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:

a=√3(√3(√3(√3(√3…….inf.)))) sq. both sides a^2=3(√3(√3(√3(√3(√3…….inf.))))) or a^2=3a ; since, a=√3(√3(√3(√3(√3…….inf.)))) a^2-3a=0 a(a-3)=0 a=0/3 0 logically doesn't fit so 3 is the ans

Interesting question. Mods, if the question needs reformatting please do so, as i was not able to properly use the sqrt symbol, given in the editor. Added a pic, pls pardon my drawing

a=√3(√3(√3(√3(√3…….inf.)))) sq. both sides a^2=3(√3(√3(√3(√3(√3…….inf.))))) or a^2=3a ; since, a=√3(√3(√3(√3(√3…….inf.)))) a^2-3a=0 a(a-3)=0 a=0/3 0 logically doesn't fit so 3 is the ans

Interesting question. Mods, if the question needs reformatting please do so, as i was not able to properly use the sqrt symbol, given in the editor.

a=√3(√3(√3(√3(√3…….inf.)))) sq. both sides a^2=3(√3(√3(√3(√3(√3…….inf.))))) or a^2=3a ; since, a=√3(√3(√3(√3(√3…….inf.)))) a^2-3a=0 a(a-3)=0 a=0/3 0 logically doesn't fit so 3 is the ans

Interesting question. Mods, if the question needs reformatting please do so, as i was not able to properly use the sqrt symbol, given in the editor. Added a pic, pls pardon my drawing

For those who are confused by formatting:

If \(a=\sqrt{3\sqrt{3\sqrt{3\sqrt{3...}}}}\), what is the value of a?

A. 0 B. \(\sqrt{3}\) C. 3 D. 2.9 E. cannot be determined.

\(a=\sqrt{3\sqrt{3\sqrt{3\sqrt{3...}}}}\) --> \(a=\sqrt{3(\sqrt{3\sqrt{3\sqrt{3...}}})}\). Now, as the expression under the square root extends infinitely, then expression in brackets would equal to \(a\) itself, so we can safely replace it with \(a\) and rewrite the given expression as \(a=\sqrt{3a}\).

Square it: \(a^2=3a\) --> \(a=0\) or \(a=3\) --> since \(a=0\) does not satisfy given expression, then we have only one solution: \(a=3\).

Re: Find the value of a. Given a=√3(√3(√3(√3(√3…….inf.)))) [#permalink]

Show Tags

30 Aug 2012, 02:40

Here a=0 or a=3, both are possible. Don't know which one to pick A or C. Should we take a as +ve by default?
_________________

Regards SD ----------------------------- Press Kudos if you like my post. Debrief 610-540-580-710(Long Journey): http://gmatclub.com/forum/from-600-540-580-710-finally-achieved-in-4th-attempt-142456.html

Re: Find the value of a. Given a=√3(√3(√3(√3(√3…….inf.)))) [#permalink]

Show Tags

20 Aug 2014, 11:39

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.
_________________

a=√3(√3(√3(√3(√3…….inf.)))) sq. both sides a^2=3(√3(√3(√3(√3(√3…….inf.))))) or a^2=3a ; since, a=√3(√3(√3(√3(√3…….inf.)))) a^2-3a=0 a(a-3)=0 a=0/3 0 logically doesn't fit so 3 is the ans

Interesting question. Mods, if the question needs reformatting please do so, as i was not able to properly use the sqrt symbol, given in the editor. Added a pic, pls pardon my drawing

For those who are confused by formatting:

If \(a=\sqrt{3\sqrt{3\sqrt{3\sqrt{3...}}}}\), what is the value of a?

A. 0 B. \(\sqrt{3}\) C. 3 D. 2.9 E. cannot be determined.

\(a=\sqrt{3\sqrt{3\sqrt{3\sqrt{3...}}}}\) --> \(a=\sqrt{3(\sqrt{3\sqrt{3\sqrt{3...}}})}\). Now, as the expression under the square root extends infinitely, then expression in brackets would equal to \(a\) itself, so we can safely replace it with \(a\) and rewrite the given expression as \(a=\sqrt{3a}\).

Square it: \(a^2=3a\) --> \(a=0\) or \(a=3\) --> since \(a=0\) does not satisfy given expression, then we have only one solution: \(a=3\).

Answer: C.

Another genius explanation from Bunuel, Kudos to you

I had in mind that \(\sqrt{3} = 1.732\) To square root it multiple times will give the value 1.00000000000xxxxxxxxxxx & so on

Now that 1 is not the option in the OA, nearest to it was 0, which I selected & went wrong

Bunuel, can you kindly explain

I opened the scientific calculator & square rooted 3 multiple times, which stuck at 1

Attachment:

calc.png [ 17.3 KiB | Viewed 1314 times ]

Now had 1 been in the OA, which would had been the answer?

(I agree that answer = 3, however the result 1 from calculator is contradicting)
_________________

For n (Infinite) number of occurrences, the equation would be built up as follows

\(a = 3^{\frac{n}{n+1}}\)

We may deem here \(n\approx{n+1}\)

\(a = 3^1 = 3\)

Answer = C

What is find here is that, if you have a calculator in mind for value of \(\sqrt{3},\)then we tend to avoid the algebraic approach (shown above) & end up wrong
_________________

Re: Find the value of a. Given a=√3(√3(√3(√3(√3…….inf.)))) [#permalink]

Show Tags

26 Aug 2016, 01:23

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.
_________________

Hey, guys, So, I’ve decided to run a contest in hopes of getting the word about the site out to as many applicants as possible this application season...

Whether you’re an entrepreneur, aspiring business leader, or you just think that you may want to learn more about business, the thought of getting your Masters in Business Administration...

Term 1 has begun. If you're confused, wondering what my post on the last 2 official weeks was, that was pre-term. What that means is that the school...