color(teal)(x^m xx x^n = x^(m+n))#

#= - cancel4^1/(3color(teal)(r^10) cancel256_64)#

#=-1/(172 r^10)#