Solve the system of equations: `(log)_a x(log)_a(x y z)=48(log)_a y log_a(x y z)=12 ,\ a >0,\ a!=1(log)_a z log_a(x y z)=84\ `

Solve the system of equations: (log)_(a)x(log)_(a)(xyz)=48(log)_(a)y log_(a)(xyz)=12,backslash a>0,backslash a!=1(log)_(a)z log_(a)(xyz)=84

Solve the system of equations : log_(a)(x)log_(a)(xyz)=48log_(a)(y)log_(a)(xyz)=12log_(a)(z)log_(a)(xyz)=84,quad a>0,a!=1

Solve: log_(a)x log_(a)(xyz)=48log_(a)y log_(a)(xyz)=12;log_(a)z log_(a)(xyz)=84

For positive numbers x ,\ y\ a n d\ z the numerical value of the determinant |1(log)_x y(log)_x z(log)_y x1(log)_y z(log)_z x(log)_z y1| is- a. 0 b. logx y z c. "log"(x+y+z) d. logx\ logy\ logz

For positive numbers x ,\ y\ a n d\ z the numerical value of the determinant |1(log)_x y(log)_x z(log)_y x1(log)_y z(log)_z x(log)_z y1| is- a. 0 b. logx y z c. "log"(x+y+z) d. logx\ logy\ logz

For positive numbers x ,\ y\ a n d\ z the numerical value of the determinant |1(log)_x y(log)_x z(log)_y x1(log)_y z(log)_z x(log)_z y1| is- a. 0 b. logx y z c. "log"(x+y+z) d. logx\ logy\ logz

If x=(log)_(2a)a , y=(log)_(3a)2a ,z=(log)_(4a)3a ,prove that 1+x y z=2y z

Prove that : log _ y^(x) . log _z^(y) . log _a^(z) = log _a ^(x)

Using properties of determinant show that |(1,log_x y,log_x z),(log_y x,1,log_y z),(log_z x,log_z y,1)|=0