Evaluate Logarithmic Expression Involving Infinite Geometric Series

To evaluate the given expression, we first compute the values of the infinite geometric series $\alpha$ and $\beta$.

Step 1: Calculate $\alpha$ The series for $\alpha$ is an infinite geometric progression with first term $a_1 = \frac{1}{4}$ and common ratio $r_1 = \frac{1}{2}$: $\alpha = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots \infty = \frac{\frac{1}{4}}{1 - \frac{1}{2}} = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2}$

Step 2: Calculate $\beta$ The series for $\beta$ is also an infinite geometric progression with first term $a_2 = \frac{1}{3}$ and common ratio $r_2 = \frac{1}{3}$: $\beta = \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \dots \infty = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}$

Step 3: Evaluate the logarithmic expression We are required to find the value of: $E = (0.2)^{\log_{\sqrt{5}}(\alpha)} + (0.04)^{\log_5(\beta)}$

Let's evaluate the first term: $(0.2)^{\log_{\sqrt{5}}(\alpha)} = (5^{-1})^{\log_{5^{1/2}}\left(\frac{1}{2}\right)}$ Using the property $\log_{b^k}(x) = \frac{1}{k}\log_b(x)$: $\log_{5^{1/2}}\left(\frac{1}{2}\right) = 2\log_5\left(\frac{1}{2}\right)$ Substituting this back into the first term: $(0.2)^{\log_{\sqrt{5}}(\alpha)} = 5^{-\left(2\log_5\left(\frac{1}{2}\right)\right)} = 5^{\log_5\left(\left(\frac{1}{2}\right)^{-2}\right)} = 5^{\log_5(4)} = 4$

Now, let's evaluate the second term: $(0.04)^{\log_5(\beta)} = (5^{-2})^{\log_5\left(\frac{1}{2}\right)} = 5^{-2\log_5\left(\frac{1}{2}\right)} = 5^{\log_5\left(\left(\frac{1}{2}\right)^{-2}\right)} = 5^{\log_5(4)} = 4$

Step 4: Sum the terms $E = 4 + 4 = 8$

Thus, the value of the given expression is $8$, which corresponds to Option C.