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Question Number 78974 by ~blr237~ last updated on 22/Jan/20
{_(3e^(2x) −y^2 =2 ) ^(e^x −y^2 =2lny−x) solve the system on R∗R
$$\left\{_{\mathrm{3e}^{\mathrm{2x}} −\mathrm{y}^{\mathrm{2}} =\mathrm{2}\:\:\:} ^{\mathrm{e}^{\mathrm{x}} −\mathrm{y}^{\mathrm{2}} =\mathrm{2lny}−\mathrm{x}} \right. \\ $$$$\mathrm{solve}\:\mathrm{the}\:\mathrm{system}\:\mathrm{on}\:\mathbb{R}\ast\mathbb{R} \\ $$
Commented by jagoll last updated on 22/Jan/20
3(e^x )^2 =2+y^2 ⇒e^x = (√((2+y^2 )/3)) x= ln((√((2+y^2 )/3))) (√((2+y^2 )/3))−y^2 = lny^2 −ln((√((2+y^2 )/3)))
$$\mathrm{3}\left(\mathrm{e}^{\mathrm{x}} \right)^{\mathrm{2}} =\mathrm{2}+\mathrm{y}^{\mathrm{2}} \Rightarrow\mathrm{e}^{\mathrm{x}} =\:\sqrt{\frac{\mathrm{2}+\mathrm{y}^{\mathrm{2}} }{\mathrm{3}}} \\ $$$$\mathrm{x}=\:\mathrm{ln}\left(\sqrt{\frac{\mathrm{2}+\mathrm{y}^{\mathrm{2}} }{\mathrm{3}}}\right) \\ $$$$\sqrt{\frac{\mathrm{2}+\mathrm{y}^{\mathrm{2}} }{\mathrm{3}}}−\mathrm{y}^{\mathrm{2}} =\:\mathrm{lny}^{\mathrm{2}} −\mathrm{ln}\left(\sqrt{\frac{\mathrm{2}+\mathrm{y}^{\mathrm{2}} }{\mathrm{3}}}\right) \\ $$
Commented by jagoll last updated on 22/Jan/20
let (√((2+y^2 )/3))=t ⇒y^2 = 3t^2 −2 t−(3t^2 −2)=ln(3t^2 −2)−ln(t)
$$\mathrm{let}\:\sqrt{\frac{\mathrm{2}+\mathrm{y}^{\mathrm{2}} }{\mathrm{3}}}=\mathrm{t}\:\Rightarrow\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{3t}^{\mathrm{2}} −\mathrm{2} \\ $$$$\mathrm{t}−\left(\mathrm{3t}^{\mathrm{2}} −\mathrm{2}\right)=\mathrm{ln}\left(\mathrm{3t}^{\mathrm{2}} −\mathrm{2}\right)−\mathrm{ln}\left(\mathrm{t}\right) \\ $$
Commented by MJS last updated on 22/Jan/20
give a few more examples where we can use this solving method (let me call it "finding a function") so we can practice and learn something. to post well constructed problems to test if somebody will find the key can be fun but I usually refuse to solve riddles.
Answered by MJS last updated on 22/Jan/20
x=0∧y=1
$${x}=\mathrm{0}\wedge{y}=\mathrm{1} \\ $$
Commented by jagoll last updated on 22/Jan/20
how to get the result sir?
$$\mathrm{how}\:\mathrm{to}\:\mathrm{get}\:\mathrm{the}\:\mathrm{result}\:\mathrm{sir}? \\ $$
Commented by mr W last updated on 22/Jan/20
such stupid equations can not be solved analytically. you should “see” the solution directly. when a value in the equation is changed, the solution can not be “seen” anymore. since the solution is not solved, but “seen”, you learn nothing through such questions. so just forget it, if you don′t “see” the solution at first glance.
$${such}\:{stupid}\:{equations}\:{can}\:{not}\:{be} \\ $$$${solved}\:{analytically}.\:{you}\:{should}\:“{see}'' \\ $$$${the}\:{solution}\:{directly}.\:{when}\:{a}\:{value}\:{in} \\ $$$${the}\:{equation}\:{is}\:{changed},\:{the}\:{solution} \\ $$$${can}\:{not}\:{be}\:“{seen}''\:{anymore}.\:{since}\:{the} \\ $$$${solution}\:{is}\:{not}\:{solved},\:{but}\:“{seen}'', \\ $$$${you}\:{learn}\:{nothing}\:{through}\:{such}\: \\ $$$${questions}.\:{so}\:{just}\:{forget}\:{it},\:{if}\:{you} \\ $$$${don}'{t}\:“{see}''\:{the}\:{solution}\:{at}\:{first}\:{glance}. \\ $$
Commented by mr W last updated on 22/Jan/20
this is the same as questions in which four numbers are given and you should find the fifth number. questions of this kind are games, not mathematics.
