Question Number 185422 by Ml last updated on 21/Jan/23
Answered by hmr last updated on 21/Jan/23
$$\mathrm{2}\sqrt{\mathrm{5}}\:=\:\sqrt{\mathrm{20}}\:=\:\left(\mathrm{20}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\rightarrow{a}=\:\sqrt[{\mathrm{5}}]{\mathrm{2}\sqrt{\mathrm{5}}}\:=\:\sqrt[{\mathrm{5}}]{\left(\mathrm{20}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }\:=\:\left(\mathrm{20}\right)^{\frac{\mathrm{1}}{\mathrm{10}}} \\ $$$$ \\ $$$$\sqrt[{\mathrm{3}}]{{a}^{\mathrm{16}} }\:=\:{a}^{\frac{\mathrm{16}}{\mathrm{3}}} \:=\:\left(\left(\mathrm{20}\right)^{\frac{\mathrm{1}}{\mathrm{10}}} \right)^{\frac{\mathrm{16}}{\mathrm{3}}} =\:\left(\mathrm{20}\right)^{\frac{\mathrm{8}}{\mathrm{15}}} \\ $$$$=\:\sqrt[{\mathrm{15}}]{\mathrm{20}^{\mathrm{8}} } \\ $$
Commented by Frix last updated on 21/Jan/23
$$\mathrm{Typo}\:\mathrm{in}\:\mathrm{line}\:\mathrm{3} \\ $$
Commented by hmr last updated on 21/Jan/23
$${yes}\:{sir},\:{thank}\:{you}. \\ $$$${I}\:{will}\:{edit}\:{it}. \\ $$
Answered by Frix last updated on 21/Jan/23
$${a}=\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{5}}} \mathrm{5}^{\frac{\mathrm{1}}{\mathrm{10}}} \\ $$$${a}^{\frac{\mathrm{16}}{\mathrm{3}}} =\mathrm{2}^{\frac{\mathrm{16}}{\mathrm{15}}} \mathrm{5}^{\frac{\mathrm{8}}{\mathrm{15}}} =\mathrm{20}^{\frac{\mathrm{8}}{\mathrm{15}}} \\ $$