Question:
The value of $(0.16)^{\log _{2.5}\left(\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\ldots \text { to } \infty\right)}$ is equal to______________
Solution:
$(0.16)^{\log _{2.5}\left(\frac{1}{3}+\frac{1}{3^{2}}+\ldots \ldots \ldots \ldots \text { to } \infty\right)}$
$=\left(\frac{4}{25}\right)^{\log \left(\frac{5}{2}\right)\left(\frac{1}{2}\right)}$
$=\left(\frac{1}{2}\right)^{\log \left(\frac{5}{2}\right)\left(\frac{4}{25}\right)}=\left(\frac{1}{2}\right)^{-2}=4$
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