Question

Solve krr diyo
chalo byee yaaro
Gdnt
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Answer 1

Answer:-

$\Large\boxed{\sf{(C)\quad\!3^{\frac{31}{32}}}}$

Solution:-

We have,

$\longrightarrow\sf{\sqrt{ab}=\sqrt a\times\sqrt b}$

Thus it is true that,

$\longrightarrow\sf{\sqrt{a\sqrt{b}}=\sqrt a\times\sqrt{\sqrt b}}$

Then,

$\longrightarrow\sf{\sqrt{3\sqrt{3\sqrt{3\sqrt{3\sqrt{3}}}}}=\sqrt{3}\times\sqrt{\sqrt3}\times\sqrt{\sqrt{\sqrt3}}\times\sqrt{\sqrt{\sqrt{\sqrt3}}}\times\sqrt{\sqrt{\sqrt{\sqrt{\sqrt3}}}}}$

The following is true that,

$\mapsto\sf{\sqrt3=3^{\frac{1}{2}}}$

$\mapsto\sf{\sqrt{\sqrt3}=\left(3^{\frac{1}{2}}\right)^{\frac{1}{2}}=3^{\frac{1}{4}}}$

$\mapsto\sf{\sqrt{\sqrt{\sqrt3}}=\left(3^{\frac{1}{4}}\right)^{\frac{1}{2}}=3^{\frac{1}{8}}}$

$\mapsto\sf{\sqrt{\sqrt{\sqrt{\sqrt3}}}=\left(3^{\frac{1}{8}}\right)^{\frac{1}{2}}=3^{\frac{1}{16}}}$

$\mapsto\sf{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt3}}}}=\left(3^{\frac{1}{16}}\right)^{\frac{1}{2}}=3^{\frac{1}{32}}}$

Then,

$\longrightarrow\sf{\sqrt{3\sqrt{3\sqrt{3\sqrt{3\sqrt{3}}}}}=3^{\frac{1}{2}}\times3^{\frac{1}{4}}\times3^{\frac{1}{8}}\times3^{\frac{1}{16}}\times3^{\frac{1}{32}}}$

Since $\sf{a^m\times a^n=a^{m+n},}$

$\longrightarrow\sf{\sqrt{3\sqrt{3\sqrt{3\sqrt{3\sqrt3}}}}=3^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}}}$

$\longrightarrow\sf{\underline{\underline{\sqrt{3\sqrt{3\sqrt{3\sqrt{3\sqrt3}}}}=3^{\frac{31}{32}}}}}$