Solve the equation. \( \begin{array}{c}\log _{6}(x-2)+\log _{6}(x+3)=2 \\ x=[?]\end{array} \)

To solve the equation \( \log_{6}(x-2) + \log_{6}(x+3) = 2 \), we can combine the logarithms using the product property of logarithms: \[ \log_{6}((x-2)(x+3)) = 2 \] This means that: \[ (x-2)(x+3) = 6^2 \] Calculating \(6^2\): \[ (x-2)(x+3) = 36 \] Now, let's expand the left side: \[ x^2 + 3x - 2x - 6 = 36 \] Simplifying it gives: \[ x^2 + x - 6 = 36 \] Next, we set the equation to zero by subtracting 36 from both sides: \[ x^2 + x - 42 = 0 \] Now we can factor this quadratic: \[ (x - 6)(x + 7) = 0 \] This gives us two possible solutions: \[ x - 6 = 0 \quad \Rightarrow \quad x = 6 \] \[ x + 7 = 0 \quad \Rightarrow \quad x = -7 \] We need to check if these solutions are valid given the logarithmic arguments must be positive: 1. **For \(x = 6\)**: \(x - 2 = 4 > 0\) and \(x + 3 = 9 > 0\) both are valid. 2. **For \(x = -7\)**: \(x - 2 = -9 < 0\) (invalid) and \(x + 3 = -4 < 0\) (invalid). Thus, the only valid solution is: \[ x = 6 \] Final answer: \(x = 6\).