PS Algrebra

[dunnec3]

Can anyone help solving this please?

[2^(x+y)^2 ]/ [2^(x-y)^2], xy=1

Ans: 2, 4, 8, 16, 32

[GMATGuruNY]

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

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.

[MartyMurray]

2^(x+y)² / 2^(x-y)² = ? xy=1

Ans: 2, 4, 8, 16, 32

2^(x+y)² / 2^(x-y)² = 2^[(x+y)² - (x-y)²]

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.

[Brent@GMATPrepNow]

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

I should mention that Mitch's technique works for ANY pair of values for x and y where xy = 1
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