\[ l(x)=\log _{4}(x+3)-\log _{4}(x-2) \] (a) Write down \( D_{f} \) and solve the inequality \( f(x) \leq 0 \) (b) Solve the equation \( g(x)=-4 \) (c) Solve the equation \( 4^{h(x)}=8 \) (d) Write down \( D_{l} \) and solve the inequality \( l(x)=\frac{1}{2} \)

Let's solve the problem step by step. ### Part (a): Find \( D_f \) and solve the inequality \( f(x) \leq 0 \) The function given is: \[ l(x) = \log_{4}(x+3) - \log_{4}(x-2) \] **Step 1: Determine the domain \( D_f \)** The logarithmic function is defined only for positive arguments. Therefore, we need: 1. \( x + 3 > 0 \) → \( x > -3 \) 2. \( x - 2 > 0 \) → \( x > 2 \) The more restrictive condition is \( x > 2 \). Thus, the domain is: \[ D_f = (2, \infty) \] **Step 2: Solve the inequality \( l(x) \leq 0 \)** Using the properties of logarithms, we can rewrite the expression: \[ l(x) = \log_{4}\left(\frac{x+3}{x-2}\right) \] We need to solve: \[ \log_{4}\left(\frac{x+3}{x-2}\right) \leq 0 \] This implies: \[ \frac{x+3}{x-2} \leq 1 \] **Step 3: Solve the inequality** Cross-multiplying (valid since \( x > 2 \)): \[ x + 3 \leq x - 2 \] This simplifies to: \[ 3 \leq -2 \] This is never true, meaning there are no solutions in the domain \( D_f \). ### Part (b): Solve the equation \( g(x) = -4 \) Since \( g(x) \) is not defined in the problem, we will assume \( g(x) = l(x) \). Thus, we need to solve: \[ l(x) = -4 \] Using the logarithmic form: \[ \log_{4}\left(\frac{x+3}{x-2}\right) = -4 \] This implies: \[ \frac{x+3}{x-2} = 4^{-4} = \frac{1}{256} \] **Step 4: Cross-multiply and solve** Cross-multiplying gives: \[ 256(x + 3) = x - 2 \] Expanding and rearranging: \[ 256x + 768 = x - 2 \] \[ 255x = -770 \] \[ x = -\frac{770}{255} = -\frac{154}{51} \] Since \( -\frac{154}{51} \) is not in the domain \( D_f \), there are no solutions. ### Part (c): Solve the equation \( 4^{h(x)} = 8 \) We can rewrite \( 8 \) as \( 4^{3/2} \): \[ 4^{h(x)} = 4^{3/2} \] This implies: \[ h(x) = \frac{3}{2} \] ### Part (d): Write down \( D_l \) and solve the inequality \( l(x) = \frac{1}{2} \) **Step 1: Determine the domain \( D_l \)** The domain \( D_l \) is the same as \( D_f \): \[ D_l = (2, \infty) \] **Step 2: Solve the equation \( l(x) = \frac{1}{2} \)** Using the logarithmic form: \[ \log_{4}\left(\frac{x+3}{x-2}\right) = \frac{1}{2} \] This implies: \[ \frac{x+3}{x-2} = 4^{1/2} = 2 \] **Step 3: Cross-multiply and solve** Cross-multiplying gives: \[ x + 3 = 2(x - 2) \] Expanding and rearranging: \[ x + 3 = 2x - 4 \] \[ 3 + 4 = 2x - x \] \[ x = 7 \] Since \( 7 \) is in the domain \( D_l \), the solution is: \[ x = 7 \] ### Summary of Solutions: - (a) \( D_f = (2, \infty) \), no solutions for \( l(x) \leq 0 \). - (b) No solutions for \( g(x) = -4 \). - (c) \( h(x) = \frac{3}{2} \). - (d) \( D_l = (2, \infty) \), solution \( x = 7 \).