Domains!

• $\quad \text{Base} \\ \quad \text{For a function } g(x) = \log_{a}{b} \\ \quad a \in (0,1) \cup (1,\infty) \\ \quad \text{In this case : } \\ \quad \sin{x}+\cos{x} = \sqrt{2} \sin{\left( x + \dfrac{\pi}{4} \right) } \in (0,1) \cup \left(1,\sqrt2 \right] \\ \quad \text{Since } \sin{x} \text{ is +ve in } x \in \left[0,\pi\right] \\ \quad \sin{\left( x + \dfrac{\pi}{4} \right) } \text{ is +ve in } x \in \left[0,\dfrac{3\pi}{4}\right] \\ \quad \text{Also , } \sqrt{2} \sin{\left( x + \dfrac{\pi}{4} \right) } \ne 1 \\ \quad x \ne 0,\dfrac{\pi}{2} \\ \quad \text{For our base : } \boxed{x \in \left(0,\dfrac{\pi}{2} \right) \cup \left(\dfrac{\pi}{2},\dfrac{3\pi}{4} \right)} \\ \\$

• $\quad 2 \ge 2\sin{x}>0 \implies 1 \ge \sin{x}>0 \\ \quad \boxed{x \in \left( 0 , \pi \right)} \\ \\$

• $\quad \text{Condition : } \log_{\sqrt{2} \sin{\left( x + \frac{\pi}{4} \right) }}{2\sin{x}} \ge 0 \\ \quad \log_{\sqrt{2} \sin{\left( x + \frac{\pi}{4} \right) }}{2\sin{x}} \ge \log_{\sqrt{2} \sin{\left( x + \frac{\pi}{4} \right) }}{1} \\ \quad \text{For } x \in \left(0,\dfrac{\pi}{2} \right) \quad , \quad 2\sin{x} \ge 1 \\ \quad x \in \left[ \dfrac{\pi}{6} , \dfrac{5\pi}{6} \right] \\ \implies \boxed{ x \in \left[ \dfrac{\pi}{6} , \dfrac{\pi}{2} \right) } \\ \quad \text{For } x \in \left(\dfrac{\pi}{2},\dfrac{3\pi}{4} \right) \quad , \quad 2\sin{x} \le 1 \\ \quad x \in \left( 0 , \dfrac{\pi}{6} \right] \cup \left[ \dfrac{5\pi}{6} , \pi \right) \\ \implies \boxed{ x \in \phi } \\ \\$

$\quad \text{Our answer : } \boxed{ x \in \left[ \dfrac{\pi}{6} , \dfrac{\pi}{2} \right) } \implies a= \dfrac{\pi}{6} , b=\dfrac{\pi}{2} \\ \quad \dfrac{24 \times \dfrac{\cancel{\pi}}{6}}{\dfrac{\cancel{\pi}}{2}} \implies \boxed{8}$

---

$\text{Nice Question , Keep posting }$