$${this}\:{is}\:{the}\:{same}\:{as}\:{questions}\:{in}\:{which} \\ $$$${four}\:{numbers}\:{are}\:{given}\:{and}\:{you}\:{should} \\ $$$${find}\:{the}\:{fifth}\:{number}.\:{questions}\:{of} \\ $$$${this}\:{kind}\:{are}\:{games},\:{not}\:{mathematics}. \\ $$
Answered by mind is power last updated on 22/Jan/20
y>0 ⇒2x=ln(((2+y^2 )/3)) x=(1/2)ln(((2+y^2 )/3)) ⇒first Equation⇔(√((2+y^2 )/3))−y^2 =2ln(y)−((ln(((2+y^2 )/3)))/2) ⇔2ln(y)+y^2 =(√((2+y^2 )/3))+ln((√((2+y^2 )/3))) ⇔ln(y^2 )+y^2 =(√((2+y^2 )/3))+ln((√((2+y^2 )/3)))...._ E let f(x)=ln(x)+x,x>0 E⇔f(y^2 )=f((√((2+y^2 )/3))) f′(x)=(1/x)+1>1 increase Function⇒Injective one ⇒f(y^2 )=f((√((2+y^2 )/3)))⇔y^2 =(√((2+y^2 )/3))⇔3y^4 −y^2 −2=0 y^2 ,root of 3X^2 −X−2=0 X∈{−(2/3),1},⇒y^2 =1⇒y=1 x=(1/2)ln(((2+y^2 )/3))=((ln((3/3)))/2)=0 ⇒(x,y)∈{(0,1)}
$${y}>\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}{x}={ln}\left(\frac{\mathrm{2}+{y}^{\mathrm{2}} }{\mathrm{3}}\right) \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{2}+{y}^{\mathrm{2}} }{\mathrm{3}}\right) \\ $$$$\Rightarrow{first}\:{Equation}\Leftrightarrow\sqrt{\frac{\mathrm{2}+{y}^{\mathrm{2}} }{\mathrm{3}}}−{y}^{\mathrm{2}} =\mathrm{2}{ln}\left({y}\right)−\frac{{ln}\left(\frac{\mathrm{2}+{y}^{\mathrm{2}} }{\mathrm{3}}\right)}{\mathrm{2}} \\ $$$$\Leftrightarrow\mathrm{2}{ln}\left({y}\right)+{y}^{\mathrm{2}} =\sqrt{\frac{\mathrm{2}+{y}^{\mathrm{2}} }{\mathrm{3}}}+{ln}\left(\sqrt{\frac{\mathrm{2}+{y}^{\mathrm{2}} }{\mathrm{3}}}\right) \\ $$$$\Leftrightarrow{ln}\left({y}^{\mathrm{2}} \right)+{y}^{\mathrm{2}} =\sqrt{\frac{\mathrm{2}+{y}^{\mathrm{2}} }{\mathrm{3}}}+{ln}\left(\sqrt{\frac{\mathrm{2}+{y}^{\mathrm{2}} }{\mathrm{3}}}\right)…._{} {E} \\ $$$${let}\:{f}\left({x}\right)={ln}\left({x}\right)+{x},{x}>\mathrm{0} \\ $$$${E}\Leftrightarrow{f}\left({y}^{\mathrm{2}} \right)={f}\left(\sqrt{\frac{\mathrm{2}+{y}^{\mathrm{2}} }{\mathrm{3}}}\right) \\ $$$${f}'\left({x}\right)=\frac{\mathrm{1}}{{x}}+\mathrm{1}>\mathrm{1}\:\:{increase}\:{Function}\Rightarrow{Injective}\:{one} \\ $$$$\Rightarrow{f}\left({y}^{\mathrm{2}} \right)={f}\left(\sqrt{\frac{\mathrm{2}+{y}^{\mathrm{2}} }{\mathrm{3}}}\right)\Leftrightarrow{y}^{\mathrm{2}} =\sqrt{\frac{\mathrm{2}+{y}^{\mathrm{2}} }{\mathrm{3}}}\Leftrightarrow\mathrm{3}{y}^{\mathrm{4}} −{y}^{\mathrm{2}} −\mathrm{2}=\mathrm{0} \\ $$$${y}^{\mathrm{2}} ,{root}\:{of}\:\mathrm{3}{X}^{\mathrm{2}} −{X}−\mathrm{2}=\mathrm{0} \\ $$$${X}\in\left\{−\frac{\mathrm{2}}{\mathrm{3}},\mathrm{1}\right\},\Rightarrow{y}^{\mathrm{2}} =\mathrm{1}\Rightarrow{y}=\mathrm{1} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{2}+{y}^{\mathrm{2}} }{\mathrm{3}}\right)=\frac{{ln}\left(\frac{\mathrm{3}}{\mathrm{3}}\right)}{\mathrm{2}}=\mathrm{0} \\ $$$$\Rightarrow\left({x},{y}\right)\in\left\{\left(\mathrm{0},\mathrm{1}\right)\right\} \\ $$$$ \\ $$
Commented by ~blr237~ last updated on 22/Jan/20
Thank sir .
