Question

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

Answer 1

Answer: 4

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