How do you evaluate #1/(3^-2x^-4y^2)# for #x=3# and #y=-1#?

Question

2161 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-24T11:47:21

#729#

We have: #(1) / (3^(- 2) x^(- 4) y^(2))#; #x = 3, y = - 1#

Let's substitute the given values of #x# and #y# into the expression:

#= (1) / (3^(- 2) times (3)^(- 4) times (- 1)^(2))#

Using the laws of exponents:

#= (1) / (3^(- 2 - 4) times 1)#

#= (1) / (3^(-6))#

#= (3^(- 6))^(- 1)#

#= 3^(- 6 times - 1)#

#= 3^(6)#

#= 729#