Correct. Since $\displaystyle{4}={{2}}^{{2}},{{2}}^{{-{{3}}}}{{4}}^{{5}}={{2}}^{{-{{3}}}}{{\left({{2}}^{{2}}\right)}}^{{5}}={{2}}^{{-{{3}}}}{{2}}^{{{2}\cdot{5}}}={{2}}^{{-{{3}}}}{{2}}^{{10}}={{2}}^{{-{3}+{10}}}={{2}}^{{7}}\usepackage{cmbright}\Huge\$ .