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Find the value of a. Given a=√3(√3(√3(√3(√3…….inf.))))

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Find the value of a. Given a=√3(√3(√3(√3(√3…….inf.)))) [#permalink] New post 29 Aug 2012, 07:51
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Find the value of a. Given a=√3(√3(√3(√3(√3…….inf.)))).. [PS: nested sq. root sequence is repeated infinite times.]

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

[Reveal] Spoiler:
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 :)
[Reveal] Spoiler: OA

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Last edited by conty911 on 30 Aug 2012, 05:10, edited 1 time in total.
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Re: Find the value of a. Given a=√3(√3(√3(√3(√3…….inf.)))) [#permalink] New post 29 Aug 2012, 08:15
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conty911 wrote:
Find the value of a. Given a=√3(√3(√3(√3(√3…….inf.)))).. [PS: nested sq. root sequence is repeated infinite times.]

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

[Reveal] Spoiler:
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.


Similar questions to practice:
tough-and-tricky-exponents-and-roots-questions-125956-40.html#p1029228
if-the-expression-x-sqrt-2-sqrt-2-sqrt-2-sqrt-2-extends-98647.html
find-the-value-of-x-75403.html

Hope it helps.
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Re: Find the value of a. Given a=√3(√3(√3(√3(√3…….inf.)))) [#permalink] New post 30 Aug 2012, 02:20
a = sqrt(3*a)

squaring both sides

a^2 = 3a

=> a = 3.

Is this correct?
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Re: Find the value of a. Given a=√3(√3(√3(√3(√3…….inf.)))) [#permalink] New post 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?
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Re: Find the value of a. Given a=√3(√3(√3(√3(√3…….inf.)))) [#permalink] New post 30 Aug 2012, 02:42
SOURH7WK wrote:
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?


Zero is neither positive nor negative number.

Also, a=0 does not satisfy the equation because the left hand side of the equation is clearly positive.
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Re: Find the value of a. Given a=√3(√3(√3(√3(√3…….inf.)))) [#permalink] New post 30 Aug 2012, 03:02
I think this will be an endless sequence of multiples of 3....

for me this should be E) cannot be determined
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Re: Find the value of a. Given a=√3(√3(√3(√3(√3…….inf.)))) [#permalink] New post 30 Aug 2012, 04:48
cracked wrote:
a = sqrt(3*a)

squaring both sides

a^2 = 3a

=> a = 3.

Is this correct?


yes it is correct :) , you can see the spoiler in the question and also checkout similar examples given by Bunnuel .
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Re: Find the value of a. Given a=√3(√3(√3(√3(√3…….inf.)))) [#permalink] New post 30 Aug 2012, 07:17
conty911 wrote:
cracked wrote:
a = sqrt(3*a)

squaring both sides

a^2 = 3a

=> a = 3.

Is this correct?


yes it is correct :) , you can see the spoiler in the question and also checkout similar examples given by Bunnuel .



your approach is correct. 3 is correct answer
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Re: Find the value of a. Given a=√3(√3(√3(√3(√3…….inf.)))) [#permalink] New post 30 Aug 2012, 07:24
conty911 wrote:
Find the value of a. Given a=√3(√3(√3(√3(√3…….inf.)))).. [PS: nested sq. root sequence is repeated infinite times.]

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

[Reveal] Spoiler:
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.
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PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. NEW!!!

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set. NEW!!!


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Re: Find the value of a. Given a=√3(√3(√3(√3(√3…….inf.)))) [#permalink] New post 02 Sep 2012, 13:54
Found another way of solving it:
We can rewrite the equation:

a=3^(1/2)*3^(1/4)*3^(1/8)*...
a=3^(1/2+1/4+1/8+1/16+...)

1/2+1/4+1/8+...=1 (ok, one has to know this equation, I think)

a=3^1=3
Re: Find the value of a. Given a=√3(√3(√3(√3(√3…….inf.))))   [#permalink] 02 Sep 2012, 13:54
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