|
Author |
Message |
|
TAGS:
|
|
|
Manager
Status: Current MBA Student
Joined: 19 Nov 2009
Posts: 89
Concentration: Finance, General Management
|
If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2?
[#permalink]
Show Tags
 02 Jan 2011, 15:33
Question Stats:
73% (01:14) correct 27% (01:46) wrong based on 1039 sessions
HideShow timer Statistics
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
Official Answer and Stats are available only to registered users. Register/ Login.
|
|
|
|
|
|
|
|
Math Expert
Joined: 02 Sep 2009
Posts: 64174
|
Re: If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2?
[#permalink]
Show Tags
02 Jan 2011, 15:41
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.
_________________
|
|
|
|
|
|
|
|
Retired Moderator
Joined: 20 Dec 2010
Posts: 1469
|
Re: if xy=1, what is the value of (2^(x+y)^2)/(2^(x+y)^2)?
[#permalink]
Show Tags
18 Jun 2011, 06:44
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"
|
|
|
|
|
|
|
|
Manager
Status: Current MBA Student
Joined: 19 Nov 2009
Posts: 89
Concentration: Finance, General Management
|
Re: If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2?
[#permalink]
Show Tags
02 Jan 2011, 16:12
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. 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: 64174
|
Re: If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2?
[#permalink]
Show Tags
02 Jan 2011, 16:20
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 \(\frac{2^{(x+y)^2}}{2^{(x-y)^2}}=2^{(x+y)^2-(x-y)^2}=2^{(x+y+x-y)(x+y-x+y)}=2^{(2x)(2y)}=2^{4xy}=2^4=16\). Answer: D. 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.
_________________
|
|
|
|
|
|
Manager
Joined: 26 Feb 2015
Posts: 108
|
Re: If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2?
[#permalink]
Show Tags
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: 64174
|
Re: If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2?
[#permalink]
Show Tags
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.htmlTips on Exponents: exponents-and-roots-on-the-gmat-tips-and-hints-174993.html
_________________
|
|
|
|
|
|
EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 16763
Location: United States (CA)
|
Re: If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2?
[#permalink]
Show Tags
07 Mar 2015, 16:44
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 Final Answer: GMAT assassins aren't born, they're made, Rich
_________________
Contact Rich at: Rich.C@empowergmat.com The Course Used By GMAT Club Moderators To Earn 750+ souvik101990 Score: 760 Q50 V42 ★★★★★ ENGRTOMBA2018 Score: 750 Q49 V44 ★★★★★
|
|
|
|
|
|
Intern
Joined: 26 Apr 2017
Posts: 5
Location: Spain
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: 64174
|
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
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: 2800
|
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
_________________
See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews
If you find one of my posts helpful, please take a moment to click on the "Kudos" button.
|
|
|
|
|
|
Intern
Joined: 10 Oct 2018
Posts: 10
|
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
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: 16763
Location: United States (CA)
|
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
_________________
Contact Rich at: Rich.C@empowergmat.comThe Course Used By GMAT Club Moderators To Earn 750+ souvik101990 Score: 760 Q50 V42 ★★★★★ ENGRTOMBA2018 Score: 750 Q49 V44 ★★★★★
|
|
|
|
|
|
Director
Affiliations: IIT Dhanbad
Joined: 13 Mar 2017
Posts: 717
Location: India
Concentration: General Management, Entrepreneurship
GPA: 3.8
WE: Engineering (Energy and Utilities)
|
If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2?
[#permalink]
Show Tags
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 \(\frac{2^{(x+y)^2}}{2^{(x-y)^2}}\) \(= 2^{(x+y)^2 - (x-y)^2}\) \(= 2^{(4xy)}\) \(= 2^{(4xy)}\) \(= 2^4\) \(= 16\)
|
|
|
|
|
|
CrackVerbal Representative
Joined: 12 May 2019
Posts: 68
|
Re: If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2?
[#permalink]
Show Tags
06 Jun 2019, 04:53
Formula to be used: \(x^a/x^b = x^{a-b}\) \(2^{(x+y)^2}\) / \(2^{(x-y)^2}\)= \(2^{(x+y)^2 - (x-y)^2}\) = \(2^{( x^2 + y^2 +2xy) - ( x^2 + y^2 -2xy) }\) = \(2^{(4xy)}\) = \(2^{(4*1)}\) = \(2^4\) = 16 The correct answer is D
_________________
|
|
|
|
|
|
Intern
Joined: 07 Nov 2019
Posts: 7
Location: Germany
GPA: 2.8
|
Re: If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2?
[#permalink]
Show Tags
12 Feb 2020, 03:56
EMPOWERgmatRichC wrote: 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 Final Answer: GMAT assassins aren't born, they're made, Rich I get the algebraic way to to do and I get number picking X=Y=1 to get 16 as well. What I don't get is why if we choose X=Y=(-1) we don't get 16.. Nor if we choose X=2 and Y=0.5. X=4 and Y=0.25.. ??? They all satisfy the original requirement of X*Y=1.. Can some one please enlighten me?
|
|
|
|
|
|
|
Re: If xy = 1, what is the value of 2^(x + y)^2/2^(x-y)^2?
[#permalink]
12 Feb 2020, 03:56
|
|
|
|
|
|
|
close
