[tex]\log_{10}(x+2)-\log_{10}9 = 1-\log_{10}(x+1) \\ \\ \log_{10}(x+2)-\log_{10}9+\log_{10}(x+1) = 1 \\ \\ \log_{10}\dfrac{(x+2)(x+1)}{9} = 1 \\ \\ \dfrac{(x+2)(x+1)}{9} = 10^1 \\ \\ (x+2)(x+1) = 90 \\ \\ (8+2)(8+1) = 90 \\ \\ \Rightarrow x = 8,\quad \text{ celalalt }x \text{ nu verifica deoarece este}\leq -1.\\ \\ \Rightarrow \boxed{S = \big\{8\big\}}[/tex]
