log(x+y)=logx-logy x+y=x/y from this equality; xy+y^2=x and xy-x=-y^2 if we take left equation with x factor, x(y-1)=-y^2 and x=(-y^2)/(y-1) y=t x=(-t^2)/(t-1) and (x,y)=((-t^2)/(t-1),t), thats it.
log(x + y) = log(x / y) x + y > 0 x / y > 0 Suppose: x + y = t and x / y = t then t > 0, t in R Solving the equations simultaneously, we find that x = t^2 / t + 1 y = t / t + 1 t > 0 I think this parametric equation represents the solutions of the equation.
It should be noted that y will always be less than 1. (given that x > 0 and y > 0)
log(x+y)=logx-logy
x+y=x/y from this equality;
xy+y^2=x and
xy-x=-y^2 if we take left equation with x factor,
x(y-1)=-y^2 and x=(-y^2)/(y-1)
y=t x=(-t^2)/(t-1) and
(x,y)=((-t^2)/(t-1),t), thats it.
Yes.
x = y²/(1−y); 0 < y < 1.
x → 0 as y → 0 and x → ∞ as y → 1. Since x is continuous and positive it means that all pozitive values for x are possible.
log(x + y) = log(x / y)
x + y > 0
x / y > 0
Suppose:
x + y = t and
x / y = t
then t > 0, t in R
Solving the equations simultaneously, we find that
x = t^2 / t + 1
y = t / t + 1
t > 0
I think this parametric equation represents the solutions of the equation.
x+y=x/y
xy + y^2 = x
y^2 + xy - x = 0
y = [-x ± √(x^2 + 4x)]/2
but y cannot be -ve so:
y = √((x/2)^2 + x) - x/2