Let x=y=1.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
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.
PS Algrebra
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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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2^(x+y)² / 2^(x-y)² = 2^[(x+y)² - (x-y)²]dunnec3 wrote:2^(x+y)² / 2^(x-y)² = ? xy=1
Ans: 2, 4, 8, 16, 32
If you know the special quadratics you recognize the following.
(x+y)² = x² + 2xy + y²
(x-y)² = x² - 2xy + y²
Subtract the second from the first. The squared terms cancel out, and you are left with the following.
(x+y)² - (x-y)² = 4xy
So 2^[(x+y)² - (x-y)²] = 2�ˣʸ.
4xy = 4 * 1 = 4
2�ˣʸ = 2� = 16
The correct answer is 16.
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Brent@GMATPrepNow
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I should mention that Mitch's technique works for ANY pair of values for x and y where xy = 1If 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
Of course, x = 1 and y = 1 are the easiest to work with, but the technique will for other values as well.
For example, let's try x = 2 and y = 1/2.
We get: 2^(x+y)² / 2^(x-y)² = 2^(2 + 1/2)² / 2^(2 - 1/2)²
= 2^(5/2)² / 2^(3/2)²
= 2^(25/4) / 2^(9/4)
= 2^(25/4 - 9/4)
= 2^(16/4)
= 2^4
= 16
Some related resources:
- Laws of exponents - part I: https://www.gmatprepnow.com/module/gmat ... video/1025
- Laws of exponents - part II: https://www.gmatprepnow.com/module/gmat ... video/1029
Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
