3e-2x-y-2-2-e-x-y-2-2lny-x-solve-the-system-on-R-R-

Question Number 78974 by ~blr237~ last updated on 22/Jan/20

$$\{{}_{{3\mathrm{e}}^{2\mathrm{x}}-{\mathrm{y}}^2=2}^{{\mathrm{e}}^{\mathrm{x}}-{\mathrm{y}}^2=2\mathrm{lny}-\mathrm{x}}$$$$\mathrm{solve}\mathrm{the}\mathrm{system}\mathrm{on}\mathbb{R}\ast \mathbb{R}$$

Commented by jagoll last updated on 22/Jan/20

$$3{\left({\mathrm{e}}^{\mathrm{x}}\right)}^2=2+{\mathrm{y}}^2\Rightarrow {\mathrm{e}}^{\mathrm{x}}=\sqrt{\frac{2+{\mathrm{y}}^2}{3}}$$$$\mathrm{x}=\ln \left(\sqrt{\frac{2+{\mathrm{y}}^2}{3}}\right)$$$$\sqrt{\frac{2+{\mathrm{y}}^2}{3}}-{\mathrm{y}}^2={\mathrm{lny}}^2-\ln \left(\sqrt{\frac{2+{\mathrm{y}}^2}{3}}\right)$$

Commented by jagoll last updated on 22/Jan/20

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

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

$$x=0\wedge y=1$$

Commented by jagoll last updated on 22/Jan/20

$$\mathrm{how}\mathrm{to}\mathrm{get}\mathrm{the}\mathrm{result}\mathrm{sir}?$$

Commented by mr W last updated on 22/Jan/20

$$suchstupidequationscannotbe$$$$solvedanalytically.youshould“{see}^{″}$$$$thesolutiondirectly.whenavaluein$$$$theequationischanged,thesolution$$$$cannotbe“{seen}^{″}anymore.sincethe$$$$solutionisnotsolved,but“{seen}^{″},$$$$youlearnnothingthroughsuch$$$$questions.sojustforgetit,ifyou$$$${don}^{\prime }t“{see}^{″}thesolutionatfirstglance.$$

Commented by mr W last updated on 22/Jan/20

$$thisisthesameasquestionsinwhich$$$$fournumbersaregivenandyoushould$$$$findthefifthnumber.questionsof$$$$thiskindaregames,notmathematics.$$

Answered by mind is power last updated on 22/Jan/20

$$y>0$$$$\Rightarrow 2x=ln\left(\frac{2+y^2}{3}\right)$$$$x=\frac{1}{2}ln\left(\frac{2+y^2}{3}\right)$$$$\Rightarrow firstEquation\iff \sqrt{\frac{2+y^2}{3}}-y^2=2ln\left(y\right)-\frac{ln\left(\frac{2+y^2}{3}\right)}{2}$$$$\iff 2ln\left(y\right)+y^2=\sqrt{\frac{2+y^2}{3}}+ln\left(\sqrt{\frac{2+y^2}{3}}\right)$$$$\iff ln\left(y^2\right)+y^2=\sqrt{\frac{2+y^2}{3}}+ln\left(\sqrt{\frac{2+y^2}{3}}\right)\dots {.}_{}E$$$$letf\left(x\right)=ln\left(x\right)+x,x>0$$$$E\iff f\left(y^2\right)=f\left(\sqrt{\frac{2+y^2}{3}}\right)$$$$f^{\prime }\left(x\right)=\frac{1}{x}+1>1increaseFunction\Rightarrow Injectiveone$$$$\Rightarrow f\left(y^2\right)=f\left(\sqrt{\frac{2+y^2}{3}}\right)\iff y^2=\sqrt{\frac{2+y^2}{3}}\iff 3y^4-y^2-2=0$$$$y^2,rootof3X^2-X-2=0$$$$X\in \{-\frac{2}{3},1\},\Rightarrow y^2=1\Rightarrow y=1$$$$x=\frac{1}{2}ln\left(\frac{2+y^2}{3}\right)=\frac{ln\left(\frac{3}{3}\right)}{2}=0$$$$\Rightarrow (x,y)\in \left\{(0,1)\right\}$$$$$$

Commented by ~blr237~ last updated on 22/Jan/20

$$\mathrm{Thank}\mathrm{sir}.$$

Commented by mind is power last updated on 22/Jan/20

$$y^{\prime }reWelcom$$

Answered by ~blr237~ last updated on 22/Jan/20

$$\mathrm{with}\mathrm{the}\mathrm{first}\mathrm{equation}\mathrm{you}\mathrm{reach}\mathrm{to}(\mathrm{on}\mathrm{the}\mathrm{strain}\mathrm{y}>0)$$$${\mathrm{e}}^{\mathrm{x}}+\mathrm{x}={\mathrm{lny}}^2+{\mathrm{e}}^{{\mathrm{lny}}^2}\iff \mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left({\mathrm{lny}}^2\right)\mathrm{with}\mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{t}}+\mathrm{t}$$$$\mathrm{f}\mathrm{is}\mathrm{continue}\mathrm{and}\mathrm{strictly}\mathrm{increasing}\mathrm{on}\mathbb{R}({\mathrm{f}}^{\prime }\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{t}}+1>0)$$$$\mathrm{so}\mathrm{f}\mathrm{is}\mathrm{injective}(\mathrm{cause}\mathrm{also}\mathrm{a}\mathrm{bijection}\mathrm{from}\mathbb{R}\to \mathbb{R})$$$$\mathrm{then}\mathrm{x}={\mathrm{lny}}^2\mathrm{or}{\mathrm{y}}^2={\mathrm{e}}^{\mathrm{x}}$$$$\mathrm{when}\mathrm{replacing}\mathrm{on}\mathrm{the}\mathrm{second}\mathrm{we}\mathrm{have}$$$$3{\left({\mathrm{e}}^{\mathrm{x}}\right)}^2-{\mathrm{e}}^{\mathrm{x}}-2=0$$$$\mathrm{then}{\mathrm{e}}^{\mathrm{x}}=1\mathrm{or}{\mathrm{e}}^{\mathrm{x}}=-\frac{2}{3}$$$$\mathrm{finally}\mathrm{x}=0\mathrm{and}\mathrm{y}=1\mathrm{cause}\mathrm{the}\mathrm{strain}\mathrm{was}\mathrm{y}>0$$$$$$

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