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Question Number 196165 by mathlove last updated on 19/Aug/23
{: ((x^(x+y) =y^(24) )),((y^(x+y) =x^6 )) }⇒(x,y)=(?,?)
$$\left.\begin{matrix}{{x}^{{x}+{y}} ={y}^{\mathrm{24}} }\\{{y}^{{x}+{y}} ={x}^{\mathrm{6}} }\end{matrix}\right\}\Rightarrow\left({x},{y}\right)=\left(?,?\right) \\ $$
Answered by liuxinnan last updated on 19/Aug/23
Answered by Frix last updated on 19/Aug/23
Obviously x+y=±12 We then get 1. x^(−12) =(x+12)^(24) ∧(x+12)^(−12) =x^6 ∨ 2. x^(12) =(x−12)^(24) ∧(x−12)^(12) =x^6 ⇒ 1.1. x^(−1) =−(x+12)^2 x^3 +24x^2 +144x+1=0 x≈−12.2853∨x≈−11.7077∨x≈−.00695250 1.2. x^(−1) =(x+12)^2 x^3 +24x^2 +144x−1=0 x≈.00693642∨x≈−12.0035±.288571i 2.1. x=−(x−12)^2 x^2 −23x+144=0 x=((23)/2)±((√(47))/2)i 2.2. x=(x−12)^2 x^2 −25x+144=0 x=9∨x=16 Solve these for x, then get y. All real and complex solution pairs are valid. The “nice” solutions are 2.2. x=9∧y=3 ∨ x=16∧y=−4
$$\mathrm{Obviously}\:{x}+{y}=\pm\mathrm{12} \\ $$$$\mathrm{We}\:\mathrm{then}\:\mathrm{get} \\ $$$$\mathrm{1}.\:{x}^{−\mathrm{12}} =\left({x}+\mathrm{12}\right)^{\mathrm{24}} \wedge\left({x}+\mathrm{12}\right)^{−\mathrm{12}} ={x}^{\mathrm{6}} \\ $$$$\vee \\ $$$$\mathrm{2}.\:{x}^{\mathrm{12}} =\left({x}−\mathrm{12}\right)^{\mathrm{24}} \wedge\left({x}−\mathrm{12}\right)^{\mathrm{12}} ={x}^{\mathrm{6}} \\ $$$$\Rightarrow \\ $$$$\mathrm{1}.\mathrm{1}.\:{x}^{−\mathrm{1}} =−\left({x}+\mathrm{12}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{3}} +\mathrm{24}{x}^{\mathrm{2}} +\mathrm{144}{x}+\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:{x}\approx−\mathrm{12}.\mathrm{2853}\vee{x}\approx−\mathrm{11}.\mathrm{7077}\vee{x}\approx−.\mathrm{00695250} \\ $$$$\mathrm{1}.\mathrm{2}.\:{x}^{−\mathrm{1}} =\left({x}+\mathrm{12}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{3}} +\mathrm{24}{x}^{\mathrm{2}} +\mathrm{144}{x}−\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:{x}\approx.\mathrm{00693642}\vee{x}\approx−\mathrm{12}.\mathrm{0035}\pm.\mathrm{288571i} \\ $$$$\mathrm{2}.\mathrm{1}.\:{x}=−\left({x}−\mathrm{12}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} −\mathrm{23}{x}+\mathrm{144}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:{x}=\frac{\mathrm{23}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{47}}}{\mathrm{2}}\mathrm{i} \\ $$$$\mathrm{2}.\mathrm{2}.\:{x}=\left({x}−\mathrm{12}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} −\mathrm{25}{x}+\mathrm{144}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:{x}=\mathrm{9}\vee{x}=\mathrm{16} \\ $$$$ \\ $$$$\mathrm{Solve}\:\mathrm{these}\:\mathrm{for}\:{x},\:\mathrm{then}\:\mathrm{get}\:{y}.\:\mathrm{All}\:\mathrm{real}\:\mathrm{and} \\ $$$$\mathrm{complex}\:\mathrm{solution}\:\mathrm{pairs}\:\mathrm{are}\:\mathrm{valid}. \\ $$$$\mathrm{The}\:“\mathrm{nice}''\:\mathrm{solutions}\:\mathrm{are} \\ $$$$\mathrm{2}.\mathrm{2}.\:{x}=\mathrm{9}\wedge{y}=\mathrm{3}\:\vee\:{x}=\mathrm{16}\wedge{y}=−\mathrm{4} \\ $$
Commented by mathlove last updated on 19/Aug/23
thanks
$${thanks} \\ $$
Answered by mr W last updated on 19/Aug/23
