Question Number 53063 by Tristan last updated on 16/Jan/19
$$\mathrm{Comment}\:\mathrm{resoudre}\:\mathrm{cette}\:\mathrm{eqution}\: \\ $$$$\mathrm{2}^{\mathrm{x}} ×\mathrm{3}−\mathrm{y}^{\mathrm{2}} =−\mathrm{1} \\ $$
Answered by MJS last updated on 18/Jan/19
![(1) y=±(√(3×2^x +1)) ⇒ x∈R (2) 2^x =((y^2 −1)/3) xln 2 =ln (y^2 −1) −ln 3 x=log_2 (y^2 −1) −log_2 3 ⇒ y^2 >1 ⇒ y∈R\[−1;1]](https://www.tinkutara.com/question/Q53116.png)
$$\left(\mathrm{1}\right) \\ $$$${y}=\pm\sqrt{\mathrm{3}×\mathrm{2}^{{x}} +\mathrm{1}} \\ $$$$\Rightarrow\:{x}\in\mathbb{R} \\ $$$$\left(\mathrm{2}\right) \\ $$$$\mathrm{2}^{{x}} =\frac{{y}^{\mathrm{2}} −\mathrm{1}}{\mathrm{3}} \\ $$$${x}\mathrm{ln}\:\mathrm{2}\:=\mathrm{ln}\:\left({y}^{\mathrm{2}} −\mathrm{1}\right)\:−\mathrm{ln}\:\mathrm{3} \\ $$$${x}=\mathrm{log}_{\mathrm{2}} \:\left({y}^{\mathrm{2}} −\mathrm{1}\right)\:−\mathrm{log}_{\mathrm{2}} \:\mathrm{3} \\ $$$$\Rightarrow\:{y}^{\mathrm{2}} >\mathrm{1}\:\Rightarrow\:{y}\in\mathbb{R}\backslash\left[−\mathrm{1};\mathrm{1}\right] \\ $$