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Could have sworn I got this right

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by NeilWatson » Sun Mar 30, 2014 4:47 pm
If xy=1, what is the value of 2^(x+y)^2/2^(x-y)^2

My Explanation:
Since they have the same base, then it would simplify to (x+y)^2-(x-y)^2

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

2+2 = 4.

For some reason this is wrong.
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by GMATGuruNY » Sun Mar 30, 2014 8:11 pm
If xy=1, then what is the value of 2^(x+y)² / 2^(x-y)²?

2
4
8
16
32
Let x=y=1.
Then:
2^(x+y)² / 2^(x-y)² = 2^(1+1)² / 2^(1-1)² = 2�/2� = 16/1 = 16.

The correct answer is D.
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by [email protected] » Sun Mar 30, 2014 10:15 pm
Hi NeilWatson,

Your thinking was almost complete, but you forgot what you were solving for.

You correctly determined that the "top exponent" divided by the "bottom exponent" = 4

You forgot that the base was 2...

The correct answer asks for the value of 2^4....

2^4 = 16

Final Answer: D

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