Question

$\sqrt{11 \sqrt{11 \sqrt{11 \sqrt{11} } } }$
Simplify

Please solve this. I will mark as Brainliest immediately

Answer 1

Solution :

Let x = $\sqrt{11 \sqrt{11 \sqrt{11 \sqrt{11} } } }$

Then, $\mathsf{{x} ^{2}}$ = 11 $\sqrt{11 \sqrt{11 \sqrt{11 \sqrt{11} } } }$

$\mathsf{{x} ^{2}}$ = 11x

Hence, x = 11.

Answer 2

$\sqrt{11 \sqrt{11 \sqrt{11 \sqrt{11} } } }$

As we know

$\sqrt[x]{a} = a {}^{ \frac{1}{x} } \\ \\ a {}^{x} \times a {}^{y} = {a}^{x + y}$

Taking Last two Terms from Series

$\sqrt{11 \sqrt{11} } \\ \\ \sqrt{11 {}^{1} \times 11 {}^{ \frac{1}{2} } } \\ \\ \sqrt{11 {}^{1 + \frac{1}{2} } } \\ \\ \sqrt{11 {}^{ \frac{3}{2} } } \\ \\ 11 {}^{ \frac{3}{2} \times \frac{1}{2} } \\ \\ {11}^{ \frac{3}{4} } \\ \\ \sqrt{11 \sqrt{11} } = {11}^{ \frac{3}{4} }$

Taking Last 3 terms

$\sqrt{11 \sqrt{11 \sqrt{11} } } \\ \\ \sqrt{11 {}^{1} \times {11}^{ \frac{3}{4} } } \\ \\ \sqrt{11 {}^{1 + \frac{3}{4} } } \\ \\ \sqrt{ {11}^{ \frac{7}{4} } } \\ \\ {11}^{ \frac{7}{4} \times \frac{1}{2} } \\ \\ {11}^{ \frac{7}{8} } \\ \\ \sqrt{11 \sqrt{11 \sqrt{11} } } = {11}^{ \frac{7}{8} }$

Taking Last 4 terms

$\sqrt{11 \sqrt{11 \sqrt{11 \sqrt{11} } } } \\ \\ \sqrt{11 {}^{1} \times 11 {}^{ \frac{7}{8} } } \\ \\ \sqrt{11 {}^{1 + \frac{7}{8} } } \\ \\ \sqrt{11 {}^{ \frac{15}{8} } } \\ \\ {11}^{ \frac{15}{8} \times \frac{1}{2} } \\ \\ {11}^{ \frac{15}{16} }$

$\underline{ \huge{ \sqrt{11 \sqrt{11 \sqrt{11 \sqrt{11} } } } = {11}^{ \frac{15}{16} } }}$