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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 817 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 _________________
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
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