Correct. If $\displaystyle{3}{\sqrt[{{5}}]{{{3}{x}}}}+{4}={5}{\sqrt[{{5}}]{{{3}{x}}}}+{3}\usepackage{cmbright}\Huge\$ , then $\displaystyle{1}={2}{\sqrt[{{5}}]{{{3}{x}}}}\usepackage{cmbright}\Huge\$ , so $\displaystyle\frac{{1}}{{2}}={\sqrt[{{5}}]{{{3}{x}}}}\usepackage{cmbright}\Huge\$ . Then $\displaystyle{{\left(\frac{{1}}{{2}}\right)}}^{{5}}={{\left({\sqrt[{{5}}]{{{3}{x}}}}\right)}}^{{5}}\usepackage{cmbright}\Huge\$ , so $\displaystyle\frac{{1}}{{32}}={3}{x}\usepackage{cmbright}\Huge\$ , and therefore $\displaystyle\frac{{1}}{{96}}={x}\usepackage{cmbright}\Huge\$ .