Solve for \( x \) : \( \log (x)+\log (x+2)=1 \) \( x=\square \) You may enter the exact value or round to 4 decimal places.

To solve the equation \( \log (x) + \log (x+2) = 1 \), we can use the property of logarithms that states \( \log(a) + \log(b) = \log(ab) \). This allows us to combine the logarithms: \[ \log(x(x+2)) = 1 \] Now, we can rewrite the logarithm in its exponential form. Since \( \log(a) = b \) means \( a = 10^b \), we have: \[ x(x+2) = 10^1 = 10 \] This simplifies to: \[ x^2 + 2x - 10 = 0 \] Next, we apply the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = 2 \), and \( c = -10 \): \[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-10)}}{2 \cdot 1} \] Calculating the discriminant: \[ b^2 - 4ac = 4 + 40 = 44 \] Now plugging back into the formula: \[ x = \frac{-2 \pm \sqrt{44}}{2} = \frac{-2 \pm 2\sqrt{11}}{2} = -1 \pm \sqrt{11} \] We get two potential solutions: \[ x = -1 + \sqrt{11} \quad \text{and} \quad x = -1 - \sqrt{11} \] Since \( -1 - \sqrt{11} \) is negative and logarithms are not defined for negative values, we only take: \[ x = -1 + \sqrt{11} \] Calculating \( \sqrt{11} \): \[ \sqrt{11} \approx 3.3166 \] Thus, \[ x \approx -1 + 3.3166 \approx 2.3166 \] So the solution rounded to four decimal places is: \[ x \approx 2.3166 \] Therefore, \( x \approx 2.3166 \) (to four decimal places).