The value of the determinant `|{:(1,log_ba),(log_ab,1):}|` is equal to

Options

A. 1
B. `log_(e)b`
C. `log_(b)a`
D. 0

Answer

To find the value of the determinant

\[ D = \begin{vmatrix} 1 & \log_b a \\ \log_a b & 1 \end{vmatrix}, \]

we will use the properties of logarithms and determinants.

### Step 1: Write down the determinant formula
The value of a 2x2 determinant is calculated using the formula:

\[ D = ad - bc, \]

where \( a, b, c, d \) are the elements of the matrix:

\[ \begin{vmatrix} a & b \\ c & d \end{vmatrix}. \]

For our determinant, we have:

- \( a = 1 \)
- \( b = \log_b a \)
- \( c = \log_a b \)
- \( d = 1 \)

### Step 2: Substitute the values into the determinant formula
Substituting the values into the determinant formula gives us:

\[ D = (1)(1) - (\log_b a)(\log_a b). \]

### Step 3: Simplify the expression
Now, we simplify the expression:

\[ D = 1 - \log_b a \cdot \log_a b. \]

### Step 4: Use the change of base formula
Using the change of base formula for logarithms, we know that:

\[ \log_a b = \frac{1}{\log_b a}. \]

Thus, we can substitute this into our expression:

\[ D = 1 - \log_b a \cdot \frac{1}{\log_b a}. \]

### Step 5: Simplify further
This simplifies to:

\[ D = 1 - 1 = 0. \]

### Conclusion
Therefore, the value of the determinant is:

\[ \boxed{0}. \]