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

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Manager
Status: Current MBA Student
Joined: 19 Nov 2009
Posts: 100
Concentration: Finance, General Management
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If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2?  [#permalink]

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02 Jan 2011, 15:33
4
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Difficulty:

15% (low)

Question Stats:

71% (00:57) correct 29% (01:09) wrong based on 799 sessions

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If xy = 1, what is the value of $$\frac{2^{(x+y)^2}}{2^{(x-y)^2}}$$ ?

A. 2
B. 4
C. 8
D. 16
E. 32
Math Expert
Joined: 02 Sep 2009
Posts: 50624
Re: If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2?  [#permalink]

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02 Jan 2011, 15:41
6
4
tonebeeze wrote:
If xy = 1, what is the value of $$\frac {2^{(x + y)^2}} {2^{(x - y)^2}$$ ?

A. 2
B. 4
C. 8
D. 16
E. 32

$$\frac{2^{(x+y)^2}}{2^{(x-y)^2}}=2^{(x+y)^2-(x-y)^2}=2^{(x+y+x-y)(x+y-x+y)}=2^{(2x)(2y)}=2^{4xy}=2^4=16$$.

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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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18 Jun 2011, 06:44
41
9
WishMasterUA wrote:
if xy=1, what is the value of (2^((x+y)^2))/(2^((x-y)^2))?
1) 2
2) 4
3) 8
4) 16
5) 32

$$\frac{2^{(x+y)^2}}{2^{(x+y)^2}}$$

$$=\frac{2^{(x^2+y^2+2xy)}}{2^{(x^2+y^2-2xy)}}$$ [:Note: $$(x+y)^2=x^2+y^2+2xy \hspace{3} & \hspace{3} (x-y)^2=x^2+y^2-2xy$$]

$$=2^{(x^2+y^2+2xy-(x^2+y^2-2xy))}$$ [:Note: $$\frac{x^m}{x^n}= x^{(m-n)}$$ ]

$$=2^{(x^2+y^2+2xy-x^2-y^2+2xy)}$$

$$=2^{(4xy)}$$

$$=2^{(4xy)}=2^4=16$$ [:Note: xy=1]

Ans: "D"
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Manager
Status: Current MBA Student
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Posts: 100
Concentration: Finance, General Management
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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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02 Jan 2011, 16:12
2
1
Bunuel wrote:
tonebeeze wrote:
If xy = 1, what is the value of $$\frac {2^{(x + y)^2}} {2^{(x - y)^2}$$ ?

a. 2

b. 4

c. 8

d. 16

e. 32

$$\frac{2^{(x+y)^2}}{2^{(x-y)^2}}=2^{(x+y)^2-(x-y)^2}=2^{(x+y+x-y)(x+y-x+y)}=2^{(2x)(2y)}=2^{4xy}=2^4=16$$.

Hey Bunuel,

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!
Math Expert
Joined: 02 Sep 2009
Posts: 50624
Re: If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2?  [#permalink]

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02 Jan 2011, 16:20
4
3
tonebeeze wrote:
Bunuel wrote:
tonebeeze wrote:
If xy = 1, what is the value of $$\frac {2^{(x + y)^2}} {2^{(x - y)^2}$$ ?

a. 2

b. 4

c. 8

d. 16

e. 32

$$\frac{2^{(x+y)^2}}{2^{(x-y)^2}}=2^{(x+y)^2-(x-y)^2}=2^{(x+y+x-y)(x+y-x+y)}=2^{(2x)(2y)}=2^{4xy}=2^4=16$$.

Hey Bunuel,

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

Hope it's clear.
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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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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?
Math Expert
Joined: 02 Sep 2009
Posts: 50624
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, 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?

$$\frac{a^n}{a^m}=a^{n-m}$$

Below might help:
Theory on Exponents: math-number-theory-88376.html
Tips on Exponents: exponents-and-roots-on-the-gmat-tips-and-hints-174993.html
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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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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)?