(x+y)ln x=24ln y ...(i) (x+y)ln y=6ln x ...(ii) (i)×(ii): (x+y)^2 =6×24 ⇒x+y=±12 case 1: x+y=12 ln x=2ln y ⇒x=y^2 ⇒y^2 +y−12=0 ⇒y=−4, 3 ⇒x=16, 9 case 2: x+y=−12 ⇒(1/x)=y^2 (1/y^2 )+y+12=0 (1/y)=((((√(257))−1)/2))^(1/3) −((((√(257))+1)/2))^(1/3) y=(1/( ((((√(257))−1)/2))^(1/3) −((((√(257))+1)/2))^(1/3) )) x=−12−(1/( ((((√(257))−1)/2))^(1/3) −((((√(257))+1)/2))^(1/3) )) summary: (x,y)=(16,−4),(9,3),(−12−(1/( ((((√(257))−1)/2))^(1/3) −((((√(257))+1)/2))^(1/3) )),(1/( ((((√(257))−1)/2))^(1/3) −((((√(257))+1)/2))^(1/3) )))
$$\left({x}+{y}\right)\mathrm{ln}\:{x}=\mathrm{24ln}\:{y}\:\:\:…\left({i}\right) \\ $$$$\left({x}+{y}\right)\mathrm{ln}\:{y}=\mathrm{6ln}\:{x}\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)×\left({ii}\right): \\ $$$$\left({x}+{y}\right)^{\mathrm{2}} =\mathrm{6}×\mathrm{24} \\ $$$$\Rightarrow{x}+{y}=\pm\mathrm{12} \\ $$$${case}\:\mathrm{1}:\:{x}+{y}=\mathrm{12} \\ $$$$\mathrm{ln}\:{x}=\mathrm{2ln}\:{y} \\ $$$$\Rightarrow{x}={y}^{\mathrm{2}} \\ $$$$\Rightarrow{y}^{\mathrm{2}} +{y}−\mathrm{12}=\mathrm{0} \\ $$$$\Rightarrow{y}=−\mathrm{4},\:\mathrm{3}\:\Rightarrow{x}=\mathrm{16},\:\mathrm{9} \\ $$$${case}\:\mathrm{2}:\:{x}+{y}=−\mathrm{12} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{x}}={y}^{\mathrm{2}} \\ $$$$\:\frac{\mathrm{1}}{{y}^{\mathrm{2}} }+{y}+\mathrm{12}=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{{y}}=\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{257}}−\mathrm{1}}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{257}}+\mathrm{1}}{\mathrm{2}}} \\ $$$${y}=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{257}}−\mathrm{1}}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{257}}+\mathrm{1}}{\mathrm{2}}}} \\ $$$${x}=−\mathrm{12}−\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{257}}−\mathrm{1}}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{257}}+\mathrm{1}}{\mathrm{2}}}} \\ $$$${summary}: \\ $$$$\left({x},{y}\right)=\left(\mathrm{16},−\mathrm{4}\right),\left(\mathrm{9},\mathrm{3}\right),\left(−\mathrm{12}−\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{257}}−\mathrm{1}}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{257}}+\mathrm{1}}{\mathrm{2}}}},\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{257}}−\mathrm{1}}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{257}}+\mathrm{1}}{\mathrm{2}}}}\right) \\ $$
Commented by Frix last updated on 19/Aug/23
1. Typo in last line: (16, −4) 2. The root of the 3^(rd) degree is less complicated like this: x=−8+(((129+(√(257)))/2))^(1/3) +(((129−(√(257)))/2))^(1/3) 3. The complex solutions are also valid.
$$\mathrm{1}.\:\mathrm{Typo}\:\mathrm{in}\:\mathrm{last}\:\mathrm{line}:\:\left(\mathrm{16},\:−\mathrm{4}\right) \\ $$$$\mathrm{2}.\:\mathrm{The}\:\mathrm{root}\:\mathrm{of}\:\mathrm{the}\:\mathrm{3}^{\mathrm{rd}} \:\mathrm{degree}\:\mathrm{is}\:\mathrm{less} \\ $$$$\mathrm{complicated}\:\mathrm{like}\:\mathrm{this}: \\ $$$${x}=−\mathrm{8}+\sqrt[{\mathrm{3}}]{\frac{\mathrm{129}+\sqrt{\mathrm{257}}}{\mathrm{2}}}+\sqrt[{\mathrm{3}}]{\frac{\mathrm{129}−\sqrt{\mathrm{257}}}{\mathrm{2}}} \\ $$$$\mathrm{3}.\:\mathrm{The}\:\mathrm{complex}\:\mathrm{solutions}\:\mathrm{are}\:\mathrm{also}\:\mathrm{valid}. \\ $$
Commented by mathlove last updated on 20/Aug/23
thanks
$${thanks} \\ $$

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