Question

The value of ${\left(0.2\right)}^{{\log }_{\sqrt{5}}(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots \infty )}$ is equal to

Answer 1

$S_{\infty }=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+....\infty =\frac{\frac{1}{4}}{1-\left(\frac{1}{2}\right)}=\frac{1}{2}$ [ sum of infinite G.P]

${\left(0.2\right)}^{{\log }_{\sqrt{5}}(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots \infty )}={\left(5\right)}^{-{\log }_{\sqrt{5}}\left(\frac{1}{2}\right)}={\left(5\right)}^{{\log }_{\sqrt{5}}2}={\left(5\right)}^{{\log }_{5}4}=4$

$[\because m\log x=\log x^m\text{\&}{a}^{{\log }_{a}b}=b]$

Ans: 4