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^[(xy)^2]}?
A. 2
B. 4
C. 8
D. 16
E. 32
Any help/advice would be much appreciated!
Practice Quant problem
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We have {2^[(x+y)^2]}/{2^[(xy)^2]}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^[(xy)^2]}?
A. 2
B. 4
C. 8
D. 16
E. 32
Any help/advice would be much appreciated!
{2^[(x+y)^2]}/{2^[(xy)^2]} = 2^[{(x+y)^2}  {(xy)^2}] = 2^[{x^2 + y^2 + 2xy}  {x^2 + y^2  2xy}] = 2^(4xy) = 2^(4*1) = 2^4 = 16
The correct answer: D
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^[(xy)^2]} = {2^4}/{2^0} = 16/1 = 16  Correct answer is D.
Hope this helps!
Jay
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2^a / 2^b is equal to 2^(ab).
So 2^(x+y)^2 / 2^(xy)^2 is equal to 2^[ (x+y)^2  (xy)^2 ]
We could expand both (x+y)^2 and (xy)^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  (xy)^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.
So 2^(x+y)^2 / 2^(xy)^2 is equal to 2^[ (x+y)^2  (xy)^2 ]
We could expand both (x+y)^2 and (xy)^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  (xy)^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.
If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

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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(x2)) = 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.
Correct answer is D.
( x + y )^2 = x^2 + 2xy + y^2
( x  y )^2 = x^2  2xy + y^2
2^x / 2^x  2 = 2^(x(x2)) = 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.
Correct answer is D.