Question

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

Answer 1

$\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\cdots \infty =\frac{\frac{1}{3}}{1-\frac{1}{3}}=\frac{1}{2}$
${\log }_{2.5}\left(\frac{1}{2}\right)\Rightarrow {\log }_{\frac{5}{2}}\frac{1}{2}$
$0.16=\frac{16}{100}=\frac{4}{25}={\left(\frac{2}{5}\right)}^2$
Now,
$\Rightarrow {\left(\frac{2}{5}\right)}^{2{\log }_{\frac{5}{2}}\left(\frac{1}{2}\right)}={\left(\frac{5}{2}\right)}^{-2{\log }_{\frac{5}{2}}\left(\frac{1}{2}\right)}$
$\Rightarrow {\left(\frac{5}{2}\right)}^{{\log }_{\frac{5}{2}}{\left(\frac{1}{2}\right)}^{-2}}=4$