$x=\frac{i\pi n_{1}}{\ln(3)}-\frac{3}{2}\text{, }n_{1}\in \mathrm{Z}$
$x=\frac{i\pi n_{2}}{\ln(3)}-1\text{, }n_{2}\in \mathrm{Z}$

$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$
$x=-1$