The value of $(0.16)$ $$ \log _{2.5}\left(\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\ldots ext {

Question

The value of $(0.16)$ $$ \log _{2.5}\left(\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\ldots 	ext { to } \inftyight) 	ext { is equal to } $$

Professor PC Thomas Classes · Answer

(4) : We have, $(0.16)^{\log _{2.5}\left(\frac{1}{3}+\frac{1}{3^{2}}+\ldots \inftyight)}$\ $=(0.16)^{\log _{2.5}\left(\frac{1 / 3}{1-1 / 3}ight)}=(0.16)^{\log _{2.5} 0.5}$\ $=\left(\frac{4}{25}ight)^{-\log _{2.5} 2}=\left(\frac{5}{2}ight)^{2 \log _{5 / 2} 2}$\ $=\left(\frac{5}{2}ight)^{\log _{5 / 2} 2^{2}}=2^{2}=4$\