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If 2^{(x+y)^2} / 2^{(x-y)^2} = 2, what is the value of xy?

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by swerve » Tue Feb 13, 2018 7:57 am
$$If\ \frac{2^{\left(x+y\right)^2}}{2^{\left(x-y\right)^2}}=2,$$ What is the value of xy?

(A) -1/4
(B) 1/2
(C) 0
(D) 1/4
(E) 1/2

The OA is D.

Would any expert please break down the steps to solve this PS question, step by step? I tried to solve it but I can't get the correct answer. I need your help. Thanks.
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by Brent@GMATPrepNow » Tue Feb 13, 2018 9:22 am
swerve wrote:$$If\ \frac{2^{\left(x+y\right)^2}}{2^{\left(x-y\right)^2}}=2,$$ What is the value of xy?

(A) -1/4
(B) 1/2
(C) 0
(D) 1/4
(E) 1/2
(x + y)² = x² + 2xy + y²
(x - y)² = x² - 2xy + y²

Given: 2^(x + y)²/2^(x - y)² = 2
Expand both exponents to get: 2^(x² + 2xy + y²)/2^(x² - 2xy + y²) = 2
Apply Quotient Law to get: 2^(4xy) = 2
In other words, 2^(4xy) = 2^1
This means that: 4xy = 1
Divide both sides by 4 to get: xy = 1/4

Answer: D

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by EconomistGMATTutor » Tue Feb 13, 2018 11:45 am
$$If\ \frac{2^{\left(x+y\right)^2}}{2^{\left(x-y\right)^2}}=2,$$ What is the value of xy?

(A) -1/4
(B) 1/2
(C) 0
(D) 1/4
(E) 1/2

The OA is D.

Would any expert please break down the steps to solve this PS question, step by step? I tried to solve it but I can't get the correct answer. I need your help. Thanks.
Hi swerve,
Let's take a look at your question.

$$\frac{2^{\left(x+y\right)^2}}{2^{\left(x-y\right)^2}}=2 ... (i)$$
Let's evaluate the square of binomials in the exponent of 2, using polynomial identities:
$$\left(x+y\right)^2=x^2+2xy+y^2$$
$$\left(x-y\right)^2=x^2-2xy+y^2$$
Eq(i) will become:
$$\frac{2^{\left(x^2+2xy+y^2\right)}}{2^{\left(x^2-2xy+y^2\right)}}=2$$
$$2^{\left(x^2+2xy+y^2\right)-\left(x^2-2xy+y^2\right)}=2$$
$$2^{x^2+2xy+y^2-x^2+2xy-y^2}=2$$
$$2^{2xy+2xy}=2$$
$$2^{4xy}=2$$
$$2^{4xy}=2^1$$
Since bases on both sides of the equation are the same, we can equate the exponents, hence,
$$4xy=1$$
$$xy=\frac{1}{4}$$
Therefore, Option D is correct.

Hope it helps.
I am available if you'd like any follow up.
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by [email protected] » Tue Feb 13, 2018 5:12 pm
Hi swerve,

Since both the numerator and denominator of the fraction are 'base 2', we ultimately need a fraction that simplifies to (2^1)/(2^0) = 2.

Algebraically, this means that (X+Y) has to be '1 greater' than (X-Y), so we can set up the following equation:

X + Y = X - Y + 1

Then combine like terms and simplify (the X's cancel out):

2Y = 1
Y = 1/2

Plugging that value into the numerator, we have:

X + 1/2 = 1
X = 1/2

Thus, (X)(Y) = (1/2)(1/2) = 1/4

Final Answer: D

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