Question

Evaluate \[ (0.2)^{\log_{\sqrt{5}}\alpha} + (0.04)^{\log_{5}\beta} \] if \[ \alpha = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots \] \[ \beta = \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \cdots \]

• A. \(4\)
• B. \(8\)
• C. \(12\)
• D. \(10\)

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The task involves summing infinite geometric series to find \(\alpha\) and \(\beta\), and then substituting these values into a logarithmic expression involving base-power relationships.
Step 2: Key Formula or Approach:
Sum of an infinite geometric progression: \(S_{\infty} = \frac{a}{1 - r}\) for \(|r|<1\).
Logarithmic properties: \(\log_{b^{k}} x = \frac{1}{k} \log_{b} x\) and \(a^{\log_{a} x} = x\).
Step 3: Detailed Explanation: First, calculate \(\alpha\): This is an infinite GP with first term \(a = 1/4\) and common ratio \(r = 1/2\).
\[ \alpha = \frac{1/4}{1 - 1/2} = \frac{1/4}{1/2} = \frac{1}{2} \] Next, calculate \(\beta\): This is an infinite GP with first term \(a = 1/3\) and common ratio \(r = 1/3\).
\[ \beta = \frac{1/3}{1 - 1/3} = \frac{1/3}{2/3} = \frac{1}{2} \] Now, substitute these into the expression \((0.2)^{\log_{\sqrt{5}} \alpha} + (0.04)^{\log_{5} \beta}\).
Term 1: \((0.2)^{\log_{\sqrt{5}} (1/2)}\)
Note that \(0.2 = 1/5 = 5^{-1}\) and \(\sqrt{5} = 5^{1/2}\).
Using log base properties: \(\log_{5^{1/2}} (1/2) = \frac{1}{1/2} \log_{5} (1/2) = 2 \log_{5} (1/2) = \log_{5} (1/2)^{2} = \log_{5} (1/4)\).
Thus, \((5^{-1})^{\log_{5} (1/4)} = (5^{\log_{5} (1/4)})^{-1} = (1/4)^{-1} = 4\).

Term 2: \((0.04)^{\log_{5} (1/2)}\)
Note that \(0.04 = 4/100 = 1/25 = 5^{-2}\).
\[ (5^{-2})^{\log_{5} (1/2)} = (5^{\log_{5} (1/2)})^{-2} = (1/2)^{-2} = 4 \] Total sum = \(4 + 4 = 8\).
Step 4: Final Answer:
The evaluated value is 8, which corresponds to option (B).