I am trying to solve the following equation for $y$, and it is a little tricky. $$-\frac{1}{2}\ln(y)-\frac{1}{2}\ln(y-2)=\ln(x+3)+c$$
I'm not sure if what I have done is correct or not; thanks for the help.
$$-\left(\ln \sqrt y + \ln\sqrt {y-2}\right)=\ln(x+3)+c$$ $$-\left(\ln \sqrt {y(y-2)}\right)=\ln(x+3)+c$$ $$e^{-\left(\ln \sqrt {y(y-2)}\right)}=e^{\ln(x+3)}\cdot e^c$$ $$\left(-\sqrt {y(y-2)}\right)^2=((x+3) \cdot e^c)^2$$ $$y^2-2y=(x+3)^2 \cdot e^{2c}$$ $$y^2-2y-(x+3)^2 \cdot e^{2c}=0$$
Then applying the quadratic formula to the above second degree polynomial yields
$$y=\frac{2\pm\sqrt {4-2e^{2c}(x+3)^2}}{2}$$
$$-\frac{1}{2}\ln(y)-\frac{1}{2}\ln(y-2)=\ln(x+3)+c$$ Multiplying both parts by -2, we get $$ln(y(y-2))=-2\ln(x+3)+c=ln(x+3)^{-2}+c$$ Now applying exponent, we obtain $$y^2-2y=\frac{e^c}{(x+3)^2}$$ Thus, $$y=\frac{2\pm\sqrt{4+\frac{4e^c}{(x+3)^2}}}{2}=1\pm\sqrt{1+\frac{e^c}{(x+3)^2}}$$