$$If\ \frac{2^{\left(x+y\right)^2}}{2^{\left(x-y\right)^2}}=2,$$ 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.