Question

If \(a, b, c\) are non-zero and different from \(1\), then the value of \[ \begin{vmatrix} \log_a 1 & \log_a b & \log_a c \\ \log_b \left(\frac{1}{b}\right) & \log_b 1 & \log_b \left(\frac{1}{c}\right) \\ \log_c \left(\frac{1}{c}\right) & \log_c c & \log_c 1 \end{vmatrix} \] is:

• A. \( 0 \)
• B. \( 1 + \log_a (a + b + c) \)
• C. \( \log_a (ab + bc + ca) \)
• D. \( 1 \)
• E. \( \log_a (a + b + c) \)

Hint:

The change of base rule $\log_a b \cdot \log_b c = \log_a c$ is essentially a "chain rule" for logarithms. It simplifies multi-base expressions into a single base instantly.

Answer 1

The Correct Option is A

Understanding the Concept: Logarithm properties:
• \( \log_x 1 = 0 \)
• \( \log_x x = 1 \)
• \( \log_x (1/y) = -\log_x y \)

Step 1: Evaluate individual log terms.
The matrix is: \[ \begin{vmatrix} 0 & \log_a b & \log_a c -1 & 0 & -\log_b c -1 & 1 & 0 \end{vmatrix} \] $\log_b(1/b) = -1$, $\log_b 1 = 0$, $\log_b(1/c) = -\log_b c$.
$\log_c(1/c) = -1$, $\log_c c = 1$, $\log_c 1 = 0$.

Step 2: Expand the determinant.
Expanding along the first row: \[ 0 - \log_a b \begin{vmatrix} -1 & -\log_b c -1 & 0 \end{vmatrix} + \log_a c \begin{vmatrix} -1 & 0 -1 & 1 \end{vmatrix} \] \[ = -\log_a b (0 - \log_b c) + \log_a c (-1 - 0) \] \[ = \log_a b \cdot \log_b c - \log_a c \]

Step 3: Use the change of base formula.
Recall that \( \log_a b \cdot \log_b c = \log_a c \). \[ = \log_a c - \log_a c = 0 \]