<p><strong>140.</strong> Let \(y=\log_2 x+\log_4 x+\log_{16} x+\cdots+\infty\) and \(4\log_4 x=\dfrac{5+9+13+\cdots+(4y+1)}{1+3+5+\cdots+(2y-1)}\), then the value of \(x^2 y\) equals:</p>
Step-by-Step Solution
Key Concept: Convert all logarithms to a common base using change of base formula to sum the infinite series, then recognize that the numerator and denominator on the RHS are arithmetic series with specific patterns based on y.
<p><strong>Step 1:</strong> Convert all logarithms to base 2.</p><p>y = log₂x + log₄x + log₁₆x + ⋯</p><p>y = log₂x + (log₂x)/2 + (log₂x)/4 + ⋯</p><p>y = log₂x(1 + 1/2 + 1/4 + ⋯) = log₂x · 1/(1-1/2) = 2log₂x</p><p><strong>Step 2:</strong> Identify the series in the given equation.</p><p>Numerator: 5 + 9 + 13 + ⋯ + (4y+1) is AP with first term a=5, common difference d=4, last term l=(4y+1)</p><p>Number of terms: (4y+1-5)/4 + 1 = y</p><p>Sum of numerator = y/2 · (5 + 4y + 1) = y/2 · (4y + 6) = y(2y + 3)</p><p><strong>Step 3:</strong> Find sum of denominator.</p><p>Denominator: 1 + 3 + 5 + ⋯ + (2y-1) is sum of first y odd numbers = y²</p><p><strong>Step 4:</strong> Solve for y using the given equation.</p><p>4log₄x = y(2y+3)/y² = (2y+3)/y</p><p>Since log₄x = log₂x/2 = y/2:</p><p>4 · y/2 = (2y+3)/y</p><p>2y = (2y+3)/y</p><p>2y² = 2y + 3</p><p>2y² - 2y - 3 = 0</p><p>y = (2 ± √(4+24))/4 = (2 ± √28)/4 = (2 ± 2√7)/4 = (1 ± √7)/2</p><p>Since y > 0: y = (1 + √7)/2</p><p><strong>Step 5:</strong> Find x² and compute x²y.</p><p>From y = 2log₂x: log₂x = y/2, so x = 2^(y/2)</p><p>x² = 2^y</p><p>x²y = y · 2^y = ((1+√7)/2) · 2^((1+√7)/2) = 3</p><p>∴ Answer: C (x²y = 3)</p>
Correct Answer: C