Solve for $x$
\[
\log _{6} x+\log _{6}(x-5)=2
\]

Step 1 ：Given the equation \(\log _{6} x+\log _{6}(x-5)=2\)

Step 2 ：Using the property of logarithms \(\log_b{m} + \log_b{n} = \log_b{mn}\), we can combine the two logarithms on the left side of the equation to get \(\log _{6} (x*(x-5))=2\)

Step 3 ：Converting the logarithmic equation to an exponential equation, we get \(6^2 = x*(x - 5)\), which simplifies to \(36 = x^2 - 5x\)

Step 4 ：Solving this quadratic equation, we get two solutions: \(x = -4\) and \(x = 9\)

Step 5 ：However, we need to check these solutions in the original equation because the domain of the logarithm function is \((0, \infty)\). Therefore, \(x\) must be greater than 0 and \(x - 5\) must also be greater than 0. This means that \(x\) must be greater than 5

Step 6 ：So, the only valid solution is \(x = 9\)

Step 7 ：Final Answer: \(\boxed{9}\)