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

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

(1) From this, x = y^2, so by substitution then x^y = (y^2)^y or y^(2y), and y^x = y^(y^2). Comparing x^y to y^x can then be done by comparing y^(2y) to y^(y^2), or simply comparing the exponents 2y to y^2. If, for example, y = 2, then 2y = 4 and y^2 = 4, and then x^y would equal y^x. If, however, y = 3, then 2y = 6 and y^2 = 9, and so x^y would be less than y^x; NOT SUFFICIENT.

(2) It is known that y > 2, but no information about x is given; NOT SUFFICIENT.

If both (1) and (2) are taken together, then 2y is compared to y^2 (1) and from (2) it is known that y > 2, so 2y will always be less than y^2. Therefore, x^y < y^x.

Re: Exponents/inequalities problem from QR 2nd DS 121 [#permalink]

Show Tags

04 Aug 2010, 05:21

This is a tricky question. I think it relies on you misapplying the rule: \((x^a)^b = x^{ab}\). Is this only valid if a and b are constants?

Example: (1) \(x = y^2\);

\(x^y < y^x\) therefore, \((y^2)^y < y^{y^2}\). How do you simplify this? The guide shows to \(y^{2y} < y^{y^2}\). The left hand side makes sense to me.

Why would \(y^{y^2}\) not simplify to \(y^{2y}\) also? Plugging in numbers, it makes sense. I just want to understand the concept.

If x and y are nonzero integers, is \(x^y < y^x\)?

(1) \(x = y^2\) (2) \(y > 2\)

If x and y are nonzero integers, is \(x^y < y^x\)?

(1) \(x = y^2\) --> if \(x=y=1\), then \(x^y=1=y^x\), so the answer would be NO BUT if \(y=3\) and \(x=9\), then \(x^y=9^3<y^x=3^9\), so the answer would be YES. Not sufficient.

(2) \(y>2\). No info about \(x\), not sufficient.

(1)+(2) From (1) \(x = y^2\), thus the question becomes: is \((y^2)^y<y^{(y^2)}\)? --> is \(y^{2y}<y^{(y^2)}\)? Now, since from (2) \(y=integer>2\), then \(2y\) will always be less than \(y^2\), therefore \(y^{2y}\) will be less than \(y^{(y^2)}\). Sufficient.

Answer: C.

jpr200012 wrote:

This is a tricky question. I think it relies on you misapplying the rule: \((x^a)^b = x^{ab}\). Is this only valid if a and b are constants?

Example: (1) \(x = y^2\);

\(x^y < y^x\) therefore, \((y^2)^y < y^{y^2}\). How do you simplify this? The guide shows to \(y^{2y} < y^{y^2}\). The left hand side makes sense to me.

Why would \(y^{y^2}\) not simplify to \(y^{2y}\) also? Plugging in numbers, it makes sense. I just want to understand the concept.

If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus: \(a^m^n=a^{(m^n)}\) and not \((a^m)^n\), which on the other hand equals to \(a^{mn}\).

Re: Exponents/inequalities problem from QR 2nd DS 121 [#permalink]

Show Tags

05 Aug 2010, 01:25

Posting this msg here even though i sent a private msg to you-for the benefit of others here.

Hi Bunuel, apprecite ur wonderful explanation. I am having trouble in DS question where x & y are termed as non-zero integers.

What is the best way to analyze instances where x & y are are NEGATIVE integers. I see that u have not analyzed this possibility. is there a trick to be sure that this is not needed as u have solved in this case?

Posting this msg here even though i sent a private msg to you-for the benefit of others here.

Hi Bunuel, apprecite ur wonderful explanation. I am having trouble in DS question where x & y are termed as non-zero integers.

What is the best way to analyze instances where x & y are are NEGATIVE integers. I see that u have not analyzed this possibility. is there a trick to be sure that this is not needed as u have solved in this case?

Please enlighten. Thanks.

On DS questions when plugging numbers, goal is to prove that the statement is not sufficient. So we should try to get a YES answer with one chosen number(s) and a NO with another.

For statement (1) I got YES answer and then NO answer with positive numbers, so my goal to prove that this statement was not sufficient was reached, hence there was no need to try negative numbers.

x=y^2 tells us that x is positive, but it tells us nothing about y.

For example if y=1 then x=1. Therefore 1^1=1^1 and x^y=y^x If y=-2 then x=4. Therefore 4^-2<-2^4 Since we cannot get a definite yes or no from this statement, it is INSUFFICIENT

Statement 2: Insufficient y>2 This tells us nothing about x. If x=-1 and y=4, then -1^4>4^-1 If x=5 and y=3, then 5^3<3^5 Since we cannot get a definite yes or no from this statement, it is INSUFFICIENT

Putting both statements together we know that y>2 and x=y^2 If y=4, then x=16, then 16^4<4^16 (16^4 = 4^8). No matter which integers you choose x^y < y^x, so Statements 1 and 2 together are SUFFICIENT. The answer is C.
_________________

The Brain Dump - From Low GPA to Top MBA(Updated September 1, 2013) - A Few of My Favorite Things--> http://cheetarah1980.blogspot.com

if x and y are nonzero integers is, x^y < y^x? (1) x = y^2 (2) y > 2

x^y < y^x Stmt1: x=y^2 \(x^y ---> y^2^y= y^(2y)\) \(y^x ---> y^(y^2)\) Is y^(2y) < y^(y^2) ? Take log both side 2y logy < y^2logy ? Canceling log y Is 2y < y^2 ? Is 2<y ? i.e Is y>2? We don't know the value of y. Hence not sufficient.

