Can't solve this equation involving the natural logarithm

Question

Solve for $y$: $\frac12\ln(\sqrt{2y+1})) = \ln(3) + \ln(y-1)$.

I got to $\sqrt{2y+1} = (y+2)^2$, but then don't know where to go from here

I've now got to $(\sqrt{2y+1})^\frac{1}{2} = 3y-3$

Answer 1

I got $$\ln(\sqrt{2y+1})=\ln(9)+\ln((y-1)^2)$$ so $$\sqrt{2y+1}=9(y-1)^2$$ Then we get $$2y+1=(3(y-1))^4$$ and this is $$81 y^4-324 y^3+486 y^2-326 y+80=0$$ The Solutions don't look nice: $$\left\{1+\frac{1}{6} \sqrt{\frac{\sqrt[3]{2 \left(1+\sqrt{1297}\right)}}{3^{2/3}}-\frac{2\ 6^{2/3}}{\sqrt[3]{1+\sqrt{1297}}}}+\frac{1}{2} \sqrt{-\frac{\sqrt[3]{2 \left(1+\sqrt{1297}\right)}}{9\ 3^{2/3}}+\frac{2\ 2^{2/3}}{3 \sqrt[3]{3 \left(1+\sqrt{1297}\right)}}+\frac{4}{27 \sqrt{\frac{\sqrt[3]{2 \left(1+\sqrt{1297}\right)}}{3^{2/3}}-\frac{2\ 6^{2/3}}{\sqrt[3]{1+\sqrt{1297}}}}}}\right\}$$