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Question Number 50064 by mondodotto@gmail.com last updated on 13/Dec/18
if y^x =x^y and y and x are both not equal to 0 and l find the value of x and y
$$\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{y}}^{\boldsymbol{\mathrm{x}}} =\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{y}}} \: \\ $$$$\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{y}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{x}}\:\boldsymbol{\mathrm{are}}\:\boldsymbol{\mathrm{both}}\:\boldsymbol{\mathrm{not}} \\ $$$$\boldsymbol{\mathrm{equal}}\:\boldsymbol{\mathrm{to}}\:\mathrm{0}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{l}} \\ $$$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{x}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{y}} \\ $$
Answered by Necxx last updated on 13/Dec/18
x=y=2=4 from omega function
$${x}={y}=\mathrm{2}=\mathrm{4}\:{from}\:{omega}\:{function} \\ $$
Answered by mr W last updated on 14/Dec/18
there are infinite solutions. x=y=a is a solution with a∈R ∪ a>0 x=a y^a =a^y y=a^(y/a) y=e^((ln a)(y/a)) ye^(−(ln a)(y/a)) =1 −(ln a)(y/a)e^(−(ln a)(y/a)) =−((ln a)/a) −(ln a)(y/a)=W(−((ln a)/a)) (Lambert W function) ⇒y=−(a/(ln a))W(−((ln a)/a)) ⇒solution is { ((x=a, a∈R and a>0)),((y=a)) :} { ((x=a, a∈R and a>1)),((y=−(a/(ln a))W(−((ln a)/a)))) :} examples: x=(1/2): y=(1/(2ln 2))W(2ln 2)=((0.6931)/(2ln 2))=0.5 x=2: y=−(2/(ln 2))W(−((ln 2)/2))= { ((−(2/(ln 2))×(−0.6931)=2)),((−(2/(ln 2))×(−1.3863)=4)) :} x=5: y=−(5/(ln 5))W(−((ln 5)/5))= { ((−(5/(ln 5))×(−0.5681)=1.7649)),((−(5/(ln 5))×(−1.6094)=5)) :}
$${there}\:{are}\:{infinite}\:{solutions}. \\ $$$$ \\ $$$${x}={y}={a}\:{is}\:{a}\:{solution}\:{with}\:{a}\in{R}\:\cup\:{a}>\mathrm{0} \\ $$$$ \\ $$$${x}={a} \\ $$$${y}^{{a}} ={a}^{{y}} \\ $$$${y}={a}^{\frac{{y}}{{a}}} \\ $$$${y}={e}^{\left(\mathrm{ln}\:{a}\right)\frac{{y}}{{a}}} \\ $$$${ye}^{−\left(\mathrm{ln}\:{a}\right)\frac{{y}}{{a}}} =\mathrm{1} \\ $$$$−\left(\mathrm{ln}\:{a}\right)\frac{{y}}{{a}}{e}^{−\left(\mathrm{ln}\:{a}\right)\frac{{y}}{{a}}} =−\frac{\mathrm{ln}\:{a}}{{a}} \\ $$$$−\left(\mathrm{ln}\:{a}\right)\frac{{y}}{{a}}={W}\left(−\frac{\mathrm{ln}\:{a}}{{a}}\right)\:\left({Lambert}\:{W}\:{function}\right) \\ $$$$\Rightarrow{y}=−\frac{{a}}{\mathrm{ln}\:{a}}{W}\left(−\frac{\mathrm{ln}\:{a}}{{a}}\right) \\ $$$$ \\ $$$$\Rightarrow{solution}\:{is} \\ $$$$\begin{cases}{{x}={a},\:{a}\in{R}\:{and}\:{a}>\mathrm{0}}\\{{y}={a}}\end{cases} \\ $$$$\begin{cases}{{x}={a},\:{a}\in{R}\:{and}\:{a}>\mathrm{1}}\\{{y}=−\frac{{a}}{\mathrm{ln}\:{a}}{W}\left(−\frac{\mathrm{ln}\:{a}}{{a}}\right)}\end{cases} \\ $$$$ \\ $$$${examples}: \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{2}}: \\ $$$${y}=\frac{\mathrm{1}}{\mathrm{2ln}\:\mathrm{2}}{W}\left(\mathrm{2ln}\:\mathrm{2}\right)=\frac{\mathrm{0}.