The value of (0.05)^{log_{sqrt{20}}(0.1 + 0.0 | vedclass

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Question

(A) Let $S = 0.1 + 0.01 + 0.001 + \dots$. This is an infinite geometric series with first term $a = 0.1$ and common ratio $r = 0.1$.$S = \frac{a}{1-r}

Vedclass · Accepted Answer

$81$

Vedclass · Answer

(A) Let $S = 0.1 + 0.01 + 0.001 + \dots$. This is an infinite geometric series with first term $a = 0.1$ and common ratio $r = 0.1$.$S = \frac{a}{1-r} = \frac{0.1}{1-0.1} = \frac{0.1}{0.9} = \frac{1}{9}$.Now,the expression is $(0.05)^{\log_{\sqrt{20}}(1/9)}$.Note that $0.05 = \frac{1}{20} = 20^{-1}$ and $\sqrt{20} = 20^{1/2}$.Using the property $\log_{a^n}(b) = \frac{1}{n} \log_a(b)$,we have $\log_{20^{1/2}}(1/9) = \frac{1}{1/2} \log_{20}(1/9) = 2 \log_{20}(1/9) = \log_{20}((1/9)^2) = \log_{20}(1/81)$.Thus,the expression becomes $20^{-1 \cdot \log_{20}(1/81)} = 20^{\log_{20}((1/81)^{-1})} = 20^{\log_{20}(81)} = 81$.

The value of $(0.05)^{\log_{\sqrt{20}}(0.1 + 0.01 + 0.001 + \dots)}$ is

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