\[ \begin{array}{l}\log x+\log (x-6)=\log 27 \\ \text { Solve the equation to find the solution set. Select the correct choice below and, if necessary, fill in the answer box to co } \\ \text { A. The solution set is }\} \text {. } \\ \text { (Simplify your answer. Use a comma to separate answers as needed.) } \\ \text { B. There are infinitely many solutions. } \\ \text { C. There is no solution. }\end{array} \text { } \]

To solve the equation \(\log x + \log (x-6) = \log 27\), we can use the properties of logarithms. By applying the product property, we can rewrite the left side: \[ \log(x(x-6)) = \log 27 \] From this, we equate the arguments since the logarithmic function is one-to-one: \[ x(x-6) = 27 \] Expanding this gives: \[ x^2 - 6x - 27 = 0 \] Now we can use the quadratic formula to find the roots: \[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot (-27)}}{2 \cdot 1} \] \[ x = \frac{6 \pm \sqrt{36 + 108}}{2} \] \[ x = \frac{6 \pm \sqrt{144}}{2} \] \[ x = \frac{6 \pm 12}{2} \] This gives us two potential solutions: 1. \(x = \frac{18}{2} = 9\) 2. \(x = \frac{-6}{2} = -3\) Since logarithms are only defined for positive numbers, we discard \(x = -3\). Thus, the only solution is \(x = 9\). So the solution set is: A. The solution set is \{9\}.