\mathrm{6931}}{\mathrm{2ln}\:\mathrm{2}}=\mathrm{0}.\mathrm{5} \\ $$$$ \\ $$$${x}=\mathrm{2}: \\ $$$${y}=−\frac{\mathrm{2}}{\mathrm{ln}\:\mathrm{2}}{W}\left(−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}\right)=\begin{cases}{−\frac{\mathrm{2}}{\mathrm{ln}\:\mathrm{2}}×\left(−\mathrm{0}.\mathrm{6931}\right)=\mathrm{2}}\\{−\frac{\mathrm{2}}{\mathrm{ln}\:\mathrm{2}}×\left(−\mathrm{1}.\mathrm{3863}\right)=\mathrm{4}}\end{cases} \\ $$$$ \\ $$$${x}=\mathrm{5}: \\ $$$${y}=−\frac{\mathrm{5}}{\mathrm{ln}\:\mathrm{5}}{W}\left(−\frac{\mathrm{ln}\:\mathrm{5}}{\mathrm{5}}\right)=\begin{cases}{−\frac{\mathrm{5}}{\mathrm{ln}\:\mathrm{5}}×\left(−\mathrm{0}.\mathrm{5681}\right)=\mathrm{1}.\mathrm{7649}}\\{−\frac{\mathrm{5}}{\mathrm{ln}\:\mathrm{5}}×\left(−\mathrm{1}.\mathrm{6094}\right)=\mathrm{5}}\end{cases} \\ $$
Commented by Pk1167156@gmail.com last updated on 14/Dec/18
you are right sir.
Commented by Pk1167156@gmail.com last updated on 14/Dec/18
but only for x=y
Commented by mr W last updated on 14/Dec/18
why did you say only for x=y? as I have proved: for 0<x≤1 there is only one value for y, and it is y=x. but for x>1, there are always^(∗)) two values for y, they are y=−(x/(ln x))W(−((ln x)/x)), this W−function delivers two results, i.e. one value of y is x, the other one is not x. for example if x=2, y=−(2/(ln 2))W(−((ln 2)/2))= { (2),(4) :} so we have two solutions: (2,2) and (2,4). you can check: 2^2 =2^2 and 2^4 =4^2 . ^(∗)) exception: when ((ln x)/x)=(1/e) or x=e, y=−(x/(ln x))W(−((ln x)/x)) has only one value: y=e=x.
$${why}\:{did}\:{you}\:{say}\:{only}\:{for}\:{x}={y}? \\ $$$${as}\:{I}\:{have}\:{proved}: \\ $$$${for}\:\mathrm{0}<{x}\leqslant\mathrm{1}\:{there}\:{is}\:{only}\:{one}\:{value}\:{for}\:{y},\: \\ $$$${and}\:{it}\:{is}\:{y}={x}. \\ $$$${but}\:{for}\:{x}>\mathrm{1},\:{there}\:{are}\:{always}^{\left.\ast\right)} \:{two}\:{values} \\ $$$${for}\:{y},\:{they}\:{are}\:{y}=−\frac{{x}}{\mathrm{ln}\:{x}}{W}\left(−\frac{\mathrm{ln}\:{x}}{{x}}\right),\:{this} \\ $$$${W}−{function}\:{delivers}\:{two}\:\:{results},\:{i}.