Comment resoudre cette eqution 2^x ×3−y^2 =−1


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Question Number 53063 by Tristan last updated on 16/Jan/19

$Comment resoudre cette eqution$ $2^{x} \times 3 - y^{2} = - 1$
	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]$
	$(1)$ $y = \pm \sqrt{3 \times 2^{x} + 1}$ $\Rightarrow x \in R$ $(2)$ $2^{x} = \frac{y^{2} - 1}{3}$ $x \ln 2 = \ln (y^{2} - 1) - \ln 3$ $x = \log_{2} (y^{2} - 1) - \log_{2} 3$ $\Rightarrow y^{2} > 1 \Rightarrow y \in R ∖ [- 1; 1]$
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