Simplify : `log (1/x+1/y)-log(x+y)+logx+logy`.

If y=x^(y^(x)) , prove that, (dy)/(dx)=(y log y(1+x logx log y))/(x logx(1-x logy)) .

if x^(2) + y^(2)= 25xy , then prove that 2 log(x + y) = 3log3 + logx + logy.

x^(2) + y^(2)= 25xy , then prove that 2 log(x + y) = 3log3 + logx + logy.

f x^(2) + y^(2)= 25xy , then prove that 2 log(x + y) = 3log3 + logx + logy.

f x^(2) + y^(2)= 25xy , then prove that 2 log(x + y) = 3log3 + logx + logy.

If y= x^(y^x) , prove that y_1 = (ylogy(1+x logx logy)/(x log x (1-xlogy))

x(logx-logy)dy-ydx=0, where logx-logy=u A) cx=log(x/y) B) log.(x)/(y)=log(log.(x)/(y))-(x)/(y)=x+c C) e^((x)/(y))+e^((x)/(y)-1)=e^(x)+c D) u=c(logu-1)

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