Given the equation:
[tex]\log _6x+\log _63=\log _6(x+1)[/tex]
Let's solve for x.
To find the solution, take the following steps:
Step 1:
Apply the product property of logarithm to the left side of the equation:
[tex]\begin{gathered} \log _6(x\ast3)=\log _6(x+1) \\ \\ \log _6(3x)=\log _6(x+1) \end{gathered}[/tex]
Step 2:
Eliminate the log on both sides
[tex]\begin{gathered} 3x=(x+1) \\ \\ 3x=x+1 \end{gathered}[/tex]
Step 3:
Subtract x from both sides
[tex]\begin{gathered} 3x-x=x-x+1 \\ \\ 2x=1 \end{gathered}[/tex]
Step 4:
Divide both sides by 2
[tex]\begin{gathered} \frac{2x}{2}=\frac{1}{2} \\ \\ x=\frac{1}{2} \end{gathered}[/tex]
Therefore, the solution to the given equation is:
[tex]x=\frac{1}{2}[/tex]
ANSWER:
[tex]\text{ B. x = }\frac{1}{2}[/tex]
