If `x= tanh^-1 (y)` then `log_e((1+y)/(1-y))`

If x = tan h^(-1)(y) then log_(e)((1+y)/(1-y))

The solution of the differential equation, dy/dx=(x-y)^(2) , when y(1)=1, is (A) log_(e) abs((2-y)/(2-x))=2(y-1) (B) -log_(e)abs((1+x-y)/(1-x+y))=x+y-2 (C) log_(e) abs((2-x)/(2-y))=x-y (D) -log_(e) abs((1-x+y)/(1+x-y))=2(x-1)

If y^(x)= e^(y-x) , then prove that (dy)/(dx)= ((1+ log y)^(2))/(log y)

for x > 1 if (2x)^(2y)=4e^(2x-2y) then (1+log_e 2x)^2 (dy)/(dx)

for x > 1 if (2x)^(2y)=4e^(2x-2y) then (1+log_e 2x)^2 (dy)/(dx)

for x > 1 if (2x)^(2y)=4e^(2x-2y) then (1+log_e 2x)^2 (dy)/(dx)

The solutions to the system of equations log_(5)x+log_(27)y=4 and log_(x)5-log_(y)(27)=1 are (x_(1),y_(1)) and (x_(2),y_(2)) then log_(15)(x_(1)x_(2)y_(1)y2) is