Question:

<p><strong>Step 1: Simplify y using logarithm properties</strong></p><p>Convert all logarithms to base 2:</p><p>$$y = \log_2 x + \log_4 x + \log_{16} x + \cdots$$</p><p>$$y = \log_2 x + \frac{\log_2 x}{2} + \frac{\log_2 x}{4} + \cdots$$</p><p>$$y = \log_2 x \left(1 + \frac{1}{2} + \frac{1}{4} + \cdots\right)$$</p><p>This is a geometric series with first term 1 and common ratio 1/2:</p><p>$$y = \log_2 x \cdot \frac{1}{1-1/2} = 2\log_2 x$$</p><p><strong>Step 2: Simplify the numerator</strong></p><p>The numerator is $5 + 9 + 13 + \cdots + (4y+1)$</p><p>This is an arithmetic sequence with first term $a=5$, common difference $d=4$, and last term $l = 4y+1$.</p><p>Number of terms: $\frac{4y+1-5}{4} + 1 = y$</p><p>Sum of numerator: $\frac{y(5 + 4y + 1)}{2} = \frac{y(4y + 6)}{2} = y(2y + 3)$</p><p><strong>Step 3: Simplify the denominator</strong></p><p>The denominator is $1 + 3 + 5 + \cdots + (2y-1)$</p><p>This is the sum of first y odd numbers:</p><p>Sum of denominator: $y^2$</p><p><strong>Step 4: Set up and solve the equation</strong></p><p>$$4\log_4 x = \frac{y(2y+3)}{y^2}$$</p><p>$$4 \cdot \frac{\log_2 x}{2} = \frac{2y+3}{y}$$</p><p>$$2\log_2 x = \frac{2y+3}{y}$$</p><p>Since $y = 2\log_2 x$, we have $\log_2 x = \frac{y}{2}$:</p><p>$$2 \cdot \frac{y}{2} = \frac{2y+3}{y}$$</p><p>$$y = \frac{2y+3}{y}$$</p><p>$$y^2 = 2y + 3$$</p><p>$$y^2 - 2y - 3 = 0$$</p><p>$$(y-3)(y+1) = 0$$</p><p>Since $y > 0$, we have $y = 3$</p><p><strong>Step 5: Find x and calculate $x^2y$</strong></p><p>From $y = 2\log_2 x$:</p><p>$$3 = 2\log_2 x$$</p><p>$$\log_2 x = \frac{3}{2}$$</p><p>$$x = 2^{3/2} = 2\sqrt{2}$$</p><p>$$x^2 = (2\sqrt{2})^2 = 8$$</p><p>$$x^2y = 8 \times 3 = 24$$</p><p><strong>∴ Answer: C</strong></p>