Solve for \( x \) : \( \log _{6} x+\log _{6}(x+1)=7 \) \( x=\square \)

To solve the equation \( \log_{6} x + \log_{6} (x + 1) = 7 \), we can use the property of logarithms that states \( \log_a b + \log_a c = \log_a (bc) \). This allows us to combine the logarithms: \[ \log_{6} (x(x + 1)) = 7 \] Next, we can exponentiate both sides to eliminate the logarithm: \[ x(x + 1) = 6^7 \] Calculating \( 6^7 \): \[ 6^7 = 279936 \] Now, we rewrite the equation: \[ x^2 + x - 279936 = 0 \] We can solve this quadratic equation using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = 1, c = -279936 \). Calculating the discriminant: \[ b^2 - 4ac = 1^2 - 4(1)(-279936) = 1 + 1119744 = 1119745 \] Now, we can find the roots: \[ x = \frac{-1 \pm \sqrt{1119745}}{2} \] Calculating \( \sqrt{1119745} \) is approximately 105.83, leading to: \[ x = \frac{-1 + 105.83}{2} \approx 52.415 \quad (\text{Valid solution since } x > 0) \] \[ x = \frac{-1 - 105.83}{2} \quad (\text{Negative, not valid}) \] Thus, the solution is \[ x \approx 52.415 \] So we can express it as: \( x \approx 52.415 \)