If the square root of a^{frac{1}{a}} cdot (2a | vedclass

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(B) Let $P = a^{\frac{1}{a}} \cdot (2a)^{\frac{1}{2a}} \cdot (4a)^{\frac{1}{4a}} \cdot (8a)^{\frac{1}{8a}} \cdots \infty$. Given $\sqrt{P} = \frac{8}{

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$\frac{1}{3}$

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(B) Let $P = a^{\frac{1}{a}} \cdot (2a)^{\frac{1}{2a}} \cdot (4a)^{\frac{1}{4a}} \cdot (8a)^{\frac{1}{8a}} \cdots \infty$. Given $\sqrt{P} = \frac{8}{27}$,so $P = \left(\frac{8}{27}\right)^2 = \frac{64}{729}$. We can write $P = a^{(\frac{1}{a} + \frac{1}{2a} + \frac{1}{4a} + \cdots)} \cdot 2^{(\frac{1}{2a} + \frac{2}{4a} + \frac{3}{8a} + \cdots)}$. The exponent of $a$ is $\frac{1}{a}(1 + \frac{1}{2} + \frac{1}{4} + \cdots) = \frac{1}{a} \cdot \frac{1}{1 - 1/2} = \frac{2}{a}$. The exponent of $2$ is an Arithmetico-Geometric Progression $(AGP)$: $S = \frac{1}{2a} + \frac{2}{4a} + \frac{3}{8a} + \cdots$. $S = \frac{1}{a}(\frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \cdots)$. Using the formula for $AGP$,$S = \frac{1}{a} \cdot \frac{1/2}{(1-1/2)^2} = \frac{1}{a} \cdot \frac{1/2}{1/4} = \frac{2}{a}$. Thus,$P = a^{2/a} \cdot 2^{2/a} = (2a)^{2/a}$. We have $(2a)^{2/a} = \frac{64}{729} = (\frac{8}{27})^2 = ((\frac{2}{3})^3)^2 = (\frac{2}{3})^6$. Comparing $(2a)^{2/a} = (\frac{2}{3})^6$,we set $\frac{2}{a} = 6 \Rightarrow a = \frac{1}{3}$. Checking the base: $2a = 2(\frac{1}{3}) = \frac{2}{3}$. This satisfies the equation.

If the square root of $a^{\frac{1}{a}} \cdot (2a)^{\frac{1}{2a}} \cdot (4a)^{\frac{1}{4a}} \cdot (8a)^{\frac{1}{8a}} \cdots \infty$ is $\frac{8}{27}$,then the value of $a$ is:

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