Question

The value of $(0.16)^{\log _{2.5}\left(\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\ldots+0 \infty\right)}$ is equal to

Answer 1

$(0.16)^{\log _{2.5}\left(\frac{1}{3}+\frac{1}{3^{2}}+\ldots \ldots \ldots . . \text { to } \infty\right)}$
$=\left(\frac{4}{25}\right)^{\log _{\left(\frac{5}{2}\right)}\left(\frac{1}{2}\right)}$
$=\left(\frac{1}{2}\right)^{\log _{\left(\frac{5}{2}\right)}\left(\frac{4}{25}\right)}$
$=\left(\frac{1}{2}\right)^{-2}=4$