SOLUTION: show that determinant {{{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))}}} = 0

Algebra ->  Matrices-and-determiminant -> SOLUTION: show that determinant {{{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))}}} = 0      Log On


   

Question 654399: show that determinant
= 0
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
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%28a%2C%28b%29%29%2Alog%28b%2C%28a%29%29 = 1

which is easily proved by the change of base formula:

log%28a%2C%28b%29%29=log%28b%2C%28b%29%29%2Flog%28b%2C%28a%29%29
log%28a%2C%28b%29%29=1%2Flog%28b%2C%28a%29%29
Multiplying both sides by the denominator on the right:
log%28a%2C%28b%29%29%2Alog%28b%2C%28a%29%29 = 1

Theorem 2:

log%28a%2C%28b%29%29%2Alog%28b%2C%28c%29%29%2Alog%28c%2C%28a%29%29 = 1

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

a%5E%28%28log%28a%2C%28b%29%29%2Alog%28b%2C%28c%29%29%2Alog%28c%2C%28a%29%29%29%29==b%5E%28%28log%28b%2C%28c%29%29%2Alog%28c%2C%28a%29%29%29%29++=%28++b%5E%28+%28+log%28b%2C%28c%29%29%29++%29%29%5E%28log%28c%2C%28a%29%29%29%29++=%28c%29%5E%28log%28c%2C%28a%29%29%29%29++=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:  

 = 1·abs%28matrix%282%2C2%2C+1%2C+log%28y%2C%28z%29%29%2Clog%28z%2C%28y%29%29%2C+1%29%29 - 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