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Fraction Question (GMAT Prep)

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Junior | Next Rank: 30 Posts
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by oex85 » Mon Aug 08, 2011 6:01 am
Tried searching, but honestly don't know how to quickly find what I'm looking for.

Question: If xy=1, what is the value of 2^(x+y)^2 )/2^(x-y)^2 ?

Alternatives: 2, 4, 8, 16, 32.


Also, any input on how to get the equation better adapted to this site? Used Microsoft Word 2010.

Cheers
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by Anurag@Gurome » Mon Aug 08, 2011 6:13 am
oex85 wrote:If xy=1, what is the value of 2^(〖(x+y)〗^2 )/2^(〖(x-y)〗^2 ) ?
Correct me if I'm wrong, I guess the expression is [2^(x + y)²]/[2^(x - y)²]...

Now, the expression [2^(x + y)²]/[2^(x - y)²] is equivalent to 2^[(x + y)²] - (x - y)²].

Hence, [2^(x + y)²]/[2^(x - y)²] = 2^[(x + y)²] - (x - y)²] = 2^[(x² + 2xy + y²) - (x² - 2xy + y²)] = 2^[4xy] = 2^4 = 16

The correct answer is D.

PS : You can easily post the equation in correct form by using brackets and parentheses properly. In this case you could've done that by posting the expression as [2^{(x + y)^2}]/[2^{(x - y)^2}]
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by gmatboost » Mon Aug 08, 2011 7:38 am
Anurag@Gurome wrote:Now, the expression [2^(x + y)²]/[2^(x - y)²] is equivalent to 2^[(x + y)²] - (x - y)²].
Just wanted to explain this step in more detail, since it is the key to the question.

The main exponent rules to know are:
(a^m)*(a^n) =(a^[m+n])
(a^m)/(a^n) =(a^[m-n])
(a^m)^n =(a^[mn])

The quoted step above is using the second rule.
a = 2
m = (x + y)²
n = (x - y)²
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by oex85 » Mon Aug 08, 2011 9:43 am
Thanks guys. You make a great team :)

Totally get it now, but definitely need some practice in this area.
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