OpenSAT

Using the property of logarithms that states $\log_a b + \log_a c = \log_a (b \cdot c)$, we can rewrite the given equation as $\log_2 [(x-3)(x+1)] = 3$. This simplifies to $\log_2 (x^2 - 2x - 3) = 3$. Converting this equation from logarithmic form to exponential form, we get $2^3 = x^2 - 2x - 3$. This simplifies to $8 = x^2 - 2x - 3$. Subtracting 8 from both sides, we have $0 = x^2 - 2x - 11$. Factoring the quadratic gives us $0 = (x-6)(x+2)$. Therefore, the possible values of $x$ are 6 and -2. Since $x-3$ must be positive for the logarithm to be defined, the correct value of $x$ is 6.