Question

$\log_{\frac{x+6}{3}}\left(\log_2{\frac{x-1}{x+2}}\right)>0$

Answer 1

x \in (-6, -5) \cup (-3, -2)

Answer 2

To solve the inequality $\log_{\frac{x+6}{3}}\left(\log_2{\frac{x-1}{x+2}}\right)>0$, we first determine the domain and then consider two cases based on the base of the logarithm.

1. Domain Determination:

• Argument of the inner logarithm: $\frac{x-1}{x+2} > 0$. This implies $x \in (-\infty, -2) \cup (1, \infty)$.
• Argument of the outer logarithm: $\log_2{\frac{x-1}{x+2}} > 0$. Since the base $2 > 1$, this means $\frac{x-1}{x+2} > 2^0 = 1$. Solving $\frac{x-1}{x+2} > 1$ gives $\frac{-3}{x+2} > 0$, which implies $x+2 < 0$, so $x < -2$.
• Base of the outer logarithm: $\frac{x+6}{3} > 0$ and $\frac{x+6}{3} \neq 1$. This means $x+6 > 0 \implies x > -6$, and $x+6 \neq 3 \implies x \neq -3$.

Combining all conditions: $x \in (-\infty, -2)$, $x > -6$, and $x \neq -3$. The overall domain is $x \in (-6, -3) \cup (-3, -2)$.

2. Solving the Inequality: Let $B = \frac{x+6}{3}$ and $A = \log_2{\frac{x-1}{x+2}}$. The inequality is $\log_B(A) > 0$.

• Case 1: Base $B > 1$ and Argument $A > 1$.

  • $B > 1 \implies \frac{x+6}{3} > 1 \implies x > -3$. Within the domain, this is $x \in (-3, -2)$.
  • $A > 1 \implies \log_2{\frac{x-1}{x+2}} > 1 \implies \frac{x-1}{x+2} > 2$. Solving $\frac{x-1}{x+2} > 2$ gives $\frac{-x-5}{x+2} > 0 \implies \frac{x+5}{x+2} < 0$, which means $x \in (-5, -2)$.
  • The intersection of $x \in (-3, -2)$ and $x \in (-5, -2)$ is $x \in (-3, -2)$.

• Case 2: $0 <$ Base $B < 1$ and $0 <$ Argument $A < 1$.

  • $0 < B < 1 \implies 0 < \frac{x+6}{3} < 1 \implies -6 < x < -3$. Within the domain, this is $x \in (-6, -3)$.
  • $0 < A < 1$. We know $A > 0$ implies $x < -2$. We need to solve $A < 1$: $\log_2{\frac{x-1}{x+2}} < 1 \implies \frac{x-1}{x+2} < 2$. Solving $\frac{x-1}{x+2} < 2$ gives $\frac{-x-5}{x+2} < 0 \implies \frac{x+5}{x+2} > 0$, which means $x \in (-\infty, -5) \cup (-2, \infty)$.
  • The intersection of $x \in (-6, -3)$ and ($x \in (-\infty, -5) \cup (-2, \infty)$) is $x \in (-6, -5)$.

3. Combining Solutions: The total solution set is the union of the solutions from Case 1 and Case 2: $x \in (-3, -2) \cup (-6, -5)$. Therefore, the solution is $x \in (-6, -5) \cup (-3, -2)$.