$\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2}}}}} = $x$

### Concept & Formula This question requires solving a finite series of nested square roots of the same number. There is a direct algebraic formula to resolve these specific structures instantly. $$ \sqrt{a\sqrt{a\sqrt{a...\text{n times}}}} = a^{\frac{2^n-1}{2^n}} $$ ### Step-by-Step Solution * **Given:** The expression is $\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2}}}}}$. * **Calculation:** 1. Identify the repeating base value ($a$) and the total number of square roots ($n$). $a = 2$ $n = 5$ 2. Calculate the denominator for the fractional exponent, which is $2^n$: $2^5 = 32$ 3. Calculate the numerator for the fractional exponent, which is $2^n - 1$: $32 - 1 = 31$ 4. Substitute these values into the standard formula: $2^{\frac{31}{32}}$ ### Exam Strategy & Shortcut Do not attempt to solve this from the inside out manually during an exam. Simply count the number of radical signs. There are $5$ roots. Calculate $2^5 = 32$. The power of the base number will always be a fraction where the denominator is this result ($32$), and the numerator is exactly one less ($31$). This immediately points to option (b). ### Common Pitfall A frequent mistake is confusing this finite nested root structure with an infinite nested root structure ($\sqrt{2\sqrt{2...}}$ to infinity). For an infinite series, the answer is simply the base number itself ($2$). Always check if the series terminates. ### Final Answer **Therefore, the correct answer is $2^{\frac{31}{32}}$.**