Solve Log₃(x²+5x+4)/Log₃x < Log_x12: Finding Valid Domain

To solve the problem:

• Given the inequality $\frac{\log_3(x^2+5x+4)}{\log_3x}<\log_x12$, apply the change-of-base formula:
• Rewrite $\log_x 12 = \frac{\log_3 12}{\log_3 x}$.
• Substitute into the inequality:
• $\frac{\log_3(x^2 + 5x + 4)}{\log_3 x} < \frac{\log_3 12}{\log_3 x}$.
• Cross multiply assuming $\log_3 x > 0$ (because $x > 1$): $\log_3(x^2 + 5x + 4) < \log_3 12$.
• The inequality $\log_3(x^2 + 5x + 4) < \log_3 12$ implies:
• $x^2 + 5x + 4 < 12$.
• Simplify: $x^2 + 5x + 4 - 12 < 0$, which gives $x^2 + 5x - 8 < 0$.
• Find roots for $x^2 + 5x - 8 = 0$ using the quadratic formula:
• $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 1, b = 5, c = -8$.
• $x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1 \cdot (-8)}}{2}$.
• $x = \frac{-5 \pm \sqrt{57}}{2}$.
• We find the intervals projecting on the inequality sign: $x^2 + 5x - 8 < 0$.
• Analyzing the sign change for $1 < x < \frac{-5 + \sqrt{57}}{2}$.
• Additionally, confirm $x^2 + 5x + 4 > 0$ for valid logarithm argument, which is naturally satisfied in previous constraints.

Therefore, the solution is: $\mathbf{1 < x < -2.5+\frac{\sqrt{57}}{2}}$.