If xy=1, what is the value of 2(x+y)^2/2(x-y)^2?

a. 2
b. 4
c. 8
d. 16 <---Correct
e. 32

Thanks for your help!

I think the equation you have is not correct. Can you recheck and confirm.

wait what?

something is seriously wrong here, as elementary number picking is quick to show.

first off, the 2's clearly cancel, so get rid of them.

if we let x = 2 and y = 1/2, then the expression is (5/2)^2 / (3/2)^2, which is 25/9 = 2.77777..., an answer that is clearly not listed. worse yet, plugging in other numbers leads to other answer choices.

incidentally, the only numbers that satisfy the initial constraint (xy = 1) and actually give the answer 16 are plus or minus (√15)/5 and (√15)/3 - two numbers that, i can confidently say, no one is going to pick anytime soon.

--

the worst is that this is not a typo, as proved by the screen shot posted on this other forum. this is the first i've seen of an official problem just being downright wrong.

in my opinion Qs is 2^(x+y)^2/2^(x-y)^2
then the ans would be 16.
so Q may be misprint otherwise by plug in we knew that Qs is misprint

you are correct; the problem is supposed to involve exponentials with 2 as the base.

the problem did, for a time, appear on gmatprep this way, though. someone apparently forgot to code for the exponentials.

So can someone please explain how to solve correctly? Thanks!

yeah. but first, a prelude.

-- PRELUDE: dance of the perfect square trinomials --

you must know the following expansions
(x + y)^2 = x^2 + 2xy + y^2
(x - y)^2 = x^2 - 2xy + y^2
.. but just as importantly, you should know how they combine and cancel when added or subtracted.
specifically:
if you ADD them, you get 2(x^2 + y^2), because the two 2xy terms cancel
if you SUBTRACT them, you get 4xy, because the squared terms cancel

this is important knowledge. not only can it shave lots of time off problems like this one, but it can allow you to see right through problems that would otherwise confound you.

-- WALTZ of the exponents --

ok, so when you have a fraction with two exponents (and the same base for both of those exponents), you have to subtract the exponents.
as established in the prelude above, when you subtract these two particular exponents, which are the perfect square trinomials, the difference is 4xy.
so this fraction becomes: 2^(4xy)

since the problem gives you that xy = 1 (note that this criterion still holds, although it hasn't been mentioned explicitly by either of the last 2 posters), this reduces to 2^4, or 16.

Ron,

I know how to solve the below problem but was interested in knowing how did you manage to get :


" incidentally, the only numbers that satisfy the initial constraint (xy = 1) and actually give the answer 16 are plus or minus (√15)/5 and (√15)/3 - two numbers that, i can confidently say, no one is going to pick anytime soon"

Did you actually substitute the values for xy=1 as x=1/y and solve it or there is some logical way to narrow down to these roots ? Thanks and you are the best !!!

Hey Ron,

Is the below question too dumb to be asked ? If yes then sorry, but I could not figure this out..

good old fashioned substitution, my friend.

It was my understanding that if an exponent is raised to another exponent, you multiply the two (e.g. (2^2)^2 = 2^4.

the above posting, however, has us actually square (x+y) and (x-y), how come?

if you see a^b^c written with no parentheses (i.e., normal size a, then smaller superscript b, then even smaller super-superscript c), then that is evaluated as a^(b^c). this is a standard mathematical rule; it is not something that is unique to the gmat.

Can we just use x=1 and y=1 and plug in for x+y and x-y? This results in 16 and I just want to make sure I'm not making a mistake.

that is a very effective way to solve the problem.

Since the question doesn't state that x and y are integer, it is better to avoid substitution. We may not know what other possibilities are possible. The question is pretty simple.

[2^(x+y)^2]/[2^(x-y)^2] = 2^([x+y]^2 - [x-y]^2) = 2^4xy = 2^4*1 (xy=1 from question) = 2^4 = 16