Solving Logarithmic Equations

To solve the equation \ln(x - 3) + \ln(x + 5) = \ln(3x + 5), we first use a key property of logarithms: \ln(a) + \ln(b) = \ln(ab). This allows us to combine the left-hand side into a single logarithm: \ln\big((x - 3)(x + 5)\big) = \ln(3x + 5). Since the natural logarithm is one-to-one, we can remove the logarithms and set the expressions equal: (x - 3)(x + 5) = 3x + 5. Expanding and simplifying gives a quadratic equation: x^2 - x - 20 = 0, which factors to (x - 5)(x + 4) = 0. So the possible solutions are x = 5 and x = -4. However, we must check the domain because logarithms are only defined for positive arguments. From the original equation, we need x - 3 greater than 0, x + 5 greater than 0, and 3x + 5 greater than 0, which together give x greater than 3. Therefore, x = 5 is valid, but x = -4 is not. The final answer is x = 5.