[tex] log_\big{5} (3x+1) = 1+log_\big{5} (x-1) \Rightarrow \\ \\ \Rightarrow log_\big{5} (3x+1) =log_\big{5} 5 +log_\big{5} (x-1) \Rightarrow \\ \\ \Rightarrow log_\big{5} (3x+1) =log_\big{5} \Big(5\cdot (x-1)\Big)\Rightarrow \\ \\ \Rightarrow 3x+1 = 5\cdot(x-1) \Rightarrow\\ \\ \Rightarrow 3x+1 = 5x-5 \Rightarrow \\ \\ \Rightarrow 3x-5x= -5-1 \Rightarrow \\ \\ \Rightarrow -2x = -6 \Rightarrow \\ \\ \Rightarrow x= \dfrac{-6}{-2} \Rightarrow \\ \\ \Rightarrow x=3[/tex]
Conditii de existenta: 3x+1 > 0 si x-1 > 0 =>
3x > -1 si x > 1 => x> -1/3 si x > 1 =>
D = (1, +infinit)
x = 3 apartine D => S = {3}
