Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Nov 2007:30 AM PST
-08:30 AM PST
Learn what truly sets the UC Riverside MBA apart and how it helps in your professional growth
Nov 2211:00 AM IST
-01:00 PM IST
Do RC/MSR passages scare you? e-GMAT is conducting a masterclass to help you learn – Learn effective reading strategies Tackle difficult RC & MSR with confidence Excel in timed test environment- Nov 23
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. 
Nov 2510:00 AM EST
-11:00 AM EST
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors.
28 Apr 2020, 07:17
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
40% (01:55) correct
60%
(02:37)
wrong
based on 10
sessions
History
Date
Time
Result
Not Attempted Yet
Hello GMAT enthusiasts.
I found this very interesting question testing your concept of exponents and roots. Pls give it a go.
6. If x>0 then what is the value of- \(y^x\)?
\((1) \frac{4^{(x+y)^2}}{4^{(x-y)^2}}=128^{xy}\)
\((2) x\neq{1}\) and \(x^y=1\)
I found this very interesting question testing your concept of exponents and roots. Pls give it a go.
6. If x>0 then what is the value of- \(y^x\)?
\((1) \frac{4^{(x+y)^2}}{4^{(x-y)^2}}=128^{xy}\)
\((2) x\neq{1}\) and \(x^y=1\)
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Data Sufficiency (DS) Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
28 Apr 2020, 11:44
Kudos
Bookmarks
Statement 1-
\((1) \frac{4^{(x+y)^2}}{4^{(x-y)^2}}=128^{xy}\)
\(4^{(x^2+y^2+2xy-x^2-y^2+2xy)} = 2^{7xy}\)
\(4^{4xy} = 2^{7xy}\)
\(2^{8xy}/2^{7xy} = 1\)
\(2^{xy}= 2^0\)
xy= 0
since x>0, y= 0
Sufficient
Statement 2-
\(x^y=1\)
3 cases possible
1. x= 1
2. x= -1 and y is even
3. y=0
Since x is positive and not equal to 1, we can reject first 2 cases.
We get y= 0; \(y^x= 0\)
Sufficient
\((1) \frac{4^{(x+y)^2}}{4^{(x-y)^2}}=128^{xy}\)
\(4^{(x^2+y^2+2xy-x^2-y^2+2xy)} = 2^{7xy}\)
\(4^{4xy} = 2^{7xy}\)
\(2^{8xy}/2^{7xy} = 1\)
\(2^{xy}= 2^0\)
xy= 0
since x>0, y= 0
Sufficient
Statement 2-
\(x^y=1\)
3 cases possible
1. x= 1
2. x= -1 and y is even
3. y=0
Since x is positive and not equal to 1, we can reject first 2 cases.
We get y= 0; \(y^x= 0\)
Sufficient
rsmalan
28 Apr 2020, 21:55
Kudos
Bookmarks
rsmalan
From prompt we know, x > 0 ---- (1)
Statement - 1: \(\frac{4^{(x+y)^2}}{4^{(x-y)^2}}=128^{xy}\)
If we take the denominator to the numerator, it becomes -
\({4^{(x+y)^2}} * {4^- {(x-y)^2}}=128^{xy}\)
Adding powers as bases are same;
\(4^{(x+y)^2 - (x-y)^2}=128^{xy}\)
Now we know, \(a^2 - b^2 = (a+b)(a-b)\). Hence using this in our equation -
\(4^{(x+y+x-y)(x+y-x+y)}=128^{xy}\)
\(4^{(2x)(2y)}=128^{xy}\)
\(4^{(4xy)}=128^{xy}\)
Now 4 can be written as \(2^2\) and 128 can be written as \(2^7\)
Hence our equation becomes -
\(2^{(8xy)}=2^{7xy}\)
Now as bases are same, we can write the equation as,
\(8xy = 7xy\\
=> xy = 0\)
Now from (1) we know \(x >0\)
Hence y will be zero and \(y^x\) = \(0^x\) = 0 [Anything raised to power zero is 0]
Hence Statement 1 is sufficient.
Statement - 2: \(x\neq{1}\) and \(x^y=1\)
\(x^y=1\)
This can only be true in two cases:
1) \(y = 0\) and x can be anynumber, OR
2) \(x = 1\) and y can be any number
Now from statement we know that \(x\neq{1}\)
Hence y will be zero and \(y^x\) = \(0^x\) = 0 [Anything raised to power zero is 0]
Hence Statement 2 is sufficient.
Answer is D
