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

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(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 $r = 0.1$. The sum $S = \frac

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 $r = 0.1$. The sum $S = \frac{a}{1-r} = \frac{0.1}{1-0.1} = \frac{0.1}{0.9} = \frac{1}{9}$. Now,the expression is $(0.05)^{\log_{\sqrt{20}}(1/9)}$. Note that $0.05 = \frac{5}{100} = \frac{1}{20}$. Also,$\sqrt{20} = 20^{1/2}$. Using the property $\log_{a^n}(b) = \frac{1}{n} \log_a(b)$,we have $\log_{20^{1/2}}(1/9) = \frac{1}{1/2} \log_{20}(1/9) = 2 \log_{20}(1/9) = \log_{20}(1/9)^2 = \log_{20}(1/81)$. So,the expression becomes $(1/20)^{\log_{20}(1/81)}$. Using the property $a^{\log_a(x)} = x$,we rewrite $(1/20)^{\log_{20}(1/81)}$ as $(20^{-1})^{\log_{20}(1/81)} = (20^{\log_{20}(1/81)})^{-1} = (1/81)^{-1} = 81$.

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

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