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

Question

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. Find the value of a. Given a=√3(√3(√3(√3(√3…….inf.)))).. A. 0 B. \sqrt{3} C. 3 D. 2.9 E. cannot be determined. 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

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Bunuel · Accepted Answer

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.

Bunuel · Answer

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.

cracked · Answer

a = sqrt(3*a)

squaring both sides

a^2 = 3a

=> a = 3.

Is this correct?

SOURH7WK · Answer

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?

Bunuel · Answer

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.

ratinarace · Answer

I think this will be an endless sequence of multiples of 3....

for me this should be E) cannot be determined

conty911 · Answer

yes it is correct

, you can see the spoiler in the question and also checkout similar examples given by Bunnuel .

Zinsch123 · Answer

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

PareshGmat · Answer

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 calc.png 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)

PareshGmat · Answer

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

Madhavi1990 · Answer

Another way to think of this: it is an infinite sequence, and the last four terms are given to us (a =root 3 of root 3 of root 3 so on)
So by multiplying last two root 3's we have 3 and two more we have 3. Under root of 3 into 3 is 3.

shashankism · Answer

a=\sqrt{3\sqrt{3\sqrt{3\sqrt{3...}}}} 
 a=\sqrt{3a} 
 a^2=3a 
 a(a-3)=0 
 a=0,3 
a can't be 0. So a = 3

Answer C

Kinshook · Answer

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

a = \sqrt{3a} 
 a^2 = 3a 
a = 3; Since a is not 0

IMO C

snowbumvic · Answer

why is the LHS side of the equation clearly positive?

Bunuel · Answer

Because even roots cannot give negative result.

Even roots cannot give negative result.

\sqrt{...}  is the square root sign, a function (called the principal square root function), which cannot give negative result. So,  this sign ( \sqrt{...} ) always means non-negative square root .

https://gmatclub.com/forum/download/file.php?id=39502 
 The graph of the function f(x) = √x

Notice that it's defined for non-negative numbers and is producing non-negative results.

TO SUMMARIZE:
  When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the non-negative root.   That is:

\sqrt{9} = 3 , NOT +3 or -3;
 \sqrt {16} = 2 , NOT +2 or -2;
Similarly  \sqrt{\frac{1}{16}} = \frac{1}{4} , NOT +1/4 or -1/4.

Notice that in contrast, the equation  x^2 = 9  has TWO solutions, +3 and -3. Because  x^2 = 9  means that  x =-\sqrt{9}=-3  or  x=\sqrt{9}=3 .

GMATinsight · Answer

Such Questions ae simplest if you just take only a few steps of teh question as other steps will be insignificant due to being very small values

Given: a=√3(√3(√3(√3(√3.......inf.))))

Let, a=√3(√3(√3(√3(√3)))) = √3(√3(√3(√3(1.7)))) = √3(√3(√3(√5.1))) = √3(√3(√3(2.3))) = √3(√3(√6.9)) = √3(√3(2.6)) = √3(√7.8)) = √3(2.8)) ≈ 3

Answer: Option C

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conty911

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.

Username1512 · Answer

Bunuel  can you explain this property: Now, as the expression under the square root extends infinitely, then expression in brackets would equal to a itself. 
I tried understanidng it but I can't. I don't want to memorise it blindly for solving such questions. Thanks!

Bunuel · Answer

The point is that after the first layer, what remains is the same infinite expression again. Since the pattern never ends, removing one repeated layer still leaves an expression with the same structure and value as a . Similar questions to practice: https://gmatclub.com/forum/tough-and-tr ... l#p1029228 https://gmatclub.com/forum/find-the-val ... 38049.html https://gmatclub.com/forum/find-the-val ... 75403.html https://gmatclub.com/forum/if-the-expre ... 98647.html https://gmatclub.com/forum/m26-184445.html https://gmatclub.com/forum/if-the-expre ... 32547.html

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.))))

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