Questions: log3(x-5) + log3(x+3) = 2

Solution Steps

To solve the equation \(\log _{3}(x-5)+\log _{3}(x+3)=2\), we can use the properties of logarithms. First, apply the product rule for logarithms, which states that \(\log_b(m) + \log_b(n) = \log_b(m \cdot n)\). This allows us to combine the logarithms on the left side. Then, we can rewrite the equation in exponential form to solve for \(x\).

We start with the equation: \[ \log_{3}(x-5) + \log_{3}(x+3) = 2 \] Using the product rule for logarithms, we can combine the left side: \[ \log_{3}((x-5)(x+3)) = 2 \]

Next, we convert the logarithmic equation to its exponential form: \[ (x-5)(x+3) = 3^2 \] This simplifies to: \[ (x-5)(x+3) = 9 \]

Expanding the left side gives: \[ x^2 + 3x - 5x - 15 = 9 \] which simplifies to: \[ x^2 - 2x - 15 = 9 \] Rearranging this leads to: \[ x^2 - 2x - 24 = 0 \]

Factoring the quadratic equation: \[ (x - 6)(x + 4) = 0 \] This gives us the potential solutions: \[ x = 6 \quad \text{or} \quad x = -4 \]

We need to check if these solutions are valid in the original logarithmic expressions. For \(x = 6\): \[ \log_{3}(6-5) + \log_{3}(6+3) = \log_{3}(1) + \log_{3}(9) = 0 + 2 = 2 \quad \text{(valid)} \] For \(x = -4\): \[ \log_{3}(-4-5) + \log_{3}(-4+3) \quad \text{(invalid, as logarithm of a negative number is undefined)} \]

Final Answer