Finding the value of a logarithmic expression involving an infinite GP Find the value of $(0.16)^{\displaystyle\log_{2.5}(\frac13+\frac1{3^2}+\frac1{3^3}+\cdots)}$.
I could solve the series. It gave $$(0.16)^{\log_{2.5}0.5}$$
Unable to proceed from here.

HINT:
$$0.16=\frac4{25}=\left(2.5\right)^{-2}$$
$$\text{Now, }\displaystyle a^{\log_am }=m$$
