Question

Vă rog frumos să mă ajutați !
A

Answer 1

$log_ab=log_ac \Rightarrow b=c \ (din \ injectivitatea \ functiei \ logaritmice) \ (*)$

$log_2(2x+5)=log_2(x^2+3x+3)\\ \\ Din \ conditia \ de \ existenta \ a \ logaritmilor \ avem:$

$\left \{ {{2x+5>0} \atop {x^2+3x+3>0}} \right. \Leftrightarrow \left \{ {{x>-\frac{5}{2}} \atop {x \in \mathbb{R}}} \right. \Leftrightarrow x\in (-\frac{5}{2}; \ +\infty)$

$( \ x^2+3x+3=x^2+2\cdot 1,5\cdot x + 1,5^2+0,75=(x+1,5)^2+1,5>0 \ )$

$Din \ (*) \Rightarrow 2x+5=x^2+3x+3 \Rightarrow x^2+x-2=0$

$\Delta=1^2-4\cdot(-2)\cdot 1=1+8=9$

$x_{1, \ 2}=\frac{-1\pm \sqrt{9}}{2\cdot 1}=\frac{-1\pm 3}{2}$

$x_1=\frac{-1+3}{2}=\frac{2}{2}=1 \in (-\frac{5}{2}; \ +\infty)$

$x_2=\frac{-1-3}{2}=\frac{-4}{2}=-2 \in (-\frac{5}{2}; \ +\infty)$

$\Rightarrow S=\{-2, \ 1 \}$