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

a = sqrt(3*a)

squaring both sides

a^2 = 3a

=> a = 3.

Is this correct?

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.

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

for me this should be E) cannot be determined

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

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

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


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)

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

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.

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

Answer C

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

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