Log(x+1)+ lg(x-8)>lg(2x-8)

To solve the inequality \( \log(x + 1) + \log(x - 8) > \log(2x - 8) \), we can use properties of logarithms. Recall that \( \log(a) + \log(b) = \log(ab) \). 1. Combine the left-hand side: \[ \log((x + 1)(x - 8)) > \log(2x - 8) \] 2. Remove the logarithm by exponentiating both sides (if the logarithmic expressions are defined): \[ (x + 1)(x - 8) > 2x - 8 \] 3. Expand the left-hand side: \[ x^2 - 8x + x - 8 > 2x - 8 \] \[ x^2 - 7x - 8 > 2x - 8 \] 4. Move all terms to one side of the inequality: \[ x^2 - 7x - 2x - 8 + 8 > 0 \] \[ x^2 - 9x > 0 \] 5. Factor the quadratic: \[ x(x - 9) > 0 \] 6. Find the critical points by setting the factors to zero: \[ x = 0 \quad \text{and} \quad x = 9 \] 7. Test intervals around the critical points \(0\) and \(9\): - For \(x < 0\), choose \(x = -1\): \((-1)(-10) > 0\) (True) - For \(0 < x < 9\), choose \(x = 1\): \((1)(-8) > 0\) (False) - For \(x > 9\), choose \(x = 10\): \((10)(1) > 0\) (True) Therefore, the solution to the inequality \(x(x - 9) > 0\) is: \[ x < 0 \quad \text{or} \quad x > 9 \] 8. Check the domain of the original logarithmic expressions. For the logarithms to be defined: - \( x + 1 > 0 \Rightarrow x > -1 \) - \( x - 8 > 0 \Rightarrow x > 8 \) - \( 2x - 8 > 0 \Rightarrow x > 4 \) The most restrictive condition is \(x > 8\). 9. Thus, considering both the interval and the domain restrictions, the final solution is: \[ x > 9 \] In interval notation, the solution is: \[ (9, \infty) \]