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Bunuel
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Let 3√5√3√5√3............∞ = Y
3√5√Y = Y
Squaring both sides
9*5√Y = Y^2
Squaring again
81*25*Y = Y^4
3*3*3*3*5*5 = Y^3
Y = 3³√75

Answer is B

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[quote="Bunuel"]What is the value of \(3\sqrt{5\sqrt{3\sqrt{5\sqrt{...}}}}\), where the given expression extends to an infinite number of roots?

\(x^2 = 9×5\sqrt{x}\)
\(x^4 = 81×25x\)
\(x^3 = 3^3×75\)
\(x = 3\sqrt[3]{75}\)

IMO B

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↧↧↧ Detailed Video Solution to the Problem Series ↧↧↧



\(3\sqrt{5\sqrt{3\sqrt{5\sqrt{...}}}}\)

Let, \(3\sqrt{5\sqrt{3\sqrt{5\sqrt{...}}}}\) = x

Note that the terms is a sequence of 3 and 5 under square roots and is extending till infinity, so we can replace the part of the term after first 3 and 5 by x itself

=> \(3\sqrt{5\sqrt{3\sqrt{5\sqrt{...}}}}\) = \(3\sqrt{5\sqrt{x}}\)

Squaring both the sides we get
=> \(3^2 * (\sqrt{5\sqrt{x}})^2 = x^2\)
=> \(3^2 * 5\sqrt{x} = x^2\)

Squaring both the sides again, we get
=> \(3^4 * 5^2 * (\sqrt{x})^2 = x^4\)
=> \(3^4 * 5^2 * x = x^4\)
=> \(x^3 = 3^4 * 5^2\)
=> \(x^3 = 3^3 * 3 * 25 = 3^3 * 75\)

Taking cube root on both the sides we get
=> x = \(3\sqrt[3]{75}\)

So, Answer will be B
Hope it helps!

Watch the following video to MASTER Roots

­
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