Question

The value of ${\left(0.2\right)}^{{\log }_{\sqrt{5}}\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{6}+....to\infty \right)}$

• A. 4
• B. 1/4
• C. 2
• D. 1/2

Answer 1

Answer: (A)

$0.2^{{\log }_{\sqrt{5}}\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{6}+....to\infty \right)}$
$={\left(\frac{1}{5}\right)}^{{\log }_{\sqrt{5}}\left(\frac{1}{2}\right)}={\left(\frac{1}{5}\right)}^{2{\log }_{\sqrt{5}}}\left(\frac{1}{2}\right)$
$={\left(5\right)}^{-2{\log }_5}\left(\frac{1}{2}\right)={\left(5\right)}^{{\log }_{5}4}=4$