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. This can be rewritten using the quotient property: \[ \log_{6}\left(\frac{x^{2}}{x+1}\right) = 6 \] Exponentiating both sides gives us: \[ \frac{x^{2}}{x+1} = 6^{6} \] Calculating \( 6^{6} \): \[ 6^{6} = 46656 \] This leads to: \[ x^{2} = 46656(x+1) \] Simplifying this equation results in: \[ x^{2} = 46656x + 46656 \] Rearranging gives: \[ x^{2} - 46656x - 46656 = 0 \] We can apply the quadratic formula \( x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \): Here, \( a = 1 \), \( b = -46656 \), and \( c = -46656 \). Calculating the discriminant: \[ b^{2} - 4ac = (-46656)^{2} - 4 \cdot 1 \cdot (-46656) \] \[ = 2176782336 + 186624 = 2176782960 \] Now calculate \( \sqrt{2176782960} \approx 46655.99999 \): Finally, substituting into the quadratic formula gives: \[ x = \frac{46656 \pm 46656}{2} \] This yields two potential solutions: \[ x = \frac{93312}{2} = 46656 \quad \text{and} \quad x = 0 \] Thus, the largest value of \( x \) that satisfies the equation is: \[ \boxed{46656} \]