Question Number 169798 by mathlove last updated on 09/May/22
$${log}_{{e}} \left({e}^{\mathrm{2}} {x}^{{lnx}} \right)={log}_{{e}} \left({x}^{\mathrm{3}} \right) \\ $$$${faind}\:\:{x}=? \\ $$
Commented by cortano1 last updated on 09/May/22
$$\:\Rightarrow\mathrm{2}+\left(\mathrm{ln}\:{x}\right)^{\mathrm{2}} \:=\:\mathrm{3}\:\mathrm{ln}\:{x} \\ $$$$\Rightarrow\left(\mathrm{ln}\:{x}\right)^{\mathrm{2}} −\mathrm{3}\:\mathrm{ln}\:{x}+\mathrm{2}\:=\:\mathrm{0} \\ $$$$\Rightarrow\begin{cases}{\mathrm{ln}\:{x}=\mathrm{2}\Rightarrow{x}={e}^{\mathrm{2}} }\\{\mathrm{ln}\:{x}=\mathrm{1}\Rightarrow{x}=\:{e}}\end{cases} \\ $$
Commented by mathlove last updated on 09/May/22
$${thanks} \\ $$