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Să se calculeze suma soluțiilor ecuatiei log in baza 2 din(x^2 +3)=2

Răspuns :

[tex]\log_{2}(x^{2}+3)=2\\
\\Conditii~de~existenta:\\
\\2>0;~2 \neq 0\\
\\x^{2}+3>0;~x^{2}>-3,~x \in \math{R}\\
\\x^{2}+3=2^{2}\\
\\x^{2}=4-3\\
\\x^{2}=1\\
\\x_{1,2}=\pm \sqrt{1}\\
\\x_{1,2}=\pm 1\\
\\S=\{-1;1\}\\
\\-1+1=0[/tex]

R:0
[tex] \log_\big 2 (x^2 +3) = 2\\ \\ $Conditii de existenta:\\ \bullet $ $ x^2+3 >0 \Rightarrow x^2 > -3 \Rightarrow x\in \mathbb_R $ $ \\ \\ \log_\big 2 (x^2 +3) = 2\\ \log_\big 2 (x^2 +3) = 2\cdot 1\\ \log_\big 2 (x^2 +3) = 2\cdot \log_\big 2 2 \\ \log_\big 2 (x^2 +3) = \log_\big 2 (2^2)\\ \\ x^2+3 = 2^2 \\ x^2 + 3= 4 \\ x^2 = 4-3 \\ x^2 = 1 \\ x = \pm 1 \\ \\ x \in \Big\{-1;1\Big\} \\ \\ \Rightarrow S = 1+(-1) \Rightarrow S = 1-1 \Rightarrow \boxed{\boxed{S = 0}}[/tex]