The equation $\displaystyle{{b}}^{{y}}={x}\usepackage{cmbright}\Huge\$ is equivalent to $\displaystyle{\log}_{{{b}}}{x}={y}\usepackage{cmbright}\Huge\$ , so $\displaystyle{b}=\frac{{1}}{{3}}\usepackage{cmbright}\Huge\$ , $\displaystyle{y}=-{3}\usepackage{cmbright}\Huge\$ , and $\displaystyle{x}={27}\usepackage{cmbright}\Huge\$ . Substitute these values to obtain $\displaystyle{\log}_{{\frac{{1}}{{3}}}}{27}=-{3}\usepackage{cmbright}\Huge\$ .