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To be more fast I think we can do this Since XY = 1 the answer for the given question must be same for any values of X and Y that satisfy XY = 1

So take X=1 and Y=1. The given value becomes 2^4 = 16 And I think the problem is killed.

What do u say?

Regards,

that's a bit dangerous, no? x=-2 and y=-1/2 could possibly give us one of the answers if the test makes wanted to catch this

Friend it's not at all dangerous.
For any given value of X and Y that satisfy XY=1, the given expression always gives only one answer. Try to understand this.........

To be more fast I think we can do this Since XY = 1 the answer for the given question must be same for any values of X and Y that satisfy XY = 1

So take X=1 and Y=1. The given value becomes 2^4 = 16 And I think the problem is killed.

What do u say?

Regards,

that's a bit dangerous, no? x=-2 and y=-1/2 could possibly give us one of the answers if the test makes wanted to catch this

Friend it's not at all dangerous. For any given value of X and Y that satisfy XY=1, the given expression always gives only one answer. Try to understand this.........

it just so happens that in the equation presented, x=1 and y=1 works. I don't agree that it will always work.....what if you had a term like x/y?
by your logic, x/y would be reduced to 1, which is clearly overstepping the given data.

To be more fast I think we can do this Since XY = 1 the answer for the given question must be same for any values of X and Y that satisfy XY = 1

So take X=1 and Y=1. The given value becomes 2^4 = 16 And I think the problem is killed.

What do u say?

Regards,

Very good point! By the wording of the question, any values of x and y whose product is 1 will yield the same value of 2^.../2^... Here's a great question for picking numbers, though the algebraic approach is quite manageable.

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

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