Question

Let \[ \alpha = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots \infty \] and \[ \beta = \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \dots \infty. \] 
Then the value of \[ (0.2)^{\log_{\sqrt{5}}(\alpha)} + (0.04)^{\log_{5}(\beta)} \] is equal to: ________

• A. 4
• B. 5
• C. 8
• D. 25

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Evaluate the sum of the infinite geometric progressions for \(\alpha\) and \(\beta\). Then substitute these fractions into the logarithmic expression, utilizing exponent and base change properties to simplify.
Step 2: Key Formula or Approach:
Infinite GP sum: \(S_\infty = \frac{a}{1 - r}\).
Logarithm properties: \(\log_{b^n}(x) = \frac{1}{n} \log_b(x)\) and \(a^{\log_a(x)} = x\).
Step 3: Detailed Explanation:
Evaluate \(\alpha\):
\(a = 1/4\), \(r = 1/2\)
\[ \alpha = \frac{1/4}{1 - 1/2} = \frac{1}{2} \] Evaluate \(\beta\):
\(a = 1/3\), \(r = 1/3\)
\[ \beta = \frac{1/3}{1 - 1/3} = \frac{1}{2} \] Simplify the first term: \((0.2)^{\log_{\sqrt{5}}(\alpha)}\)
\(0.2 = 5^{-1}\) and \(\sqrt{5} = 5^{1/2}\).
\[ \log_{\sqrt{5}}(1/2) = \frac{1}{1/2} \log_5(1/2) = 2 \log_5(1/2) = \log_5(1/4) \] \[ (5^{-1})^{\log_5(1/4)} = 5^{-\log_5(1/4)} = 5^{\log_5(4)} = 4 \] Simplify the second term: \((0.04)^{\log_5(\beta)}\)
\(0.04 = 5^{-2}\).
\[ (5^{-2})^{\log_5(1/2)} = 5^{-2\log_5(1/2)} = 5^{\log_5(4)} = 4 \] Total sum = \(4 + 4 = 8\).
Step 4: Final Answer:
The final value is 8.