[tex]\displaystyle \sf \log_x(x+6) = 2\\\\ \text{Condi\c c\~ao de exist\^encia (C.E) }:\\\\ \underline{\text{logaritmando maior 0 e base maior 0 e diferente de 1}}\\\\ x+6 > 0 \to x > -6 \\\\ 1\neq x > 0\\\\ \boxed{\sf C.E : 1\neq x > 0} \\\\\\ \log_x(x+6) = 2 \\\\ x+6 = x^2 \\\\ x^2-x-6 = 0 \\\\\ x^2+2x-3x-6=0 \\\\ x(x+2)-3(x+2)=0 \\\\ (x-3)(x+2) = 0 \\\\ x-3=0 \to x= 3 \ \in C.E\ \checkmark \\\\ x+2 = 0 \to x=-2 \to \not\in C.E \\\\\ \text{Solu\c c\~ao}: \\\\ \Large\boxed{\sf \ x = 3 \ }\checkmark[/tex]