Question-99410

Question Number 99410 by Harasanemanabrandah last updated on 20/Jun/20

Commented by PRITHWISH SEN 2 last updated on 20/Jun/20

one of the soln. set is for every positive real number greater than 0 , x=y

$$\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{soln}.\:\mathrm{set}\:\mathrm{is}\:\mathrm{for}\:\mathrm{every}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{number} \\ $$$$\mathrm{greater}\:\mathrm{than}\:\mathrm{0}\:,\:\mathrm{x}=\mathrm{y} \\ $$

Commented by Rasheed.Sindhi last updated on 20/Jun/20

$$\mathrm{2}^{\mathrm{4}} =\mathrm{4}^{\mathrm{2}} ;\:\left\{{x},{y}\right\}=\left\{\mathrm{2},\mathrm{4}\right\} \\ $$

Answered by mr W last updated on 20/Jun/20

for x,y∈R there are infinite many solutions. x=y=t∈R^+ is always a solution. for x≠y: METHOD I let x=t ∈R^+ x^y =y^t y=t^(y/t) =e^((y/t)ln t) ye^(−(y/t)ln t) =1 (−(y/t)ln t)e^(−(y/t)ln t) =−((ln t)/t) −(y/t)ln t=W(−((ln t)/t)) y=−(t/(ln t))W(−((ln t)/t)) ⇒general solution { ((x=t)),((y=−(t/(ln t))W(−((ln t)/t)))) :} METHOD II let y=tx with t∈R^+ x^(tx) =(tx)^x x^t =tx x^(t−1) =t ⇒x=t^(1/(t−1)) ⇒y=t^(1+(1/(t−1))) =t^(t/(t−1)) ⇒general solution { ((x=t^(1/(t−1)) )),((y=t^(t/(t−1)) )) :}

$${for}\:{x},{y}\in\mathbb{R}\:{there}\:{are}\:{infinite}\:{many} \\ $$$${solutions}. \\ $$$$ \\ $$$${x}={y}={t}\in{R}^{+} \:{is}\:{always}\:{a}\:{solution}. \\ $$$${for}\:{x}\neq{y}: \\ $$$$ \\ $$$${METHOD}\:{I} \\ $$$${let}\:{x}={t}\:\in{R}^{+} \\ $$$${x}^{{y}} ={y}^{{t}} \\ $$$${y}={t}^{\frac{{y}}{{t}}} ={e}^{\frac{{y}}{{t}}\mathrm{ln}\:{t}} \\ $$$${ye}^{−\frac{{y}}{{t}}\mathrm{ln}\:{t}} =\mathrm{1} \\ $$$$\left(−\frac{{y}}{{t}}\mathrm{ln}\:{t}\right){e}^{−\frac{{y}}{{t}}\mathrm{ln}\:{t}} =−\frac{\mathrm{ln}\:{t}}{{t}} \\ $$$$−\frac{{y}}{{t}}\mathrm{ln}\:{t}=\mathbb{W}\left(−\frac{\mathrm{ln}\:{t}}{{t}}\right) \\ $$$${y}=−\frac{{t}}{\mathrm{ln}\:{t}}\mathbb{W}\left(−\frac{\mathrm{ln}\:{t}}{{t}}\right) \\ $$$$\Rightarrow{general}\:{solution}\:\begin{cases}{{x}={t}}\\{{y}=−\frac{{t}}{\mathrm{ln}\:{t}}\mathbb{W}\left(−\frac{\mathrm{ln}\:{t}}{{t}}\right)}\end{cases} \\ $$$${METHOD}\:{II} \\ $$$${let}\:{y}={tx}\:{with}\:{t}\in{R}^{+} \\ $$$${x}^{{tx}} =\left({tx}\right)^{{x}} \\ $$$${x}^{{t}} ={tx} \\ $$$${x}^{{t}−\mathrm{1}} ={t} \\ $$$$\Rightarrow{x}={t}^{\frac{\mathrm{1}}{{t}−\mathrm{1}}} \\ $$$$\Rightarrow{y}={t}^{\mathrm{1}+\frac{\mathrm{1}}{{t}−\mathrm{1}}} ={t}^{\frac{{t}}{{t}−\mathrm{1}}} \\ $$$$\Rightarrow{general}\:{solution}\:\begin{cases}{{x}={t}^{\frac{\mathrm{1}}{{t}−\mathrm{1}}} }\\{{y}={t}^{\frac{{t}}{{t}−\mathrm{1}}} }\end{cases} \\ $$

Commented by 1549442205 last updated on 21/Jun/20

Thank you,sir.I like second way,it is explicit

$$\mathrm{Thank}\:\mathrm{you},\mathrm{sir}.\mathrm{I}\:\mathrm{like}\:\mathrm{second}\:\mathrm{way},\mathrm{it}\:\mathrm{is} \\ $$$$\mathrm{explicit} \\ $$

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