(2^4)/(2^0) =
16/1 =
16

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Special Offer: Save $75 + GMAT Club Tests Free Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/ *****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***** Intern Joined: 26 Apr 2017 Posts: 5 Location: Spain GMAT 1: 600 Q47 V25 WE: Consulting (Consulting) Re: If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2? [#permalink] ### Show Tags 26 Jul 2017, 06:21 Bunuel wrote: tonebeeze wrote: If xy = 1, what is the value of $$\frac {2^{(x + y)^2}} {2^{(x - y)^2}$$ ? A. 2 B. 4 C. 8 D. 16 E. 32 $$\frac{2^{(x+y)^2}}{2^{(x-y)^2}}=2^{(x+y)^2-(x-y)^2}=2^{(x+y+x-y)(x+y-x+y)}=2^{(2x)(2y)}=2^{4xy}=2^4=16$$. Answer: D. Hi Bunuel, Could you help me with the following question? There is a rule for exponents that: $$(n^a)^b= n^{a*b}$$ So If we have: $$\frac{2^{(x+y)^2}}{2^{(x-y)^2}}$$ Why woudn't it be the following? $$\frac{2^{2(x+y)}}{2^{2(x-y)}}=\frac{2^{2x+2y}}{2^{2x-2y}}=2^{2x-2x+2y+2y}=2^{4y}$$ I'm confused because if we have $$2^{x^2}$$ the value when $$x=3$$, is the value $$2^{3·2}=2^6$$ or $$2^{2^3}=2^8$$? Thank you in advance! Math Expert Joined: 02 Sep 2009 Posts: 50624 Re: If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2? [#permalink] ### Show Tags 26 Jul 2017, 06:31 1 guillemgc wrote: Bunuel wrote: tonebeeze wrote: If xy = 1, what is the value of $$\frac {2^{(x + y)^2}} {2^{(x - y)^2}$$ ? A. 2 B. 4 C. 8 D. 16 E. 32 $$\frac{2^{(x+y)^2}}{2^{(x-y)^2}}=2^{(x+y)^2-(x-y)^2}=2^{(x+y+x-y)(x+y-x+y)}=2^{(2x)(2y)}=2^{4xy}=2^4=16$$. Answer: D. Hi Bunuel, Could you help me with the following question? There is a rule for exponents that: $$(n^a)^b= n^{a*b}$$ So If we have: $$\frac{2^{(x+y)^2}}{2^{(x-y)^2}}$$ Why woudn't it be the following? $$\frac{2^{2(x+y)}}{2^{2(x-y)}}=\frac{2^{2x+2y}}{2^{2x-2y}}=2^{2x-2x+2y+2y}=2^{4y}$$ I'm confused because if we have $$2^{x^2}$$ the value when $$x=3$$, is the value $$2^{3·2}=2^6$$ or $$2^{2^3}=2^8$$? Thank you in advance! Because it's $$2^{(x+y)^2}$$ and not $$(2^{(x+y)})^2$$ $$(a^m)^n=a^{mn}$$ $$a^m^n=a^{(m^n)}$$ and not $$(a^m)^n$$ (if exponentiation is indicated by stacked symbols, the rule is to work from the top down). _________________ Target Test Prep Representative Status: Head GMAT Instructor Affiliations: Target Test Prep Joined: 04 Mar 2011 Posts: 2830 Re: If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2? [#permalink] ### Show Tags 09 Aug 2017, 12:27 tonebeeze wrote: If xy = 1, what is the value of $$\frac {2^{(x + y)^2}} {2^{(x - y)^2}$$ ? A. 2 B. 4 C. 8 D. 16 E. 32 We can simplify the given expression: [2^(x+y)^2]/[2^(x-y)^2] Expanding the exponents in both the numerator and the denominator, we have: [2^(x^2+y^2+2xy)]/[2^(x^2+y^2-2xy] We subtract the denominator’s exponent from the numerator’s exponent: 2^(x^2 + y^2 + 2xy - x^2 - y^2 + 2xy) 2^(2xy + 2xy) = 2^(4xy) Since xy = 1, 2^4xy = 2^4 = 16. Answer: D _________________ Jeffery Miller Head of GMAT Instruction GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions Intern Joined: 10 Oct 2018 Posts: 6 Re: If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2? [#permalink] ### Show Tags 05 Nov 2018, 23:58 1 HI! I got the right answer but with different approach... I assumed that xy=1 if |x|=1 and |y|=1; So if we substitute x as 1 or -1 and y as 1 or -1 both lead to 16. Can this be a correct solution? EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 12877 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2? [#permalink] ### Show Tags 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. GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
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If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2?  [#permalink]

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14 Nov 2018, 21:24
what the xxxx

my lord

i just took prep1 ...spent like 8 min on this question ..eventually i had to give up ,,

It was the very first question !! can you believe it ???? 8 min wasted on this one .......

what i saw is different from this lol

2(x+y)^2 / 2(x−y)^2

no wonder i couldnt get an answer ....my lord ..very disgusting

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

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14 Nov 2018, 22:12
tonebeeze wrote:
If xy = 1, what is the value of $$\frac{2^{(x+y)^2}}{2^{(x-y)^2}}$$ ?

A. 2
B. 4
C. 8
D. 16
E. 32

$$\frac{2^{(x+y)^2}}{2^{(x-y)^2}}$$
$$= 2^{(x+y)^2 - (x-y)^2}$$
$$= 2^{(4xy)}$$
$$= 2^{(4xy)}$$
$$= 2^4$$
$$= 16$$
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If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2? &nbs [#permalink] 14 Nov 2018, 22:12
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