{e}. \\ $$$${one}\:{value}\:{of}\:{y}\:{is}\:{x},\:{the}\:{other}\:{one}\:{is}\:{not}\:{x}. \\ $$$${for}\:{example}\:{if}\:{x}=\mathrm{2},\:{y}=−\frac{\mathrm{2}}{\mathrm{ln}\:\mathrm{2}}{W}\left(−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}\right)=\begin{cases}{\mathrm{2}}\\{\mathrm{4}}\end{cases} \\ $$$${so}\:{we}\:{have}\:{two}\:{solutions}:\:\left(\mathrm{2},\mathrm{2}\right)\:{and}\: \\ $$$$\left(\mathrm{2},\mathrm{4}\right).\:{you}\:{can}\:{check}: \\ $$$$\mathrm{2}^{\mathrm{2}} =\mathrm{2}^{\mathrm{2}} \:{and}\:\mathrm{2}^{\mathrm{4}} =\mathrm{4}^{\mathrm{2}} . \\ $$$$ \\ $$$$\:^{\left.\ast\right)} {exception}:\:{when}\:\frac{\mathrm{ln}\:{x}}{{x}}=\frac{\mathrm{1}}{{e}}\:{or}\:{x}={e}, \\ $$$${y}=−\frac{{x}}{\mathrm{ln}\:{x}}{W}\left(−\frac{\mathrm{ln}\:{x}}{{x}}\right)\:{has}\:{only}\:{one}\:{value}: \\ $$$${y}={e}={x}. \\ $$
Commented by mondodotto@gmail.com last updated on 14/Dec/18
thank you
$$\mathrm{thank}\:\mathrm{you} \\ $$
Answered by mr W last updated on 15/Dec/18
an other way without using Lambert function the solution x=y=any real number >0 is clear. we only need to find the solution for x≠y. let y=λx (λx)^x =x^((λx)) λx=x^λ λ=x^(λ−1) ⇒x=λ^(1/(λ−1)) y=λ×λ^(1/(λ−1)) =λ^(1+(1/(λ−1))) ⇒y=λ^(λ/(λ−1)) i.e. the general solution is { ((x=λ^(1/(λ−1)) )),((y=λ^(λ/(λ−1)) )) :} with λ>1 examples: λ=2⇒x=2, y=4 λ=3⇒x=(√3), y=3(√3) λ=4⇒x=^3 (√4), y=4^3 (√4) ......
$${an}\:{other}\:{way}\:{without}\:{using}\:{Lambert} \\ $$$${function} \\ $$$${the}\:{solution}\:{x}={y}={any}\:{real}\:{number}\:>\mathrm{0} \\ $$$${is}\:{clear}.\:{we}\:{only}\:{need}\:{to}\:{find}\:{the} \\ $$$${solution}\:{for}\:{x}\neq{y}. \\ $$$${let}\:{y}=\lambda{x} \\ $$$$\left(\lambda{x}\right)^{{x}} ={x}^{\left(\lambda{x}\right)} \\ $$$$\lambda{x}={x}^{\lambda} \\ $$$$\lambda={x}^{\lambda−\mathrm{1}} \\ $$$$\Rightarrow{x}=\lambda^{\frac{\mathrm{1}}{\lambda−\mathrm{1}}} \\ $$$${y}=\lambda×\lambda^{\frac{\mathrm{1}}{\lambda−\mathrm{1}}} =\lambda^{\mathrm{1}+\frac{\mathrm{1}}{\lambda−\mathrm{1}}} \\ $$$$\Rightarrow{y}=\lambda^{\frac{\lambda}{\lambda−\mathrm{1}}} \\ $$$${i}.{e}.\:{the}\:{general}\:{solution}\:{is} \\ $$$$\begin{cases}{{x}=\lambda^{\frac{\mathrm{1}}{\lambda−\mathrm{1}}} }\\{{y}=\lambda^{\frac{\lambda}{\lambda−\mathrm{1}}} }\end{cases}\:\:\:{with}\:\lambda>\mathrm{1} \\ $$$${examples}: \\ $$$$\lambda=\mathrm{2}\Rightarrow{x}=\mathrm{2},\:{y}=\mathrm{4} \\ $$$$\lambda=\mathrm{3}\Rightarrow{x}=\sqrt{\mathrm{3}},\:{y}=\mathrm{3}\sqrt{\mathrm{3}} \\ $$$$\lambda=\mathrm{4}\Rightarrow{x}=\:^{\mathrm{3}} \sqrt{\mathrm{4}},\:{y}=\mathrm{4}\:^{\mathrm{3}} \sqrt{\mathrm{4}} \\ $$$$…… \\ $$

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