The right way to writte it down is this:[MATH]\log{y} = \dfrac{1}{2}\log{x} + (x^2-x) + \dfrac{3}{2}\log(x+1)[/MATH]
\(\displaystyle \log{y} = \dfrac{1}{2}\log{x} +(x^2-x)loge+ \dfrac{3}{2}\log(x+1)\)
Because: \(\displaystyle lne=1\) and \(\displaystyle loge\neq 1\)