If x , y and z be greater than 1, then the value of ` |{:(1, log_(x)y, log_(x) z),(log_(y)x , 1 ,log_(y)z),(log_(z)x , log_z y , 1 ):}|` =

Options

A. log x.logy.logz
B. logx + logy + logz
C. 0
D. 1 - {(logx).(logy).(logz)}

Answer

To solve 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 use properties of logarithms and determinants.

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

\[ \log_x y = \frac{\log y}{\log x}, \quad \log_x z = \frac{\log z}{\log x}, \quad \log_y z = \frac{\log z}{\log y}, \quad \log_y x = \frac{\log x}{\log y}, \quad \log_z x = \frac{\log x}{\log z}, \quad \log_z y = \frac{\log y}{\log z} \]

Substituting these into the determinant gives:

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

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

- Multiply Row 1 by \(\log x\)
- Multiply Row 2 by \(\log y\)
- Multiply Row 3 by \(\log z\)

This gives:

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

### Step 3: Identify identical rows
Now we can see that all three rows of the determinant are identical.

### Step 4: Conclusion
Since a determinant with two or more identical rows is equal to zero, we conclude that:

\[ D = 0 \]

Thus, the value of the determinant is:

\[ \boxed{0} \]