## Practice Quant problem

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### Practice Quant problem

by myspecialtie » Mon Mar 25, 2019 11:17 am
I'm having difficulty figuring out the solution to this problem from one of the GMAT practice exams:

If xy=1, what is the value of {2^[(x+y)^2]}/{2^[(x-y)^2]}?

A. 2
B. 4
C. 8
D. 16
E. 32

Any help/advice would be much appreciated!

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by [email protected] » Mon Apr 22, 2019 11:00 pm
myspecialtie wrote:I'm having difficulty figuring out the solution to this problem from one of the GMAT practice exams:

If xy=1, what is the value of {2^[(x+y)^2]}/{2^[(x-y)^2]}?

A. 2
B. 4
C. 8
D. 16
E. 32

Any help/advice would be much appreciated!
We have {2^[(x+y)^2]}/{2^[(x-y)^2]}

{2^[(x+y)^2]}/{2^[(x-y)^2]} = 2^[{(x+y)^2} - {(x-y)^2}] = 2^[{x^2 + y^2 + 2xy} - {x^2 + y^2 - 2xy}] = 2^(4xy) = 2^(4*1) = 2^4 = 16

Alternate approach:

Since we can choose any value for x or y given that xy = 1, say x = y = 1

Thus, (x + y)^2 = (1 + 1)^2 = 2^2 = 4 and (x - y)^2 = (1 - 1)^2 = 0^2 = 0

{2^[(x+y)^2]}/{2^[(x-y)^2]} = {2^4}/{2^0} = 16/1 = 16 -- Correct answer is D.

Hope this helps!

-Jay
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by Ian Stewart » Mon Jun 03, 2019 6:38 am
2^a / 2^b is equal to 2^(a-b).

So 2^(x+y)^2 / 2^(x-y)^2 is equal to 2^[ (x+y)^2 - (x-y)^2 ]

We could expand both (x+y)^2 and (x-y)^2 and subtract, but it's faster just to use the difference of squares factorization immediately, since we're subtracting one square from another in the exponent. So our exponent is equal to

(x+y)^2 - (x-y)^2 = (x + y + x - y) ( x + y - x + y) = (2x)(2y) = 4xy

and since xy = 1, our exponent is equal to 4, so the overall value of the expression is 2^4 = 16.
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### Re: Practice Quant problem

by PhillipZhang87 » Fri Sep 18, 2020 7:03 am
For this question we have to know the following:
( x + y )^2 = x^2 + 2xy + y^2
( x - y )^2 = x^2 - 2xy + y^2
2^x / 2^x - 2 = 2^(x-(x-2)) = 2^2
Moving to the problem
2^(x^2 + 2xy + y^2) / 2^(x^2 - 2xy + y^2) = 2^4xy
We know xy = 1
So, 2^4(1) = 2^4 = 16.