The value of (0.16)^{log _{2.5}left(frac{1}{3 | vedclass

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(A) Let $S = \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \ldots \infty$. This is an infinite geometric series with first term $a = \frac{1}{3}$ and

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$4$

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(A) Let $S = \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \ldots \infty$. This is an infinite geometric series with first term $a = \frac{1}{3}$ and common ratio $r = \frac{1}{3}$.Using the formula $S = \frac{a}{1-r}$,we get $S = \frac{1/3}{1 - 1/3} = \frac{1/3}{2/3} = \frac{1}{2}$.Now,the expression is $(0.16)^{\log_{2.5}(1/2)}$.Note that $0.16 = \frac{16}{100} = \frac{4}{25} = (2.5)^{-2}$.So,the expression becomes $((2.5)^{-2})^{\log_{2.5}(1/2)}$.Using the property $(a^b)^c = a^{bc}$,we have $(2.5)^{-2 \log_{2.5}(1/2)} = (2.5)^{\log_{2.5}((1/2)^{-2})}$.Since $a^{\log_a(x)} = x$,the expression simplifies to $(1/2)^{-2} = 2^2 = 4$.

The value of $(0.16)^{\log _{2.5}\left(\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\ldots \infty\right)}$ is equal to

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