What is #(5r^-2)^-2/(2r^3)^2#?

Question

829 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-14T07:23:32

#((5r^(-2))^(-2))/((2r^3)^2)=1/(100r^2)#

We can use here the identity #a^(-m)=1/a^m#.

Hence, #((5r^(-2))^(-2))/((2r^3)^2)#

= #(1/(5r^(-2))^2)/((2r^3*2 r^3)#

= #(1/(5/r^2)^2)/(4r^6)#

= #(1/(25/r^4))/(4r^6)#

= #(r^4/(25))/(4r^6)#

= #r^4/(25*4r^6)#

= #1/(100*r^((6-4)))=1/(100r^2)#

Alternatively one can also use #(a^m)^n=a^(mn)# for all integers.

Hence, #((5r^(-2))^(-2))/((2r^3)^2)#

= #(5^(-2)*r^((-2)×(-2)))/((2r^3*2 r^3)#

= #r^4/(25*4r^6)#

= #1/(100*r^((6-4)))=1/(100r^2)#