We of course must require that x,y, and z are positive and not equal to 1

This one uses two theorems on logarithms, where a,b, and c are positive and
not equal to 1:

Theorem 1:

 = 1

which is easily proved by the change of base formula:



Multiplying both sides by the denominator on the right:
 = 1

Theorem 2:

 = 1

We consider a raised the the power of the left side:

=====

Since "a" raised to that power equals "a" raised to the 1 power, that
power must be 1.

We expand the determinant about the 1st row:  

 = 1· - logx(y)· + logx(z)· =

1·[1·1 - logy(z)·logz(y)] - logx(y)·[logy(x)·1 - logy(z)·logz(x)] + logx(z)·[logy(x)·logz(y) - 1·logz(x)] =

1·[1 - logy(z)·logz(y)] - logx(y)·[logy(x) - logy(z)·logz(x)] + logx(z)·[logy(x)·logz(y) - logz(x)] =

1 - logy(z)·logz(y) - logx(y)·logy(x) + logx(y)·logy(z)·logz(x) + logx(z)·logy(x)·logz(y) - logx(z)·logz(x) =

[using theorem 1 above on the 2nd, 3rd, and 6th terms, and theorem 2 on 
the 4th and 5th terms]:

1 - 1 - 1 + 1 + 1 - 1 = 0

Edwin