Find the largest value of \( x \) that satisfies: \( \log _{6}\left(x^{2}\right)-\log _{6}(x+1)=6 \) \( x=\square \) Question Help: \( \square \) Video

To solve the equation \[ \log_{6}(x^{2}) - \log_{6}(x+1) = 6, \] we can use the properties of logarithms. Specifically, we can use the property that states \[ \log_{a}(b) - \log_{a}(c) = \log_{a}\left(\frac{b}{c}\right). \] Applying this property, we rewrite the left side of the equation: \[ \log_{6}\left(\frac{x^{2}}{x+1}\right) = 6. \] Next, we can convert the logarithmic equation into an exponential form. The equation \[ \log_{6}\left(\frac{x^{2}}{x+1}\right) = 6 \] can be rewritten as: \[ \frac{x^{2}}{x+1} = 6^{6}. \] Now, we calculate \( 6^{6} \): \[ 6^{6} = 46656. \] Thus, we have: \[ \frac{x^{2}}{x+1} = 46656. \] To eliminate the fraction, we multiply both sides by \( x + 1 \): \[ x^{2} = 46656(x + 1). \] Expanding the right side gives: \[ x^{2} = 46656x + 46656. \] Rearranging this equation leads to: \[ x^{2} - 46656x - 46656 = 0. \] Now, we can solve this quadratic equation using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}, \] where \( a = 1 \), \( b = -46656 \), and \( c = -46656 \). Plugging in these values, we get: \[ x = \frac{46656 \pm \sqrt{(-46656)^{2} - 4 \cdot 1 \cdot (-46656)}}{2 \cdot 1}. \] Calculating the discriminant: \[ (-46656)^{2} = 2176782336, \] \[ 4 \cdot 1 \cdot (-46656) = -186624. \] Thus, the discriminant becomes: \[ 2176782336 + 186624 = 2176968960. \] Now we calculate the square root of the discriminant: \[ \sqrt{2176968960} \approx 46656.5. \] Now substituting back into the quadratic formula: \[ x = \frac{46656 \pm 46656.5}{2}. \] This gives us two potential solutions: 1. \( x = \frac{46656 + 46656.5}{2} \approx 46656.25 \) 2. \( x = \frac{46656 - 46656.5}{2} \approx -0.25 \) Since \( x \) must be positive (as logarithms are only defined for positive arguments), we take the positive solution: \[ x \approx 46656.25. \] Thus, the largest value of \( x \) that satisfies the equation is: \[ \boxed{46656}. \]