Let $\alpha = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \ldots\infty$ and$\beta = \frac{1}{3} + \fr

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Let $\alpha = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \ldots\infty$ and

$\beta = \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots\infty$. Then the value of $(0.2)^{\log_{\sqrt{5}}(\alpha)} + (0.04)^{\log_{5}(\beta)}$ is equal to :

Accepted Answer

Sol. [Key Concept: Both series are GPs summing to $\frac{1}{2}$; simplify the logarithmic exponents $\log_{\sqrt{5}}\alpha$ and $\log_{5}\beta$.]$${\alpha = \frac{1/4}{1 - \frac{1}{2}} = \frac{1}{2} }{\beta = \frac{1/3}{1 - \frac{1}{3}} = \frac{1}{2} }{(0.2)^{\log_{\sqrt{5}}\alpha} = \alpha^{\log_{\sqrt{5}\frac{1}{5}}} = \left( \frac{1}{2} \right)^{- 2} = 4 }{(0.04)^{\log_{5}\beta} = 5^{- 2\log_{5}\beta} = \left( \frac{1}{2} \right)^{- 2} = 4 }{\Rightarrow 4 + 4 = 8 }$$

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