Exponent values-GMATPREPI

medea66 wrote:How do you solve for this?

i am sure the question shud read like this ...

2^(x + Y)^2)/2^(x - Y)^2 ...

anyways here we go

2^((x+y)^2 - (x-y)^2)

2^(x^2 + 2xy + y^2 - x^2 + 2xy - y^2)

2^(4xy)

2^4*1 = 16..

hope this helps.. i am sure after u see the question in the right way ..its not at all difficult

sudhir3127 wrote:

medea66 wrote:How do you solve for this?

i am sure the question shud read like this ...

2^(x + Y)^2)/2^(x - Y)^2 ... GMAT PREP we dont except this!!!!!!

anyways here we go

2^((x+y)^2 - (x-y)^2)

2^(x^2 + 2xy + y^2 - x^2 + 2xy - y^2)

2^(4xy)

2^4*1 = 16..

hope this helps.. i am sure after u see the question in the right way ..its not at all difficult

What is different from the problem in GMATPrep and the way you wrote it? I think it is the corrected form of the problem.

Also, shouldn't that second 2xy be a -2xy in [2^(x^2 + 2xy + y^2 - x^2 + 2xy - y^2) ]? That way it is 2^ (2xy - (-2xy)) to become 2^(4xy)?

(x+y)^2 = x^2 + 2xy + y^2
(x-y)^2 = x^2 -2xy + y^2


Hope that is right...

uncbeers wrote:

sudhir3127 wrote:

medea66 wrote:How do you solve for this?

i am sure the question shud read like this ...

2^(x + Y)^2)/2^(x - Y)^2 ...

anyways here we go

2^((x+y)^2 - (x-y)^2)

2^(x^2 + 2xy + y^2 - x^2 + 2xy - y^2)

2^(4xy)

2^4*1 = 16..

hope this helps.. i am sure after u see the question in the right way ..its not at all difficult

What is different from the problem in GMATPrep and the way you wrote it? I think it is the corrected form of the problem.

Also, shouldn't that second 2xy be a -2xy in [2^(x^2 + 2xy + y^2 - x^2 + 2xy - y^2) ]? That way it is 2^ (2xy - (-2xy)) to become 2^(4xy)?

(x+y)^2 = x^2 + 2xy + y^2
(x-y)^2 = x^2 -2xy + y^2


Hope that is right...

ur right on 2xy -(-2xy) .. If u see i have already factored in that step and done the calculation...

So how do you get 2^4*1 = 16 from 2^(4xy)? Where does the 1 come from?

xy=1 comes from the question

Oh, duh, that was a stupid question on my part.