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

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

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

y log y (dx)/(dy) + x - log y =0

If x^2 + y^2 = 10xy , prove that 2 log (x + y) = log x + log y + 2 log 2 + log 3 .

Solve y log y dx + (x-log y) dy = 0.

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

If y=a^(x^(a^x..oo)) then prove that dy/dx=(y^2 log y )/(x(1-y log x log y))

If y=a^(x^(a^x..oo)) then prove that dy/dx=(y^2 log y )/(x(1-y log x log y))

If y=a^(x^(a^x..oo)) then prove that dy/dx=(y^2 log y )/(x(1-y log x log y))