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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]
20 Aug 2014, 10:39

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).

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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 46 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) _________________

Find the value of a. Given a=√3(√3(√3(√3(√3…….inf.)))) [#permalink]
20 Aug 2014, 19:29

Just take 4 occurrences of square root

a = \sqrt{3\sqrt{3\sqrt{3\sqrt{3}}}}

a = \sqrt{3\sqrt{3\sqrt{3^1 * 3^{\frac{1}{2}}}}}

a = \sqrt{3\sqrt{3\sqrt{3^{\frac{3}{2}}}}}

a = \sqrt{3\sqrt{3 * 3^{\frac{3}{2} * \frac{1}{2}}}}

a = \sqrt{3\sqrt{3^1 * 3^{\frac{3}{4}}}}

a = \sqrt{3\sqrt{3^{\frac{7}{4}}}}

a = \sqrt{3 * 3^{\frac{7}{4} * \frac{1}{2}}}

a = \sqrt{3^{\frac{15}{8}}}

a = 3^{\frac{15}{8} * \frac{1}{2}}

a = 3^{\frac{15}{16}}

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 _________________

Kindly press "+1 Kudos" to appreciate

gmatclubot

Find the value of a. Given a=√3(√3(√3(√3(√3…….inf.))))
[#permalink]
20 Aug 2014, 19:29