Math question from the GMATprep

Question

If $xy=1$ what is the value of: $2^{(x+y)^2}/2^{(x-y)^2}$

A 1
B 2
C 4
D 16
E 19

$(x+y)^2/(x-y)^2$ because $2$ just cancels out from numerator and denominator, right?

Answer 1

This isn't as hard as you think.

xy = 1*1 or -1*-1

If xy = 1*1 then:

2^(x+y)^2/2^(x-y)^2 = (2^(2)^2)/(2^0^2) = 16/1 = 16

If xy = -1 *-1 then:

2^(x+y)^2/2^(x-y)^2 = (2^(-2)^2)/2^(-1 - (-1) )^2 = 16/1 = 16.

So, D is the answer. Let me know if you're confused

Answer 2

Using exponentiation rules, it can be simplified.

$$ \frac{2^{(x+y)^2}}{2^{(x-y)^2}} = 2^{(x+y)^² - (x-y)^2} = 2^{x^2 + 2xy + y^2 - (x^2 - 2xy + y^2)} = 2^{4xy} = \left(2^{4}\right)^{xy} = 16 $$