When 12 is substituted in for x and 1 for y, you get:

$\displaystyle{1}={\log}_{{{12}}}{12}\usepackage{cmbright}\Huge\$

This is equivalent to $\displaystyle{{12}}^{{1}}={12}\usepackage{cmbright}\Huge\$ , which is true. Therefore, the point $\displaystyle{\left({12},{1}\right)}\usepackage{cmbright}\Huge\$ is on the graph.