Menu Close

x-y-y-x-x-y-




Question Number 113954 by Khalmohmmad last updated on 16/Sep/20
x^y =y^x        { (x),(y) :}=?
$${x}^{{y}} ={y}^{{x}} \:\:\:\: \\ $$$$\begin{cases}{\mathrm{x}}\\{\mathrm{y}}\end{cases}=? \\ $$
Commented by Dwaipayan Shikari last updated on 16/Sep/20
when x=y  x,y∈Z
$${when}\:{x}={y} \\ $$$${x},{y}\in\mathbb{Z} \\ $$
Answered by $@y@m last updated on 16/Sep/20
2,4
$$\mathrm{2},\mathrm{4} \\ $$
Answered by Dwaipayan Shikari last updated on 16/Sep/20
(x/(logx))=(y/(logy))=k  (k≠0)   (x^y =y^x ⇒xlogy=ylogx)  x=klogx  e^(logx) =klogx  e^(−logx) =(1/(klogx))  −logxe^(−logx) =−(1/k)  −logx=W_0 (−(1/k))  x=e^(−W_0 (−(1/k)))   y=e^(−W_0 (−(1/k)))   k∈(any number)(k≠0)
$$\frac{{x}}{{logx}}=\frac{{y}}{{logy}}={k}\:\:\left({k}\neq\mathrm{0}\right)\:\:\:\left({x}^{{y}} ={y}^{{x}} \Rightarrow{xlogy}={ylogx}\right) \\ $$$${x}={klogx} \\ $$$${e}^{{logx}} ={klogx} \\ $$$${e}^{−{logx}} =\frac{\mathrm{1}}{{klogx}} \\ $$$$−{logxe}^{−{logx}} =−\frac{\mathrm{1}}{{k}} \\ $$$$−{logx}={W}_{\mathrm{0}} \left(−\frac{\mathrm{1}}{{k}}\right) \\ $$$${x}={e}^{−{W}_{\mathrm{0}} \left(−\frac{\mathrm{1}}{{k}}\right)} \\ $$$${y}={e}^{−{W}_{\mathrm{0}} \left(−\frac{\mathrm{1}}{{k}}\right)} \\ $$$${k}\in\left({any}\:{number}\right)\left({k}\neq\mathrm{0}\right) \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *