`|[lnx,lny,lnz] , [ln2x,ln2y,ln2z] , [ln3x,ln3y,ln3z]|=0`

Prove that the value of each the following |log x,log y,log zlog2x,log2y,log2zlog3x,log3y,log3z]|

Show that |(log x, log y, logz),(log 2x, log2y, log2z),(log3x, log3y,log3z)|=0

Delta=|[log x ,log y, log z],[log 2 x, log 2 y, log 2 z],[log 3 x, log 3 y, log 3 z ]|=

If x, y, z being positive |[1, log _x y,log _x z],[log _y x,1,log _y z],[log _z x,log _z y, 1]|=

If ln(x+z)+ln(x-2y+z)=2ln(x-z) then

Let (x_(0),y_(0)) be the solution of the following equations (2x)^(ln2)=(3y)^(ln3) and 3^(lnx)=2^(lny) then 2x_(0)=

If x,y,z are in G.P then show that log x,log y,log z are in A.P. (x>0,y>0,z>0) .

If log, y, log, x, log, z are in G.P. xyz 64 and x y, z' are in A.P., then

If log x, log y and log z are in AP, then