For positive real numbers $x,$ $y,$ and $z,$ none of which are equal to 1, find
\[\begin{vmatrix} 1 & \log_x y & \log_x z \\ \log_y x & 1 & \log_y z \\ \log_z x & \log_z y & 1 \end{vmatrix}.\]
Alternatively, I can use the cofactor expansion along any row or column, but I notice that the rule of Sarrus might be simpler in this case, since the matrix has a lot of 1s in the diagonal. So, using the rule of Sarrus, I get
\[\begin{vmatrix} 1 & \log_x y & \log_x z \\ \log_y x & 1 & \log_y z \\ \log_z x & \log_z y & 1 \end{vmatrix} = 1 \cdot 1 \cdot 1 + \log_x y \cdot \log_y z \cdot \log_z x + \log_x z \cdot \log_y x \cdot \log_z y - \log_x z \cdot 1 \cdot \log_z x - \log_x y \cdot \log_y x \cdot 1 - 1 \cdot \log_y z \cdot \log_z y.\] This looks complicated, but I can simplify it by using some properties of logarithms. First, I recall that $\log_a b \cdot \log_b c = \log_a c,$ since I can use the change of base formula to write $\log_a b = \frac{\log_c b}{\log_c a}$ and then cancel out the $\log_c b$ terms. This means that $\log_x y \cdot \log_y z \cdot \log_z x = \log_x x = 1,$ and similarly for the other terms involving three different bases. Second, I recall that $\log_a a = 1$ and $\log_a b \cdot \log_b a = 1,$ since they are inverse functions of each other. This means that $\log_x z \cdot \log_z x = 1,$ and similarly for the other terms involving the same base twice. So, applying these properties, I can rewrite the determinant as
\[\begin{vmatrix} 1 & \log_x y & \log_x z \\ \log_y x & 1 & \log_y z \\ \log_z x & \log_z y & 1 \end{vmatrix} = 1 \cdot 1 \cdot 1 + 1 + 1 - 1 - 1 - 1.\] This simplifies to 1 + 1 + 1 - 1 - 1 - 1 = 0. 