Logarithmic inequality

• $\quad \text{ Domain }$

$f(x) = \log_{\left(\dfrac{x+6}{3}\right)}\left({\log_{2}{\left(\dfrac{x-1}{x+2}\right)} }\right) \\ 1) \quad \log_{2}{\left( \dfrac{x-1}{x+2} \right) } > 0 \implies \log_{2}{\left( \dfrac{x-1}{x+2} \right) } > \log_{2}{1} \\ \quad \dfrac{x-1}{x+2} > 1 \implies \dfrac{1}{x+2}<0 \\ \quad x \in (-\infty,-2)$

$2) \quad \dfrac{x+6}{3} \in (0,1)\cup(1,\infty) \\ \quad x+6 \in (0,3) \cup (3,\infty) \\ \quad x \in (-6,-3) \cup (-3,\infty) \\ \quad \text{Domain of function}(D_f) = (-6,-3) \cup (-3,-2)$

• $\quad \text{Condition}$

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

$1) \quad \dfrac{x+6}{3} > 1 \implies x \in (-3,\infty) \\ \quad \log_{\left(\dfrac{x+6}{3}\right)}\left({\log_{2}{\left(\dfrac{x-1}{x+2}\right)} }\right)>\log_{\left(\dfrac{x+6}{3}\right)}{1} \\ \quad \log_{2}{\left( \dfrac{x-1}{x+2} \right) } > 1 \\ \quad \dfrac{x-1}{x+2} > 2 \implies \dfrac{x+5}{x+2} < 0 \\ \quad x \in (-5,-2) \\ \quad \implies \boxed{x\in (-3,-2)}$

$2) \quad 0< \dfrac{x+6}{3} < 1 \implies x \in (-6,-3) \\ \quad \log_{\left(\dfrac{x+6}{3}\right)}\left({\log_{2}{\left(\dfrac{x-1}{x+2}\right)} }\right)>\log_{\left(\dfrac{x+6}{3}\right)}{1} \\ \quad \log_{2}{\left( \dfrac{x-1}{x+2} \right) } < 1 \\ \quad \dfrac{x-1}{x+2} < 2 \implies \dfrac{x+5}{x+2} > 0 \\ \quad x \in (-\infty,-5) \cup (-2,\infty) \\ \quad \implies \boxed{x \in (-6,-5)}$

$\text{Our answer : } D_f \cap \left((-6,-5) \cup (-3,-2)\right) \implies \boxed {(-6,-5) \cup (-3,-2)} \\ -6-5-3-2 = \boxed{-16}$