[tex]log_\big3 (x+2) - log_\big3 (x-4) = 1 \\ \\ $Conditii de existenta:
\left\{ \begin{array}{c} x+2 \ \textgreater \ 0 \\ x-4 \ \textgreater \ 0 \end{array} \right \Rightarrow
\left\{ \begin{array}{c} x\ \textgreater \ -2 \\ x\ \textgreater \ 4 \end{array} \right| \Rightarrow D = (4,+\infty) \\ \\ log_\big3 (x+2) - log_\big3 (x-4) = 1 \Rightarrow log_\big3 \dfrac{x+2}{x-4} = log_\big3 3 \Rightarrow \dfrac{x+2}{x-4} = 3 \Rightarrow [/tex]
[tex]\Rightarrow \dfrac{x+2}{x-4} -3 = 0 \Rightarrow \dfrac{x+2-3(x-4)}{x-4} = 0 \Rightarrow \dfrac{x+2-3x+12}{x-4} = 0 \Rightarrow \\ \\ \Rightarrow \dfrac{-2x+14}{x-4} = 0 \Big|\cdot(x-4) \Rightarrow -2x+14 = 0 \Rightarrow -2x = -14 \Rightarrow \\ \\ \Rightarrow x = \dfrac{-14}{-2} \Rightarrow x = 7\in D \Rightarrow \boxed{S = \Big\{ 7\Big\}}[/tex]
