Question 654399
<pre>
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:

{{{log(a,(b))*log(b,(a))}}} = 1

which is easily proved by the change of base formula:

{{{log(a,(b))=log(b,(b))/log(b,(a))}}}
{{{log(a,(b))=1/log(b,(a))}}}
Multiplying both sides by the denominator on the right:
{{{log(a,(b))*log(b,(a))}}} = 1

Theorem 2:

{{{log(a,(b))*log(b,(c))*log(c,(a))}}} = 1

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

{{{a^((log(a,(b))*log(b,(c))*log(c,(a))))}}}={{{(  a^( ( log(a,(b)))  ))^((log(b,(c))*log(c,(a))))  }}}={{{b^((log(b,(c))*log(c,(a))))  }}}={{{(  b^( ( log(b,(c)))  ))^(log(c,(a))))  }}}={{{(c)^(log(c,(a))))  }}}={{{a}}}

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:  

{{{abs(matrix(3,5,

1, "",log(x,(y)), "",log(x,(z)),
log(y,(x)),"", 1, "",log(y,(z)),
log(z,(x)),"", log(z,(y)),"", 1))}}} = 1·{{{abs(matrix(2,2, 1, log(y,(z)),log(z,(y)), 1))}}} - log<sub>x</sub>(y)·{{{abs(matrix(2,2, log(y,(x)),log(y,(z)),log(z,(x)), 1))}}} + log<sub>x</sub>(z)·{{{abs(matrix(2,2, log(y,(x)),1,log(z,(x)),log(z,(y))))}}} =

1·[1·1 - log<sub>y</sub>(z)·log<sub>z</sub>(y)] - log<sub>x</sub>(y)·[log<sub>y</sub>(x)·1 - log<sub>y</sub>(z)·log<sub>z</sub>(x)] + log<sub>x</sub>(z)·[log<sub>y</sub>(x)·log<sub>z</sub>(y) - 1·log<sub>z</sub>(x)] =

1·[1 - log<sub>y</sub>(z)·log<sub>z</sub>(y)] - log<sub>x</sub>(y)·[log<sub>y</sub>(x) - log<sub>y</sub>(z)·log<sub>z</sub>(x)] + log<sub>x</sub>(z)·[log<sub>y</sub>(x)·log<sub>z</sub>(y) - log<sub>z</sub>(x)] =

1 - log<sub>y</sub>(z)·log<sub>z</sub>(y) - log<sub>x</sub>(y)·log<sub>y</sub>(x) + log<sub>x</sub>(y)·log<sub>y</sub>(z)·log<sub>z</sub>(x) + log<sub>x</sub>(z)·log<sub>y</sub>(x)·log<sub>z</sub>(y) - log<sub>x</sub>(z)·log<sub>z</sub>(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</pre>