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Thanks for the response. I totally understand the key logic of this problem. My method: \(\frac{2^{(x+y)^2}}{2^{(x-y)^2}}=2^{(x+y)^2-(x-y)^2}=2^{(x^2+2xy+y^2) - (x^2-2xy + y^2)} =2^{(x^2+2xy+y^2 - x^2+ 2xy - y^2)}=2^{(4xy)}=2^4=16\)
Your method looks more simple and I would like to understand it. I just got a little lost during your factoring-out transition from \(2^{(x+y)^2-(x-y)^2}=2^{(x+y+x-y)(x+y-x+y)}\) . Can u please explain. Thanks!
Thanks for the response. I totally understand the key logic of this problem. My method: \(\frac{2^{(x+y)^2}}{2^{(x-y)^2}}=2^{(x+y)^2-(x-y)^2}=2^{(x^2+2xy+y^2) - (x^2-2xy + y^2)} =2^{(x^2+2xy+y^2 - x^2+ 2xy - y^2)}=2^{(4xy)}=2^4=16\)
Your method looks more simple and I would like to understand it. I just got a little lost during your factoring-out transition from \(2^{(x+y)^2-(x-y)^2}=2^{(x+y+x-y)(x+y-x+y)}\) . Can u please explain. Thanks!
\(a^2-b^2=(a+b)*(a-b)\), so \((x+y)^2-(x-y)^2=((x+y)+(x-y))*((x+y)-(x-y))=2x*2y=4xy\).
Re: If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2?
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04 Mar 2015, 08:36
\(2^{(x+y)^2}/2^{(x-y)^2}=2^{(x+y)^2-(x-y)^2}\)
This part is very unclear to me. I do understand the basics of negative square is just "one over", but I dont understand this. The math compendium barely touches on this either. Can someone explain this please?
Re: If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2?
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04 Mar 2015, 08:50
erikvm wrote:
\(2^{(x+y)^2}/2^{(x-y)^2}=2^{(x+y)^2-(x-y)^2}\)
This part is very unclear to me. I do understand the basics of negative square is just "one over", but I dont understand this. The math compendium barely touches on this either. Can someone explain this please?
Re: If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2?
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07 Mar 2015, 16:44
6
Hi All,
Most Quant questions can be approached in a variety of ways, so it's useful to practice more than one method during your studies. In this question, it appears that all of the posters took the same Algebraic approach (which is fine), but was that approach really the fastest and easiest way to get to the solution.....?
Watch what happens when we TEST VALUES....
We're told that XY = 1. Since the answer choices are all numbers, one of them MUST be the solution to the equation, so I should be able to use ANY combination of X and Y that I choose (as long as the product of those values = 1).
Let's try... X = 1 Y = 1
The question then becomes...what is the value of (2^4)/(2^0)?
Re: If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2?
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06 Nov 2018, 12:00
Hi aaliyahkhalifa,
YES - your approach (TESTing VALUES) works perfectly on this question; compared with some of the longer 'math-heavy' approaches, it's considerably easier and faster. That's something to keep in mind as you continue to study for the GMAT. Most questions can be approached in more than one way - and it's possible that "your way" of dealing with a question might not actually be the most efficient option. Learning multiple approaches/Tactics can make the overall GMAT a lot easier to deal with and is essential to maximizing your performance on Test Day.
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71% (00:57) correct 
