Question

The value of $(0.16{)}^{{\log }_{2.5}(\frac{1}{3}+\frac{1}{3^2}+\cdots \infty )}$ is

Answer 1

$\frac{1}{3}+\frac{1}{3^2}+\cdots \infty$
$=\frac{1/3}{1-1/3}=\frac{1}{2}$

$(0.16{)}^{{\log }_{2.5}(\frac{1}{3}+\frac{1}{3^2}+\cdots \infty )}$
$=\left[\right(0.4{)}^2{]}^{{\log }_{5/2}\frac{1}{2}}$
$=(0.4{)}^{{\log }_{5/2}\frac{1}{4}}$
$={\left(\frac{1}{4}\right)}^{{\log }_{5/2}\frac{2}{5}}(\because a^{{\log }_{b}c}=c^{{\log }_{b}a})$
$={\left(4\right)}^{-{\log }_{5/2}\frac{2}{5}}$
$={\left(4\right)}^{{\log }_{2/5}\frac{2}{5}}$
$=4$