Question Number 143906 by mathdanisur last updated on 19/Jun/21
Answered by TheHoneyCat last updated on 19/Jun/21
$$\Leftrightarrow{x}^{{x}} −{y}^{{y}} ={e}.\mathrm{ln}\left(\frac{{y}}{{x}}\right)={e}\left(\mathrm{ln}\left({y}\right)−\mathrm{ln}\left({x}\right)\right) \\ $$$$\Leftrightarrow{x}^{{x}} +{e}\mathrm{ln}{x}={y}^{{y}} +{e}\mathrm{ln}{y} \\ $$$$ \\ $$$$\mathrm{so}\:\mathrm{oviously}\:\left\{\left({t},{t}\right),\:{t}\in\mathbb{R}_{+} ^{\ast} \right\}\:\mathrm{is}\:\mathrm{part}\:\mathrm{of}\:\mathrm{the}\:\mathrm{solutions} \\ $$$$\mathrm{but}\:\mathrm{are}\:\mathrm{there}\:\mathrm{any}\:\mathrm{other} \\ $$$$ \\ $$$$\mathrm{considering}\:{f}:\begin{cases}{\mathbb{R}_{+} ^{\ast} }&{\rightarrow}&{\mathbb{R}}\\{{x}}&{ }&{{x}^{{x}} +{e}\mathrm{ln}{x}}\end{cases} \\ $$$$\mathrm{if}\:{x}\neq{y}\:\mathrm{are}\:\mathrm{a}\:\mathrm{solution},\:{f}\:\left({x}\right)={f}\left({y}\right)\: \\ $$$$\mathrm{thus}\:{f}\:\mathrm{is}\:\mathrm{not}\:\mathrm{injective} \\ $$$$ \\ $$$$\mathrm{but}\:{f}\:\mathrm{is}\:\mathrm{actualy}\:\mathrm{strictly}\:\mathrm{increasing} \\ $$$$\left({by}\:{a}\:{painful}\:{study}\:{of}\:{the}\:{derivative}\right) \\ $$$$\left({if}\:{anyone}\:{has}\:{a}\:{better}\:{method},\:{please}\:{send}\:{it}\right. \\ $$$$\left.{I}'{ve}\:{had}\:{to}\:{check}\:{derivatives}\:{too}\:{far}\:{to}\:{show}\:{it}\:{here}\right) \\ $$$$ \\ $$$$\mathrm{so}\:\mathrm{it}\:\mathrm{is}\:\mathrm{injective}\:\left({f}\left({x}\right)={f}\left({y}\right)\Rightarrow{x}={y}\right) \\ $$$$ \\ $$$$\mathrm{The}\:\mathrm{solutions}\:\mathrm{are}\:: \\ $$$$\left({x},{y}\right)\in\left\{\left({t},{t}\right)\:{t}\in\mathbb{R}_{+} ^{\ast} \right\} \\ $$$$ \\ $$$${the}\:{equation}\:{is}\:{equivalent}\:{to}\:''{x}={y}'' \\ $$
Commented by mathdanisur last updated on 19/Jun/21
$${thank}\:{you}\:{Sir} \\ $$