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:

Rewrite as \(32^{\frac{1}{x}}=2^{-1}\); Since \(32=2^5\), then we'd have that \(2^{\frac{5}{x}}=2^{-1}\); Equate the powers of 2 on both sides: \(\frac{5}{x}=-1\). Solving gives \(x=-5\).

Now, if \(y\) is even, then \(x^y=(-5)^{even}>0\) but if \(y\) is not even, say if it's 1, then \(x^y=(-5)^1<0\). Not sufficient.

(2) \(\sqrt[-y]{16}=\frac{1}{2}\)

Rewrite as \(16^{\frac{1}{-y}}=2^{-1}\); Since \(16=2^4\), then we'd have that \(2^{\frac{4}{-y}}=2^{-1}\); Equate the powers of 2 on both sides: \(\frac{4}{-y}=-1\). Solving gives \(y=4\).

Since given that \(x\) is not zero (from \(xy\neq{0}\)), then \(x^y=x^{4}>0\). Sufficient.

Answer: B.

Try NEW Exponents and Roots PS question. _________________

Now from statement 2, we can find out the value of Y i.e. 4 No matter X is positive or negative, the value of x^y would always be greater than 0. Hence Statement B is sufficient to answer the question.

Statement 1 says that x is a positive number and x cannot be negative. Hence a positive number raised to any will always be positive. Hence sufficient.

Statement 2 says that y is negative. x^y sign still depends on sign of x or on whether y is even or not. Hence insufficient.

Rewrite as \(32^{\frac{1}{x}}=2^{-1}\); Since \(32=2^5\), then we'd have that \(2^{\frac{5}{x}}=2^{-1}\); Equate the powers of 2 on both sides: \(\frac{5}{x}=-1\). Solving gives \(x=-5\).

Now, if \(y\) is even, then \(x^y=(-5)^{even}>0\) but if \(y\) is not even, say if it's 1, then \(x^y=(-5)^1<0\). Not sufficient.

(2) \(\sqrt[-y]{16}=\frac{1}{2}\)

Rewrite as \(16^{\frac{1}{-y}}=2^{-1}\); Since \(16=2^4\), then we'd have that \(2^{\frac{4}{-y}}=2^{-1}\); Equate the powers of 2 on both sides: \(\frac{4}{-y}=-1\). Solving gives \(y=4\).

Since given that \(x\) is not zero (from \(xy\neq{0}\)), then \(x^y=x^{4}>0\). Sufficient.

Answer: B.

Kudos points given to correct solutions.

Try NEW Exponents and Roots PS question. _________________

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.
_________________

After days of waiting, sharing the tension with other applicants in forums, coming up with different theories about invites patterns, and, overall, refreshing my inbox every five minutes to...

I was totally freaking out. Apparently, most of the HBS invites were already sent and I didn’t get one. However, there are still some to come out on...

In early 2012, when I was working as a biomedical researcher at the National Institutes of Health , I decided that I wanted to get an MBA and make the...