Question Number 222697 by gabthemathguy25 last updated on 05/Jul/25
Answered by gregori last updated on 05/Jul/25
$$\:\sqrt[{\mathrm{3}}]{{e}^{\mathrm{2ln}\:\left(\mathrm{81}\right)} }\:=\:\left(\mathrm{81}^{\frac{\mathrm{2}}{\mathrm{3}}} \right)=\:\mathrm{9}^{\frac{\mathrm{4}}{\mathrm{3}}} \\ $$$$\:\sqrt[{\mathrm{4}}]{\left(\mathrm{3}^{\mathrm{log}\:_{\mathrm{9}} \left(\mathrm{9}^{\mathrm{2}} \right)} \right)^{\mathrm{2}} }\:=\:\sqrt{\mathrm{9}}=\:\mathrm{9}^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\:\Rightarrow\:\mathrm{log}\:_{\mathrm{9}^{\frac{\mathrm{4}}{\mathrm{3}}} } \:\left(\mathrm{9}^{\frac{\mathrm{1}}{\mathrm{2}}} \right)=\:\frac{\mathrm{1}}{\mathrm{2}}\:.\frac{\mathrm{3}}{\mathrm{4}}\:=\:\frac{\mathrm{3}}{\mathrm{8}} \\ $$
Commented by gabthemathguy25 last updated on 05/Jul/25
$${close}.. \\ $$
Commented by A5T last updated on 05/Jul/25
$$\mathrm{6561}=\mathrm{9}^{\mathrm{4}} \:\:\neq\:\mathrm{9}^{\mathrm{2}} \:,\:\mathrm{so}\:\sqrt[{\mathrm{4}}]{\left(\mathrm{3}^{\mathrm{log}_{\mathrm{9}} \left(\mathrm{9}^{\mathrm{4}} \right)} \right)^{\mathrm{2}} }=\mathrm{9}\:\mathrm{not}\:\sqrt{\mathrm{9}} \\ $$