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.

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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