{_(3e^(2x) −y^2 =2 ) ^(e^x −y^2 =2lny−x) solve the system on R∗R


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Question Number 78974 by ~blr237~ last updated on 22/Jan/20
${_(3e^(2x) −y^2 =2 ) ^(e^x −y^2 =2lny−x) solve the system on R∗R$
${_{{3 e}^{2 x} - y^{2} = 2}^{e^{x} - y^{2} = 2 lny - x}$ $solve the system on R * R$
Commented by jagoll last updated on 22/Jan/20

$3 {(e^{x})}^{2} = 2 + y^{2} \Rightarrow e^{x} = \sqrt{\frac{2 + y^{2}}{3}}$ $x = \ln (\sqrt{\frac{2 + y^{2}}{3}})$ $\sqrt{\frac{2 + y^{2}}{3}} - y^{2} = {lny}^{2} - \ln (\sqrt{\frac{2 + y^{2}}{3}})$
Commented by jagoll last updated on 22/Jan/20

$let \sqrt{\frac{2 + y^{2}}{3}} = t \Rightarrow y^{2} = {3 t}^{2} - 2$ $t - ({3 t}^{2} - 2) = \ln ({3 t}^{2} - 2) - \ln (t)$
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

	$x = 0 \land y = 1$
	Commented by jagoll last updated on 22/Jan/20

	$how to get the result sir ?$
	Commented by mr W last updated on 22/Jan/20

	$s u c h s t u p i d e q u a t i o n s c a n n o t b e$ $s o l v e d a n a l y t i c a l l y . y o u s h o u l d ‘ ‘ {s e e}^{″}$ $t h e s o l u t i o n d i r e c t l y . w h e n a v a l u e i n$ $t h e e q u a t i o n i s c h a n g e d, t h e s o l u t i o n$ $c a n n o t b e ‘ ‘ {s e e n}^{″} a n y m o r e . s i n c e t h e$ $s o l u t i o n i s n o t s o l v e d, b u t ‘ ‘ {s e e n}^{″},$ $y o u l e a r n n o t h i n g t h r o u g h s u c h$ $q u e s t i o n s . s o j u s t f o r g e t i t, i f y o u$ ${d o n}^{'} t ‘ ‘ {s e e}^{″} t h e s o l u t i o n a t f i r s t g l a n c e .$
	Commented by mr W last updated on 22/Jan/20

	$t h i s i s t h e s a m e a s q u e s t i o n s i n w h i c h$ $f o u r n u m b e r s a r e g i v e n a n d y o u s h o u l d$ $f i n d t h e f i f t h n u m b e r . q u e s t i o n s o f$ $t h i s k i n d a r e g a m e s, n o t m a t h e m a t i c s .$
	Answered by mind is power last updated on 22/Jan/20
	$y>0 ⇒2x=ln(((2+y^2 )/3)) x=(1/2)ln(((2+y^2 )/3)) ⇒first Equation⇔(√((2+y^2 )/3))−y^2 =2ln(y)−((ln(((2+y^2 )/3)))/2) ⇔2ln(y)+y^2 =(√((2+y^2 )/3))+ln((√((2+y^2 )/3))) ⇔ln(y^2 )+y^2 =(√((2+y^2 )/3))+ln((√((2+y^2 )/3)))...._ E let f(x)=ln(x)+x,x>0 E⇔f(y^2 )=f((√((2+y^2 )/3))) f′(x)=(1/x)+1>1 increase Function⇒Injective one ⇒f(y^2 )=f((√((2+y^2 )/3)))⇔y^2 =(√((2+y^2 )/3))⇔3y^4 −y^2 −2=0 y^2 ,root of 3X^2 −X−2=0 X∈{−(2/3),1},⇒y^2 =1⇒y=1 x=(1/2)ln(((2+y^2 )/3))=((ln((3/3)))/2)=0 ⇒(x,y)∈{(0,1)}$
	$y > 0$ $\Rightarrow 2 x = l n (\frac{2 + y^{2}}{3})$ $x = \frac{1}{2} l n (\frac{2 + y^{2}}{3})$ $\Rightarrow f i r s t E q u a t i o n \Leftrightarrow \sqrt{\frac{2 + y^{2}}{3}} - y^{2} = 2 l n (y) - \frac{l n (\frac{2 + y^{2}}{3})}{2}$ $\Leftrightarrow 2 l n (y) + y^{2} = \sqrt{\frac{2 + y^{2}}{3}} + l n (\sqrt{\frac{2 + y^{2}}{3}})$ $\Leftrightarrow l n (y^{2}) + y^{2} = \sqrt{\frac{2 + y^{2}}{3}} + l n (\sqrt{\frac{2 + y^{2}}{3}}) . . . ._{} E$ $l e t f (x) = l n (x) + x, x > 0$ $E \Leftrightarrow f (y^{2}) = f (\sqrt{\frac{2 + y^{2}}{3}})$ $f^{'} (x) = \frac{1}{x} + 1 > 1 i n c r e a s e F u n c t i o n \Rightarrow I n j e c t i v e o n e$ $\Rightarrow f (y^{2}) = f (\sqrt{\frac{2 + y^{2}}{3}}) \Leftrightarrow y^{2} = \sqrt{\frac{2 + y^{2}}{3}} \Leftrightarrow 3 y^{4} - y^{2} - 2 = 0$ $y^{2}, r o o t o f 3 X^{2} - X - 2 = 0$ $X \in {- \frac{2}{3}, 1}, \Rightarrow y^{2} = 1 \Rightarrow y = 1$ $x = \frac{1}{2} l n (\frac{2 + y^{2}}{3}) = \frac{l n (\frac{3}{3})}{2} = 0$ $\Rightarrow (x, y) \in {(0, 1)}$
	Commented by ~blr237~ last updated on 22/Jan/20

	$Thank sir .$
	Commented by mind is power last updated on 22/Jan/20

	$y^{'} r e W e l c o m$
	Answered by ~blr237~ last updated on 22/Jan/20

	$with the first equation you reach to (on the strain y > 0)$ $e^{x} + x = {lny}^{2} + e^{{lny}^{2}} \Leftrightarrow f (x) = f ({lny}^{2}) with f (t) = e^{t} + t$ $f is continue and strictly increasing on R (f^{'} (t) = e^{t} + 1 > 0)$ $so f is injective (cause also a bijection from R \to R)$ $then x = {lny}^{2} or y^{2} = e^{x}$ $when replacing on the second we have$ $3 {(e^{x})}^{2} - e^{x} - 2 = 0$ $then e^{x} = 1 or e^{x} = - \frac{2}{3}$ $finally x = 0 and y = 1 cause the strain was y > 0$
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