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

$\begin{array}{c}0.{16}^{{\log }_2\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\dots \infty \right)}=0.{16}^{{\log }_{2.5}\left(\frac{{\displaystyle \frac{1}{3}}}{1-{\displaystyle \frac{1}{3}}}\right)}\\ =0.{16}^{{\log }_{2.5}\left(\frac{1}{2}\right)}=0.{16}^{-{\log }_{2.5}2}\\ ={\left(2.5\right)}^{2{\log }_{2.5}2}\\ ={\left(2.5\right)}^{{\log }_{2.5}4}=4\end{array}$