2) \( \ln (x-1)+\ln (x-3)<\ln 3 \)

To solve the inequality \( \ln (x-1) + \ln (x-3) < \ln 3 \), you can combine the logarithms on the left side using the property that \( \ln a + \ln b = \ln (ab) \). This gives you: \[ \ln((x-1)(x-3)) < \ln 3 \] Exponentiating both sides to eliminate the logarithm leads to: \[ (x-1)(x-3) < 3 \] This can be rewritten as a standard quadratic inequality. Expanding the left side results in: \[ x^2 - 4x + 3 < 3 \] Simplifying this, we have: \[ x^2 - 4x < 0 \] Factoring gives: \[ x(x - 4) < 0 \] To find where this inequality holds, analyze the intervals defined by the roots \( x = 0 \) and \( x = 4 \). The test intervals are: \( (-\infty, 0) \), \( (0, 4) \), and \( (4, \infty) \). Testing a point from each interval: - For \( x = -1 \): \( (-1)(-5) > 0 \) (not valid) - For \( x = 1 \): \( (1)(-3) < 0 \) (valid) - For \( x = 5 \): \( (5)(1) > 0 \) (not valid) Thus, \( x(x - 4) < 0 \) holds for \( 0 < x < 4 \). Remembering the domain restrictions from the logarithmic expressions, \( x - 1 > 0 \) and \( x - 3 > 0 \) gives \( x > 3 \). Combining this restriction with the valid range \( 0 < x < 4 \), the only valid solution for the original inequality is: \[ x \in (3, 4) \]