4) \( \log _{\sqrt{6}}(x-1)+\log _{\sqrt{6}}(x+4)=\log _{\sqrt{6}} 6 \)

To solve the equation \( \log _{\sqrt{6}}(x-1)+\log _{\sqrt{6}}(x+4)=\log _{\sqrt{6}} 6 \), we can use the property of logarithms that states \( \log_a b + \log_a c = \log_a(bc) \). This allows us to combine the left-hand side: \[ \log_{\sqrt{6}}((x-1)(x+4)) = \log_{\sqrt{6}} 6 \] Since the logarithms are equal, we can drop the logarithm and set the arguments equal to each other: \[ (x-1)(x+4) = 6 \] Expanding the left side: \[ x^2 + 4x - x - 4 = 6 \] Which simplifies to: \[ x^2 + 3x - 4 = 6 \] Subtracting 6 from both sides gives: \[ x^2 + 3x - 10 = 0 \] Now, we can factor this quadratic equation: \[ (x + 5)(x - 2) = 0 \] Setting each factor to zero results in: \[ x + 5 = 0 \quad \Rightarrow \quad x = -5 \] \[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \] Now we must check if these solutions are within the domain of the original logarithm expressions. The expressions \( x - 1 \) and \( x + 4 \) must both be positive. 1. For \( x = -5 \): - \( x - 1 = -5 - 1 = -6 \) (not valid) - \( x + 4 = -5 + 4 = -1 \) (not valid) 2. For \( x = 2 \): - \( x - 1 = 2 - 1 = 1 \) (valid) - \( x + 4 = 2 + 4 = 6 \) (valid) Thus, the only valid solution is: \[ \boxed{2} \]