$$\mathrm{Thank}\:\mathrm{sir}\:. \\ $$
Commented by mind is power last updated on 22/Jan/20
y′re Welcom
$${y}'{re}\:{Welcom} \\ $$
Answered by ~blr237~ last updated on 22/Jan/20
with the first equation you reach to ( on the strain y>0) e^x +x=lny^2 +e^(lny^2 ) ⇔ f(x)=f(lny^2 ) with f(t)=e^t +t f is continue and strictly increasing on R ( f′(t)=e^t +1 >0 ) so f is injective ( cause also a bijection from R→R ) then x=lny^2 or y^2 =e^x when replacing on the second we have 3(e^x )^2 −e^x −2=0 then e^x =1 or e^x =−(2/3) finally x=0 and y=1 cause the strain was y>0
$$\mathrm{with}\:\mathrm{the}\:\mathrm{first}\:\mathrm{equation}\:\mathrm{you}\:\mathrm{reach}\:\mathrm{to}\:\left(\:\mathrm{on}\:\mathrm{the}\:\mathrm{strain}\:\mathrm{y}>\mathrm{0}\right) \\ $$$$\mathrm{e}^{\mathrm{x}} +\mathrm{x}=\mathrm{lny}^{\mathrm{2}} +\mathrm{e}^{\mathrm{lny}^{\mathrm{2}} } \:\:\Leftrightarrow\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{lny}^{\mathrm{2}} \right)\:\:\mathrm{with}\:\mathrm{f}\left(\mathrm{t}\right)=\mathrm{e}^{\mathrm{t}} +\mathrm{t} \\ $$$$\mathrm{f}\:\mathrm{is}\:\mathrm{continue}\:\mathrm{and}\:\mathrm{strictly}\:\mathrm{increasing}\:\mathrm{on}\:\mathbb{R}\:\left(\:\mathrm{f}'\left(\mathrm{t}\right)=\mathrm{e}^{\mathrm{t}} +\mathrm{1}\:>\mathrm{0}\:\right) \\ $$$$\mathrm{so}\:\mathrm{f}\:\mathrm{is}\:\mathrm{injective}\:\left(\:\mathrm{cause}\:\mathrm{also}\:\mathrm{a}\:\mathrm{bijection}\:\mathrm{from}\:\mathbb{R}\rightarrow\mathbb{R}\:\:\right) \\ $$$$\mathrm{then}\:\:\mathrm{x}=\mathrm{lny}^{\mathrm{2}} \:\:\mathrm{or}\:\mathrm{y}^{\mathrm{2}} =\mathrm{e}^{\mathrm{x}} \\ $$$$\mathrm{when}\:\mathrm{replacing}\:\mathrm{on}\:\mathrm{the}\:\mathrm{second}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{3}\left(\mathrm{e}^{\mathrm{x}} \right)^{\mathrm{2}} −\mathrm{e}^{\mathrm{x}} −\mathrm{2}=\mathrm{0}\: \\ $$$$\mathrm{then}\:\:\mathrm{e}^{\mathrm{x}} =\mathrm{1}\:\mathrm{or}\:\mathrm{e}^{\mathrm{x}} =−\frac{\mathrm{2}}{\mathrm{3}}\: \\ $$$$\mathrm{finally}\:\:\mathrm{x}=\mathrm{0}\:\mathrm{and}\:\mathrm{y}=\mathrm{1}\:\mathrm{cause}\:\mathrm{the}\:\mathrm{strain}\:\mathrm{was}\:\mathrm{y}>\mathrm{0} \\ $$$$ \\ $$

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