If `y` is a function of `x` given by `2log (y-1)-log x-log (y-2)=0`, then

For relation 2 log y - log x - log ( y-1) =0

For relation 2 log y - log x - log ( y-1) =0

For relation 2 log y - log x - log ( y-1) =0

1+log_(x)y=log_(2)y

If (x-y)^(2)=121xy and 2log(x+y)=k+log x+log y then k=

If x and y satisfy the following system of equations, log x – log y = -2 , log x + 2 log y=1 then the value of xy is

If (1)/(2)log x+(1)/(2)log y+log2=log(x+y) then :

log (x + y) = log2 + (1) / (2) log x + (1) / (2) log y, then

Let the solution of y log ydx+(x-log y)dy=0 is x.log y=A(log y)^(2)+c where c is constant then A=?