The value of (0.2)^ log (1/4 +1/8 +1/16 +....) , where base square root 5

Solution:       We  have ,

$(0.2)^{\log_{\sqrt{5}}(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+..........\infty )}$

$\Rightarrow$        $\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+.........+\infty =\frac{\frac{1}{4}}{1-\frac{1}{2}}=\frac{1}{2}=2^{-1}$

$\therefore$                $(0.2)^{\log_{\sqrt{5}}(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+..........\infty )}=(\frac{1}{5})^{\log_{\sqrt5}}2^{-1}$

$\Rightarrow$                                                                    $=(5^{-1})^{\frac{-1}{\frac{1}{2}}}\log_{5}2$

$\Rightarrow$                                                                    $=5^{2\log_{5}2}=2^{2}=4.$

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