Lg x +lg(x-3)=1

To solve the equation \( \log x + \log(x - 3) = 1 \), we can use the properties of logarithms. Specifically, we can combine the logarithmic terms on the left-hand side using the product rule: \[ \log x + \log(x - 3) = \log(x(x - 3)) \] Substituting this back into the equation gives us: \[ \log(x(x - 3)) = 1 \] Now, we can convert the logarithmic equation to its exponential form: \[ x(x - 3) = 10^1 \] This simplifies to: \[ x(x - 3) = 10 \] Expanding the left side, we get: \[ x^2 - 3x = 10 \] Now, rearranging this into standard form gives us: \[ x^2 - 3x - 10 = 0 \] Next, we can solve this quadratic equation using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1 \), \( b = -3 \), and \( c = -10 \). Calculating the discriminant: \[ b^2 - 4ac = (-3)^2 - 4(1)(-10) = 9 + 40 = 49 \] Now substitute back into the quadratic formula: \[ x = \frac{-(-3) \pm \sqrt{49}}{2(1)} = \frac{3 \pm 7}{2} \] This gives us two possible solutions: 1. \( x = \frac{3 + 7}{2} = \frac{10}{2} = 5 \) 2. \( x = \frac{3 - 7}{2} = \frac{-4}{2} = -2 \) Since the logarithm is only defined for positive arguments, we discard \( x = -2 \). Thus, the only valid solution is: \[ x = 5 \] Lastly, we can verify the solution by substituting \( x = 5 \) back into the original equation: \[ \log 5 + \log(5 - 3) = \log 5 + \log 2 = \log(5 \cdot 2) = \log 10 = 1 \] Therefore, the solution satisfies the original equation. The final answer is: \[ \boxed{5} \]