Question:

<p><strong>Step 1:</strong> Expand f(x,n) using logarithm properties.</p><p>$$f(x,n) = \sum_{k=1}^{n} \log_x \left( \frac{k}{x} \right) = \sum_{k=1}^{n} \left[ \log_x k - \log_x x \right]$$</p><p>Since $\log_x x = 1$:</p><p>$$f(x,n) = \sum_{k=1}^{n} \log_x k - n$$</p><p><strong>Step 2:</strong> Calculate f(x,10) and f(x,11).</p><p>$$f(x,10) = \sum_{k=1}^{10} \log_x k - 10$$</p><p>$$f(x,11) = \sum_{k=1}^{11} \log_x k - 11$$</p><p><strong>Step 3:</strong> Set f(x,10) = f(x,11) and simplify.</p><p>$$\sum_{k=1}^{10} \log_x k - 10 = \sum_{k=1}^{11} \log_x k - 11$$</p><p>$$\sum_{k=1}^{10} \log_x k - 10 = \sum_{k=1}^{10} \log_x k + \log_x 11 - 11$$</p><p>$$-10 = \log_x 11 - 11$$</p><p>$$\log_x 11 = 1$$</p><p><strong>Step 4:</strong> Solve for x.</p><p>$$\log_x 11 = 1$$</p><p>$$x^1 = 11$$</p><p>$$x = 11$$</p><p><strong>Verification:</strong> When x = 11, we have $\log_{11} 11 = 1$, confirming our solution.</p><p>$$\therefore \text{Answer: C}$$</p>