Ian Stewart wrote:Or simpler still, if, as the answer choices indicate, there is an exact numerical answer here *whenever* xy = 1 (there's no answer choice which says 'cannot be determined'), we must get the right answer when we plug in *any* values for x and y that make xy = 1 true. So we can simply plug in x = 1 and y = 1 to get

[2^(1+1)^2] / 2^(1-1)^2 = 2^4/2^0 = 16

Or, algebraically, you might notice that, using our exponent rules:

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

what we have now in the exponent is a difference of squares:

= 2^[(x + y + x - y)(x + y - x + y)]
= 2^[(2x)(2y)]
= 2^(4xy)
= 2^4
= 16

Of course, the algebraic solutions above are very good as well - I'm just suggesting alternative ways to approach the problem.

Or simpler still, if, as the answer choices indicate, there is an exact numerical answer here *whenever* xy = 1 (there's no answer choice which says 'cannot be determined'), we must get the right answer when we plug in *any* values for x and y that make xy = 1 true. So we can simply plug in x = 1 and y = 1 to get