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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 \((x+y)^2-(x-y)^2=((x+y)+(x-y))*((x+y)-(x-y))=2x*2y=4xy\).

2^(2.5)^2/2^(1.5)^2 I keep getting 4 as answer for this; the answer is 16 can someone explain what I am missing original question DS x*y=1 2^(x+Y)^2/2^(x-y)^2 THanks

You were doing this (highlighted in Red) which is wrong 2^(2.5)^2/2^(1.5)^2 2^5/ 2^3 2^(5-3) 2^(2) 4

Correct solution is this. 2^(2.5)^2/2^(1.5)^2 2^6.25/ 2^2.25 2^(6.25-2.25) 2^(4) 16

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Re: If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2? [#permalink]

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27 Feb 2014, 21:43

Can some explain why (x-y)^2 is not x^2-y^2 which then translates into (x+y)(x-y)? When I put it as (x-y)(x-y) I can see how it translates out to x^2+y^2-2xy but I just don't understand why it doesn't originally translate to the squared form when you take the exponent and then give it to each variable in the paranthesis.

Re: If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2? [#permalink]

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27 Feb 2014, 22:34

amjet12 wrote:

Can some explain why (x-y)^2 is not x^2-y^2 which then translates into (x+y)(x-y)? When I put it as (x-y)(x-y) I can see how it translates out to x^2+y^2-2xy but I just don't understand why it doesn't originally translate to the squared form when you take the exponent and then give it to each variable in the paranthesis.

Thanks for the help!

Hi amjet12, I think this will be better understood by an example than the theory behind it

(5 - 2)^2 = (3)^2 = 9

According to BODMAS (also refereed as PEMDAS) , you solve first anything in the brackets/parenthesis and proceed with exponents. Hence, subtraction and then exponents. (PEMDAS = Paranthesis, Exponents, Multiplication, Divisibility, Addition, Subtraction in the same order). Try to search more about it if you are not aware of it.

5^2 - 2^2 = 25 -4 (Note we don't have exponents out of paranthesis in fact they are inside. so order will be exponents, then subtraction = 21

Now, the expansion (x-y)^2 = x^2 - 2xy +y^2 and x^2-y^2 = (x-y)(x+y) are just algebraic rules which all the numbers have to follow. You can check that with above example.

Re: If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2? [#permalink]

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04 Mar 2015, 09:36

\(2^{(x+y)^2}/2^{(x-y)^2}=2^{(x+y)^2-(x-y)^2}\)

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?

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?

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)?

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

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30 May 2016, 03:22

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Although I'm pretty much familiar with exponential rules, and got this question right in the Gmatprep, I've a doubt.

The rules of exponentials tell us that whenever we have \((n^x)^y\) the result is n elevated at the x*y, so why aren't we multiplying (x+y) and (x-y) for 2?