Resposta :
Lembrando que:
[tex]\log_ax=\log_a y \to x = y[/tex]
Assim:
[tex]\log_6(3x-1) = \log_6(x+7)\\3x - 1 = x + 7\\3x - x = 7 + 1\\2x = 8\\x = \dfrac{8}{2}\\x = 4\\\\\boxed{\boxed{d) \ S=\{4 \}}}[/tex]
Lembrando que:
[tex]\log_ax=\log_a y \to x = y[/tex]
Assim:
[tex]\log_6(3x-1) = \log_6(x+7)\\3x - 1 = x + 7\\3x - x = 7 + 1\\2x = 8\\x = \dfrac{8}{2}\\x = 4\\\\\boxed{\boxed{d) \ S=\{4 \}}}[/tex]