Question:
The value of $(0.16) \log _{2.5}\left(\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\ldots\right.$ to $\left.\infty\right)$ is equal to________.
Solution:
$(0.16)^{\log _{2.5}\left(\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\ldots . \infty\right)}$
$=0.16$ $\log _{2.5}\left(\frac{\frac{1}{3}}{1-\frac{1}{3}}\right)$$\left[\because S_{\infty}=\frac{a}{1-r}\right]$
$=0.16^{\log _{2}}$ $16^{\log _{2.5}\left(\frac{1}{2}\right)}$
$=(2.5)^{-2 \log _{2.5}\left(\frac{1}{2}\right)}=\left(\frac{1}{2}\right)^{-2}=4$
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