You may have correctly rewritten the equation as $\displaystyle{\log{{16}}}{x}={\log{{32}}}+{\log{{2}}}\usepackage{cmbright}\Huge\$ . But it appears that you tried to simplify by dividing both sides by $\displaystyle{\log{{16}}}\usepackage{cmbright}\Huge\$ . However, $\displaystyle{\log{{16}}}{x}\usepackage{cmbright}\Huge\$ is not the same as $\displaystyle{\left({\log{{16}}}\right)}\cdot{x}\usepackage{cmbright}\Huge\$ . The expression $\displaystyle{\log{{16}}}{x}\usepackage{cmbright}\Huge\$ tells you to multiply x by 16 and then compute a logarithm. So $\displaystyle{\log{{16}}}\usepackage{cmbright}\Huge\$ is not a factor of $\displaystyle{\log{{16}}}{x}\usepackage{cmbright}\Huge\$ .