Question Number 172029 by Mikenice last updated on 23/Jun/22
$${solve}: \\ $$$${x}^{\mathrm{2}} =\mathrm{2}^{{x}} \\ $$
Commented by Rasheed.Sindhi last updated on 23/Jun/22
$${x}=\mathrm{2},\mathrm{4} \\ $$
Commented by Mikenice last updated on 23/Jun/22
$${please}\:{show}\:{the}\:{solution} \\ $$
Commented by mr W last updated on 23/Jun/22
$${x}=−\mathrm{0}.\mathrm{76666}\:{as}\:{well} \\ $$
Commented by puissant last updated on 23/Jun/22
$${x}^{\mathrm{2}} =\mathrm{2}^{{x}\:} \:\Rightarrow\:\:\mathrm{2}{lnx}\:=\:{xln}\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\frac{{lnx}}{{x}}\:=\:\frac{{ln}\mathrm{2}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\frac{{lnx}}{{x}}\:=\:\frac{\mathrm{2}{ln}\mathrm{2}}{\mathrm{4}}\:=\:\frac{{ln}\mathrm{4}}{\mathrm{4}}. \\ $$$$\:\:\:\:\:{S}=\left\{\mathrm{2}\:;\:\mathrm{4}\right\} \\ $$
Commented by mr W last updated on 23/Jun/22
$${from}\:\frac{{a}}{{b}}=\frac{{c}}{{d}}\:{we}\:{get}\:{not}\:{only} \\ $$$${a}={c},\:{b}={d}.\: \\ $$$${generally}\:{we}\:{get} \\ $$$${a}={kc},\:{b}={kd}. \\ $$
Answered by mr W last updated on 25/Jun/22
$${x}^{\mathrm{2}} =\mathrm{2}^{{x}} \\ $$$${x}=\pm\mathrm{2}^{\frac{{x}}{\mathrm{2}}} =\pm{e}^{\frac{{x}\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}} \\ $$$$\left(−\frac{{x}\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}\right){e}^{−\frac{{x}\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}} =\pm\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}} \\ $$$$−\frac{{x}\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}={W}\left(\pm\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}\right) \\ $$$$\Rightarrow{x}=−\frac{\mathrm{2}}{\mathrm{ln}\:\mathrm{2}}{W}\left(\pm\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:=\begin{cases}{−\frac{\mathrm{2}}{\mathrm{ln}\:\mathrm{2}}{W}\left(−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}\right)=\begin{cases}{\mathrm{2}}\\{\mathrm{4}}\end{cases}}\\{−\frac{\mathrm{2}}{\mathrm{ln}\:\mathrm{2}}{W}\left(\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}\right)=−\mathrm{0}.\mathrm{766665}}\end{cases} \\ $$