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?
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parulmahajan89
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Some additional brackets would help understand the original question. Plus I think you added an extra 2.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?
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
Cheers,
Brent
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Let x=y=1.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
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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sanju09
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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 haveparulmahajan89 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?
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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Mathsbuddy
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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.
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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bnpetteway
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Mathsbuddy
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A nice and sweet solution in its simplicity. Thanks.GMATGuruNY wrote:Let x=y=1.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
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.
