Mathematics - Sequence & Series Question with Solution | TestHub

$0.{16}^{{\log }_{2.5}\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\dots .\infty \right)}$

$=0.{16}^{{\log }_{2.5}\left(\frac{{\displaystyle \frac{1}{3}}}{1-{\displaystyle \frac{1}{3}}}\right)}={\left(\frac{4}{25}\right)}^{{\log }_{\frac{5}{2}}\left(\frac{1}{2}\right)}$

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