Correct. If $\displaystyle{\sqrt[{{5}}]{{{3}{x}-{1}}}}={2}\usepackage{cmbright}\Huge\$ , then $\displaystyle{{\left({\sqrt[{{5}}]{{{3}{x}-{1}}}}\right)}}^{{5}}={{2}}^{{5}}={32}\usepackage{cmbright}\Huge\$ , so $\displaystyle{3}{x}-{1}={32}\usepackage{cmbright}\Huge\$ . Therefore, $\displaystyle{3}{x}={33}\usepackage{cmbright}\Huge\$ , so $\displaystyle{x}={11}\usepackage{cmbright}\Huge\$ .