Question

Va rog sa ma ajutati la exercitiu 3 (b)

Answer 1

   
[tex]\displaystyle\\ \log_{0,1}(x^2-x-1)=\log_{0,1}(x+4)\\\\ \text{Logaritmii au aceeasi baza.}\\\\ \Longrightarrow~~~(x^2-x-1)= (x+4)\\\\ x^2-x-1 -x-4 = 0\\\\ x^2-2x-5 = 0\\\\ x_{12}= \frac{-b\pm \sqrt{b^2-4ac} }{2a}=\frac{2\pm \sqrt{4+4\cdot 5} }{2}=\\\\ =\frac{2\pm \sqrt{24} }{2}= \frac{2\pm 2\sqrt{6} }{2}= 1\pm \sqrt{6}\\\\ \boxed{x_1 = 1-\sqrt{6}} \\\\ \boxed{x_2= 1+\sqrt{6}} [/tex]