Easiest approach to solve this problem is to note that the answer choices are independent of x and y. So, we can pick any value for x and y such that xy = 1.

Obvious numbers to pick are x = 1 and y = 1

So, (2^(x + y)^2)/(2^(x - y)^2) = (2^(1 + 1)^2)/(2^(1 - 1)^2) = (2^(2^2))/(2^(0^2)) = (2^4)/(2^0) = 16/1 = 16

Answer : D

Thank you. I like both approaches. I tend to choose the algebriac way over the pick numbers way just because by doing that I get better understanding of the concept. But the pick numbers makes a lot of sense too.

One question though.

(2^(x + y)^2)/(2^(x - y)^2) = (2^(1 + 1)^2)/(2^(1 - 1)^2) = ((2^2)^2)/((2^0)^2) = (2^4)/(1^2) = 2^4 = 16

As far as order of operations goes, wouldn't this be (2^(2^2))/(2^(0^2))? For this example,I guess it doesn't matter; either way you get 16. But wouldn't you square the exponent before you square the base?

Sorry about the yellow. I didn't realize how bright that would be.

Rudy414 wrote:Thank you. I like both approaches. I tend to choose the algebriac way over the pick numbers way just because by doing that I get better understanding of the concept. But the pick numbers makes a lot of sense too.

Yes, I also prefer algebraic approach to always sticking with plugging numbers. But that doesn't mean I always stick with Algebra. Plugging numbers and other non-conventional methods has their own place. For example, this one.

Rudy414 wrote:As far as order of operations goes, wouldn't this be (2^(2^2))/(2^(0^2))? For this example,I guess it doesn't matter; either way you get 16. But wouldn't you square the exponent before you square the base?

Yes. My mistake.
Editing it right away.

PS : Although the way you have posted the problem "(2^(x+y)^2)" can be interpreted as either [(2^(x+y))^2] or [2^((x+y)^2)]