How do you simplify #((2^-1x^-3y^-1)(4x^-2)^0)/(2x^-4y^-3)^2#?

Question

1143 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-18T18:23:10

#(xy)^5/8#

#((2^-1x^-3y^-1)(4x^-2)^0)/(2x^-4y^-3)^2#

We should first notice the power of #0#. Anything to the power of #0# returns #1#.

#((2^-1x^-3y^-1)cancel((4x^-2)^0))/(2x^-4y^-3)^2#

#=>(2^-1x^-3y^-1)/(2x^-4y^-3)^2#

Now, let's square all of the terms in the bottom:

#(2^-1x^-3y^-1)/(2x^-4y^-3)^2#

#=>(2^-1x^-3y^-1)/(2^2x^-8y^-6)#

Now we compare terms and combine like-terms:

#(color(blue)(2^-1)color(red)(x^-3)color(orange)(y^-1))/(color(blue)(2^2)color(red)(x^-8)color(orange)(y^-6))#

#=>(color(red)(x^5)color(orange)(y^5))/(color(blue)(2^3))#

We can finally simplify the way it is written:

#(x^5y^5)/(2^3)#

#=>color(green)( (xy)^5/8)#