<p>The value of \(0.2^{\log_{\sqrt{5}}\!\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots\right)}\) is</p>
Step-by-Step Solution
Key Concept: First, sum the infinite geometric series to get a concrete value, then convert the exponential expression using logarithm properties and base conversions to simplify.
<p><strong>Step 1: Find the infinite geometric series sum</strong></p><p>The series is: 1/4 + 1/8 + 1/16 + ⋯</p><p>This is a geometric series with first term a = 1/4 and common ratio r = 1/2</p><p>Sum = a/(1-r) = (1/4)/(1-1/2) = (1/4)/(1/2) = 1/2</p><p><strong>Step 2: Simplify the logarithm</strong></p><p>We need: 0.2^(log_{√5}(1/2))</p><p>Let y = log_{√5}(1/2), which means (√5)^y = 1/2</p><p><strong>Step 3: Convert to base 5</strong></p><p>Since √5 = 5^(1/2), we have: (5^(1/2))^y = 1/2</p><p>This gives: 5^(y/2) = 1/2</p><p>Taking log₅ of both sides: y/2 = log₅(1/2)</p><p>Therefore: y = 2log₅(1/2)</p><p><strong>Step 4: Evaluate the final expression</strong></p><p>0.2^y = (1/5)^(2log₅(1/2)) = (5^(-1))^(2log₅(1/2)) = 5^(-2log₅(1/2))</p><p>Using the property a^(log_a(x)) = x:</p><p>5^(-2log₅(1/2)) = 5^(log₅((1/2)^(-2))) = (1/2)^(-2) = 4</p><p>∴ Answer: A</p>
Correct Answer: A