31. \( \log _{2}\left(x^{2}-x-6\right)=2 \)

To solve the equation \( \log _{2}\left(x^{2}-x-6\right)=2 \), we first rewrite it in exponential form. This gives us: \[ x^{2}-x-6 = 2^2 \] which simplifies to: \[ x^{2}-x-6 = 4 \] Next, we move the 4 to the left-hand side: \[ x^{2}-x-10 = 0 \] Now, we can factor this quadratic equation. We look for two numbers that multiply to -10 and add to -1. Those numbers are -5 and 2: \[ (x - 5)(x + 2) = 0 \] Setting each factor to zero yields the solutions: \[ x - 5 = 0 \quad \Rightarrow \quad x = 5 \] \[ x + 2 = 0 \quad \Rightarrow \quad x = -2 \] However, we must ensure that both solutions are valid by checking them in the original logarithmic expression, since the argument of a logarithm must be positive. For \( x = 5 \): \[ x^{2} - x - 6 = 5^{2} - 5 - 6 = 25 - 5 - 6 = 14 > 0 \] For \( x = -2 \): \[ x^{2} - x - 6 = (-2)^{2} - (-2) - 6 = 4 + 2 - 6 = 0 \] Since the logarithm of zero is undefined, \( x = -2 \) is not a valid solution. Thus, the only solution to the equation is: \[ \boxed{5} \]