<p>The value of determinant \(\begin{vmatrix} \log_a\left(\frac{x}{y}\right) & \log_a\left(\frac{y}{z}\right) & \log_a\left(\frac{z}{x}\right) \\ \log_{a^2}\left(\frac{y}{z}\right) & \log_{a^2}\left(\frac{z}{x}\right) & \log_{a^2}\left(\frac{x}{y}\right) \\ \log_{a^3}\left(\frac{z}{x}\right) & \log_{a^3}\left(\frac{x}{y}\right) & \log_{a^3}\left(\frac{y}{z}\right) \end{vmatrix}\) is</p>
Step-by-Step Solution
Key Concept: Convert all logarithms to a common base using the change of base formula: log_{a^n}(m) = (1/n)·log_a(m), then recognize that each row is a cyclic permutation of the same three quantities, making rows linearly dependent.
<p><strong>Step 1:</strong> Apply change of base formula. Let log_a(x/y) = p, log_a(y/z) = q, log_a(z/x) = r.</p><p>Note that p + q + r = log_a(x/y) + log_a(y/z) + log_a(z/x) = log_a(1) = 0.</p><p><strong>Step 2:</strong> Convert the determinant using log_{a^n}(m) = (1/n)·log_a(m):</p><p>Row 1: [p, q, r]</p><p>Row 2: [(1/2)q, (1/2)r, (1/2)p]</p><p>Row 3: [(1/3)r, (1/3)p, (1/3)q]</p><p><strong>Step 3:</strong> Factor out constants: Det = (1/2)·(1/3)·|Row 1 matrix with cyclic permutations|</p><p><strong>Step 4:</strong> Notice Row 2 = (1/2)·[q, r, p] and Row 3 = (1/3)·[r, p, q]. Each row is a cyclic permutation of [p, q, r].</p><p><strong>Step 5:</strong> Since p + q + r = 0, we have r = -(p+q). The three rows become linearly dependent (any row can be expressed as a linear combination of the others), making the determinant equal to 0.</p><p>∴ Answer: <strong>C</strong> (0)</p>
Correct Answer: C