Question 9 Solve the equation: \( \log _{4}(x+2)-\log _{4}(x-3)=2 \)

To solve the equation \( \log _{4}(x+2) - \log _{4}(x-3) = 2 \), we can use the properties of logarithms. The left side can be combined using the quotient rule: \[ \log_{4} \left( \frac{x+2}{x-3} \right) = 2 \] Now, we can rewrite the equation in exponential form: \[ \frac{x+2}{x-3} = 4^2 \] This simplifies to: \[ \frac{x+2}{x-3} = 16 \] Next, we cross-multiply to solve for \(x\): \[ x + 2 = 16(x - 3) \] Expanding the right side gives: \[ x + 2 = 16x - 48 \] Now, we can move all terms involving \(x\) to one side and the constant terms to the other side: \[ 2 + 48 = 16x - x \] Simplifying yields: \[ 50 = 15x \] Now, divide both sides by 15: \[ x = \frac{50}{15} = \frac{10}{3} \] Finally, we need to confirm that this value satisfies the domain of the logarithm. The arguments \(x + 2\) and \(x - 3\) must be positive: \[ x + 2 > 0 \quad \Rightarrow \quad \frac{10}{3} + 2 > 0 \quad \Rightarrow \quad \frac{16}{3} > 0 \] \[ x - 3 > 0 \quad \Rightarrow \quad \frac{10}{3} - 3 > 0 \quad \Rightarrow \quad \frac{10}{3} - \frac{9}{3} > 0 \quad \Rightarrow \quad \frac{1}{3} > 0 \] Both conditions are satisfied, so the solution to the equation is: \[ \boxed{\frac{10}{3}} \]