\[ l(x)=\log _{4}(x+3)-\log _{4}(x-2) \] (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 \( l(x)=\frac{1}{2} \)
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To solve the given problem, we will break it down into parts. ### Part (a) First, let's determine the domain \( D_{f} \) for the function \( f(x) = \log_{4}(x+3) - \log_{4}(x-2) \). The logarithm functions are defined only for positive arguments, which gives us the conditions: 1. \( x + 3 > 0 \) → \( x > -3 \) 2. \( x - 2 > 0 \) → \( x > 2 \) Therefore, the domain \( D_{f} \) is \( x > 2 \). Now, let's solve the inequality \( f(x) \leq 0 \). We can apply the properties of logarithms: \[ \log_{4}(x + 3) - \log_{4}(x - 2) = \log_{4}\left(\frac{x + 3}{x - 2}\right) \leq 0 \] This inequality implies: \[ \frac{x + 3}{x - 2} \leq 1 \] Multiplying both sides by \( x - 2 \) (valid since \( x > 2 \) ensures \( x - 2 > 0 \)): \[ x + 3 \leq x - 2 \] This simplifies to \( 3 \leq -2 \), which is never true. Thus, there are no solutions for \( f(x) \leq 0 \) in the domain \( D_{f} \). ### Part (b) Now, to solve the equation \( g(x) = -4 \), we first note that \( g(x) \) must be defined. However, you didn't provide the definition of \( g(x) \). Assuming \( g(x) \) is another logarithmic function related to \( f(x) \), we might approach it in a similar manner, but we need more information to continue. ### Part (c) To solve \( 4^{h(x)} = 8 \), we can express 8 as a power of 4: \[ 8 = 4^{3/2} \] Thus, we set \( h(x) = \frac{3}{2} \). ### Part (d) Next, we address the function \( l(x) = \log_{4}(x + 3) - \log_{4}(x - 2) \) again. The domain \( D_{l} \) is the same as \( D_{f} \), so \( D_{l} = (2, \infty) \). Now, we solve the equation \( l(x) = \frac{1}{2} \): \[ \log_{4}\left(\frac{x + 3}{x - 2}\right) = \frac{1}{2} \] This can be rewritten in exponential form: \[ \frac{x + 3}{x - 2} = 4^{1/2} = 2 \] Cross-multiplying gives us: \[ x + 3 = 2(x - 2) \] Expanding and simplifying: \[ x + 3 = 2x - 4 \] Thus, \[ 3 + 4 = 2x - x \\ 7 = x \\ \] The solution is \( x = 7 \). We should check if it lies within the domain \( D_{l} = (2, \infty) \). Yes, it does! ### Summary of Solutions: - (a) \( D_f = (2, \infty) \) and no solutions for \( f(x) \leq 0 \). - (b) More information needed for \( g(x) \). - (c) \( h(x) = \frac{3}{2} \). - (d) \( x = 7 \) is the solution for \( l(x) = \frac{1}{2} \).
