For positive numbers x, y and z, the numerical value of the determinant `|{:(1,"log"_(x)y, "log"_(x)z),("log"_(y)x, 1, "log"_(y)z),("log"_(z)x, "log"_(z)y, 1):}|` is……

Answer

To find the numerical value of the determinant

\[ D = \begin{vmatrix} 1 & \log_x y & \log_x z \\ \log_y x & 1 & \log_y z \\ \log_z x & \log_z y & 1 \end{vmatrix} \]

we will follow these steps:

### Step 1: Rewrite the logarithms
Using the change of base formula for logarithms, we can express the logarithms in terms of natural logarithms (or any common base). Specifically, we have:

\[ \log_a b = \frac{\log b}{\log a} \]

Thus, we can rewrite the determinant as:

\[ D = \begin{vmatrix} 1 & \frac{\log y}{\log x} & \frac{\log z}{\log x} \\ \frac{\log x}{\log y} & 1 & \frac{\log z}{\log y} \\ \frac{\log x}{\log z} & \frac{\log y}{\log z} & 1 \end{vmatrix} \]

### Step 2: Multiply each row by the logarithm of the respective base
To simplify the determinant, we can multiply each row by the logarithm of the base of the logarithms in that row:

- Multiply the first row by \(\log x\)
- Multiply the second row by \(\log y\)
- Multiply the third row by \(\log z\)

This gives us:

\[ D = \begin{vmatrix} \log x & \log y & \log z \\ \log x & \log y & \log z \\ \log x & \log y & \log z \end{vmatrix} \]

### Step 3: Identify the rows
Notice that all three rows are now identical. When a determinant has two or more identical rows, the value of the determinant is zero.

### Step 4: Conclude the result
Thus, we conclude:

\[ D = 0 \]

### Final Answer:
The numerical value of the determinant is \(0\).

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