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 \( \log(a) + \log(b) = \log(ab) \).
Combine the left-hand side: \[ \log((x + 1)(x - 8)) > \log(2x - 8) \]
Remove the logarithm by exponentiating both sides (if the logarithmic expressions are defined): \[ (x + 1)(x - 8) > 2x - 8 \]
Expand the left-hand side: \[ x^2 - 8x + x - 8 > 2x - 8 \] \[ x^2 - 7x - 8 > 2x - 8 \]
Move all terms to one side of the inequality: \[ x^2 - 7x - 2x - 8 + 8 > 0 \] \[ x^2 - 9x > 0 \]
Factor the quadratic: \[ x(x - 9) > 0 \]
Find the critical points by setting the factors to zero: \[ x = 0 \quad \text{and} \quad x = 9 \]
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: \[ x < 0 \quad \text{or} \quad x > 9 \]
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 \(x > 8\).
Thus, considering both the interval and the domain restrictions, the final solution is: \[ x > 9 \]
In interval notation, the solution is: \[ (9, \infty) \]
