A logarithimic simplification How could I simplify this? It must give $\dfrac{9}{2}+\ln (2)$.

$$\left( \dfrac{4}{2}+6+\ln (2)\right) - \left( \dfrac{1}{2}+3+\ln (1)\right) =8+\ln (2)-\dfrac{7}{2}-\ln (1)=\dfrac{9}{2}+\ln (2)-\ln (1)$$
We have $$8-\frac{7}{2}+\ln(2)-\ln(1)=\frac{16-7}{2}+\ln(2)$$ since $$\ln(1)=0$$
We have $$8-\frac{7}{2}+\ln(2)-\ln(1)=\frac{16-7}{2}+\ln(2)$$ since $$\ln(1)=0$$