Hello, I need some assistance with this precalculus homework question, please?HW Q2

Answer:
A. The solution set is {4096}.
Explanation:
Given the logarithmic equation:
[tex]\frac{1}{2} \log _{6} x=3 \log _{6} 4[/tex]Multiply both sides of the equation by 2:
[tex]\begin{gathered} \frac{1}{2}\times2\log_6x=3\times2\log_64 \\ \log_6x=6\log_64 \end{gathered}[/tex]Next, apply the power law of logarithms to the right side of the equation.
[tex]\begin{gathered} \log_6x=\log_64^6 \\ \implies\operatorname{\log}_6x=\operatorname{\log}_64096 \end{gathered}[/tex]Since the bases are the same, equate the numbers:
[tex]x=4096[/tex]The solution set is {4096}.