Solve \( \ln x+\ln (x-2)=5 \). Round your answer to the nearest hundredth. Check for extraneous solutions.

To solve the equation \( \ln x + \ln (x-2) = 5 \), we can use the property of logarithms that states \( \ln a + \ln b = \ln (a \cdot b) \). Hence, we can rewrite the equation as: \[ \ln (x(x-2)) = 5 \] Next, we can exponentiate both sides to eliminate the logarithm: \[ x(x-2) = e^5 \] This simplifies to: \[ x^2 - 2x = e^5 \] Rearranging this gives us a quadratic equation: \[ x^2 - 2x - e^5 = 0 \] We can solve for \(x\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -2\), and \(c = -e^5\): \[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-e^5)}}{2 \cdot 1} \] This simplifies to: \[ x = \frac{2 \pm \sqrt{4 + 4e^5}}{2} \] \[ x = 1 \pm \sqrt{1 + e^5} \] Calculating \(e^5\) using its approximate value of \(148.413\): \[ 1 + e^5 \approx 149.413 \] Taking the square root: \[ \sqrt{149.413} \approx 12.22 \] Thus the possible solutions for \(x\) are: \[ x = 1 + 12.22 \approx 13.22 \] \[ x = 1 - 12.22 \approx -11.22 \quad \text{(not valid since } x \text{ must be greater than 2)} \] This gives us one valid solution: \[ x \approx 13.22 \] Lastly, we check for extraneous solutions. We need to ensure \(x > 2\). Since \( x = 13.22 \), this is greater than 2, which is valid. Rounding to the nearest hundredth, the final answer is: \[ \boxed{13.22} \]