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Source: — Problem Solving |

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by Brent@GMATPrepNow » Mon Dec 02, 2013 1:58 pm
parulmahajan89 wrote:if XY=1 then what is the value of 2^(X+Y)²/ 2^(X-Y)²

1)2
2) 4
3) 8
4) 16
5) 32

NOT sure what I did wrong here I think answer is 8. But correct answer is 16?
Some additional brackets would help understand the original question. Plus I think you added an extra 2.
Based on the official answer, I've edited the question above.

2^(X+Y)²/ 2^(X-Y)² = 2^(X² + 2XY + Y²)/ 2^(X² - 2XY + Y²)
= 2^(4XY) [after we subtract exponents]
= 2^4 [since XY = 1]
= 16

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by GMATGuruNY » Mon Dec 02, 2013 10:05 pm
parulmahajan89 wrote:if XY=1 then what is the value of 2^(X+Y)^2/ 2^(X-Y)2^2

1)2
2) 4
3) 8
4) 16
5) 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 sanju09 » Mon Dec 02, 2013 10:57 pm
parulmahajan89 wrote:if XY=1 then what is the value of 2^(X+Y)^2/ 2^(X-Y)2^2

1)2
2) 4
3) 8
4) 16
5) 32

NOT sure what I did wrong here I think answer is 8. But correct answer is 16?
If xy = 1, then (x + y)^2 = x^2 + y^2 + 2, and (x - y)^2 = x^2 + y^2 - 2. Using Rules of Exponents, we can have

2^(x + y)^2/ 2^(x - y)^2 = 2^{( x^2 + y^2 + 2) - (x^2 + y^2 - 2)} = [spoiler]2^4 = 16; and this is no Arithmetic.[/spoiler].
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by Mathsbuddy » Tue Dec 03, 2013 10:06 am
At first I thought there was a typo in the question. So I solved it with a "2" removed.

Try this:

Firstly (2^a)/(2^b) = 2^(a - b)

So 2^(X+Y)^2/ 2^(X-Y)^2 = 2^((X+Y)^2 - (X-Y)^2) = 2^C

Let's look at C = (X+Y)^2 - (X-Y)^2

If we expand it we get:

C = (X^2 + 2XY + Y^2) - (X^2 - Y^2) = 2XY = 2 (because XY = 1 is given us in the question)

So the answer is 2^C = 2^2 = 4

If however there isn't a typo in the question, we need to square our answer (because we replace the power of 2 that I removed) to get 16.
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by bnpetteway » Tue Dec 03, 2013 7:32 pm
I came up with 16 and my way was close to how Mitch did his.
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by Mathsbuddy » Wed Dec 04, 2013 9:24 am
GMATGuruNY wrote:
parulmahajan89 wrote:if XY=1 then what is the value of 2^(X+Y)^2/ 2^(X-Y)2^2

1)2
2) 4
3) 8
4) 16
5) 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.
A nice and sweet solution in its simplicity. Thanks.
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