\[ \begin{aligned} h(x) & =-\frac{1}{2} x+3 \\ l(x) & =\log _{4}(x+3)-\log _{4}(x-2)\end{aligned} \] (a) Write down \( D_{f} \) and solve the inequality \( f(x) \leq 0 \) (b) Solve the equation \( g(x)=-4 \).

To solve the problem, let’s first analyze the given functions \( h(x) \) and \( l(x) \). ### Part (a) **Finding the Domain \( D_f \) of \( h(x) \):** The function \( h(x) = -\frac{1}{2} x + 3 \) is a linear function, which means it is defined for all real numbers. Thus, \( D_h = (-\infty, \infty) \). **Solving the Inequality \( h(x) \leq 0 \):** We need to solve: \[ -\frac{1}{2} x + 3 \leq 0 \] Subtract 3 from both sides: \[ -\frac{1}{2} x \leq -3 \] Multiplying both sides by -2 (remember to flip the inequality sign): \[ x \geq 6 \] The solution to the inequality \( h(x) \leq 0 \) is \( x \in [6, \infty) \). ### Part (b) **Solving the equation \( l(x) = -4 \):** We need to solve: \[ \log_{4}(x+3) - \log_{4}(x-2) = -4 \] Using the logarithmic property \( \log_a(b) - \log_a(c) = \log_a\left(\frac{b}{c}\right) \): \[ \log_{4}\left(\frac{x+3}{x-2}\right) = -4 \] This can be rewritten as: \[ \frac{x+3}{x-2} = 4^{-4} = \frac{1}{256} \] Now, we cross multiply: \[ 256(x + 3) = (x - 2) \] Expanding gives: \[ 256x + 768 = x - 2 \] Rearranging leads to: \[ 255x = -770 \] Thus, solving for \( x \) yields: \[ x = -\frac{770}{255} \approx -3.02 \] ### Final Summary - For part (a), the inequality \( h(x) \leq 0 \) is satisfied for \( x \in [6, \infty) \). - For part (b), solving \( l(x) = -4 \) gives us \( x = -\frac{770}{255} \).