Stmt2: y>2 Not sufficient.

Combining, from Stmt2: we know that y>2 . Hence Stmt1: Is y>2 can be answered taking Stmt1 and Stmt2 together.

OA C.
_________________

My dad once said to me: Son, nothing succeeds like success.

if x and y are nonzero integers is, x^y < y^x? (1) x = y^2 (2) y > 2

x^y < y^x Stmt1: x=y^2 \(x^y ---> y^2^y= y^(2y)\) \(y^x ---> y^(y^2)\) Is y^(2y) < y^(y^2) ? Take log both side 2y logy < y^2logy ? Canceling log y Is 2y < y^2 ? Is 2<y ? i.e Is y>2? We don't know the value of y. Hence not sufficient.

Stmt2: y>2 Not sufficient.

Combining, from Stmt2: we know that y>2 . Hence Stmt1: Is y>2 can be answered taking Stmt1 and Stmt2 together.

OA C.

Nice solve. However, GMAT Math does not require the knowledge of logarithms. Definitely can help on the test, but for people who haven't touched a logarithm since high school it's not necessary to relearn them to answer this question.
_________________

The Brain Dump - From Low GPA to Top MBA(Updated September 1, 2013) - A Few of My Favorite Things--> http://cheetarah1980.blogspot.com

************************ If y>0; y-2>0; Is y>2 OR y<0 ***********************

However, "y<0" actually doesn't hold true for x^y<y^x (for y=-1) *********************

The only point I am trying to make here is that solving through logarithm may give us undesired results. What if x^y or y^x is negative. Then, taking the logarithms would be wrong!!!

On another look, From stmt1: \(x=y^2\), so\(y=\sqrt{x}\). Since \(\sqrt{x}\)will always be a positive number, y will always be positive. Substituting, in \(x^y < y^x\), \(x^sqrt(x)\) < \(\sqrt{x} ^ x\) Now we can safely take log as both sides are positive. \(sqrt(x)logx < xlog sqrt(x)\) \(sqrt(x) logx < x/2 log x\) Is \(sqrt(x) < x/2\) ? Is \(2 < sqrt(x)\) ? Cannot be determined. Not sufficient.

Stmt2: y>2. Not sufficient.

Combining, from stmt1 \(y=\sqrt{x}\) From stmt2: y>2. i.e \(\sqrt{x}\) >2. Hence \(2 < sqrt(x)\) ? . Yes.

OA C.
_________________

My dad once said to me: Son, nothing succeeds like success.

On another look, From stmt1: \(x=y^2\), so\(y=\sqrt{x}\). Since \(\sqrt{x}\)will always be a positive number, y will always be positive. Substituting, in \(x^y < y^x\), \(x^sqrt(x)\) < \(\sqrt{x} ^ x\) Now we can safely take log as both sides are positive. \(sqrt(x)logx < xlog sqrt(x)\) \(sqrt(x) logx < x/2 log x\) Is \(sqrt(x) < x/2\) ? Is \(2 < sqrt(x)\) ? Cannot be determined. Not sufficient.

Stmt2: y>2. Not sufficient.

Combining, from stmt1 \(y=\sqrt{x}\) From stmt2: y>2. i.e \(\sqrt{x}\) >2. Hence \(2 < sqrt(x)\) ? . Yes.

OA C.

x=y^2, Just because x is positive doesn't mean y has to be positive. Even powers always yield a positive number, even if the base is negative. For example, if x=4 then y can be 2 or -2. The only thing we can determine from statement 1 is that x is positive. y can be either positive or negative. If y is negative then x^y would be 1/x^y. 1/x^y may or may not be greater than y^x. y=-3, x=9 vs y=3, x=9. 1/9^3>-3^9. 9^3<3^9.
_________________

The Brain Dump - From Low GPA to Top MBA(Updated September 1, 2013) - A Few of My Favorite Things--> http://cheetarah1980.blogspot.com

Re: If x and y are nonzero integers, is x^y < y^x? (1) x = [#permalink]

Show Tags

20 Jul 2014, 10:19

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Since my last post, I’ve got the interview decisions for the other two business schools I applied to: Denied by Wharton and Invited to Interview with Stanford. It all...

Marketing is one of those functions, that if done successfully, requires a little bit of everything. In other words, it is highly cross-functional and requires a lot of different...