Discussion

Explanation:

Given, $0.{04}^{{\log }_{\sqrt{5}}\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...\right)}$ 

= $0.{04}^{{\log }_{\sqrt{5}}\left(\frac{1/4}{1-1/2}\right)}$

= $0.{04}^{{\log }_{\sqrt{5}}\left(\frac{1}{2}\right)}$

= ${\left(5^{-2}\right)}^{{\log }_{\sqrt{5}}\left(\frac{1}{2}\right)}$

= ${\left(5^{-2}\right)}^{{\log }_{5^{1/2}}\left(\frac{1}{2}\right)}$

= ${\left(5^{-2}\right)}^{2{\log }_5\left(\frac{1}{2}\right)}$

= $5^{-4{\log }_5\left(\frac{1}{2}\right)}$

= $5^{{\log }_5{\left(\frac{1}{2}\right)}^{-4}}$

= $5^{{\log }_{5}16}$

= ${16}^{{\log }_{5}5}$

= 16.

Hence, 16.

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