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

Concept:
Logarithm properties: 
• \( \log_x 1 = 0 \) 
• \( \log_x x = 1 \) 
• \( \log_x \left(\frac{1}{y}\right) = -\log_x y \) 

Step 1: Evaluate individual logarithmic terms.
The determinant becomes: \[ \begin{vmatrix} 0 & \log_a b & \log_a c \\ -1 & 0 & -\log_b c \\ -1 & 1 & 0 \end{vmatrix} \] Since: \[ \log_b \left(\frac{1}{b}\right)=-1,\quad \log_b 1=0,\quad \log_b \left(\frac{1}{c}\right)=-\log_b c \] and \[ \log_c \left(\frac{1}{c}\right)=-1,\quad \log_c c=1,\quad \log_c 1=0 \] 
Step 2: Expand the determinant.
Expanding along the first row: \[ \Delta= 0 -\log_a b \begin{vmatrix} -1 & -\log_b c \\ -1 & 0 \end{vmatrix} + \log_a c \begin{vmatrix} -1 & 0 \\ -1 & 1 \end{vmatrix} \] Evaluating the minors: \[ = -\log_a b \left[ (-1)(0)-(-\log_b c)(-1) \right] + \log_a c \left[ (-1)(1)-(0)(-1) \right] \] \[ = -\log_a b(-\log_b c) - \log_a c \] \[ = \log_a b \cdot \log_b c - \log_a c \] 
Step 3: Use the change of base property.
Recall: \[ \log_a b \cdot \log_b c = \log_a c \] Therefore: \[ \Delta= \log_a c-\log_a c=0 \] 
Final Answer: \[ \boxed{0} \]