Answer:
[tex]\textsf{B.} \quad x=\dfrac{1}{2}[/tex]
Step-by-step explanation:
Log laws
[tex]\textsf{Product law}: \quad \log_ax + \log_ay=\log_axy[/tex]
[tex]\textsf{Equality law}: \quad \textsf{if }\: \log_ax=\log_ay\:\textsf{ then }\:x=y[/tex]
Therefore:
[tex]\large \begin{aligned}\log_6 x+ \log_6 3 & =\log_6 (x+1)\\\log_6 (3x) & =\log_6 (x+1)\\3x & =x+1\\2x & =1\\x & =\dfrac{1}{2}\end{aligned}[/tex]
