Given the equation:
![\log _6x+\log _63=\log _6(x+1)](https://img.qammunity.org/2023/formulas/mathematics/college/ms66rz8d59qulnwdwn9kustdynccbqib81.png)
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:
![\begin{gathered} \log _6(x\ast3)=\log _6(x+1) \\ \\ \log _6(3x)=\log _6(x+1) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/600jbhdczbvg9rmq2kb52ai3lt0mlbr9nb.png)
Step 2:
Eliminate the log on both sides
![\begin{gathered} 3x=(x+1) \\ \\ 3x=x+1 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/7b1jb60x3k1k7menm57tst9b6fjqtxbp4s.png)
Step 3:
Subtract x from both sides
![\begin{gathered} 3x-x=x-x+1 \\ \\ 2x=1 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/19n9tyu9zhfby0g8u3ltbvl7rg29dgbwdk.png)
Step 4:
Divide both sides by 2
![\begin{gathered} (2x)/(2)=(1)/(2) \\ \\ x=(1)/(2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/rczq0pq2cm1ggczhc8xmmtd572l692dvqi.png)
Therefore, the solution to the given equation is:
![x=(1)/(2)](https://img.qammunity.org/2023/formulas/mathematics/high-school/pqut627712ixsd2szaa62k7ajbq97gcvu4.png)
ANSWER:
![\text{ B. x = }(1)/(2)](https://img.qammunity.org/2023/formulas/mathematics/college/apfbrmznrikzuqvmgde8a82lo5qbev85zh.png)