Find the largest value of \( x \) that satisfies: \( \log _{3}\left(x^{2}\right)-\log _{3}(x+3)=5 \) \( x=\square \)

To solve the equation \( \log _{3}\left(x^{2}\right) - \log _{3}(x + 3) = 5 \), we can apply the properties of logarithms. First, we use the quotient property of logarithms: \[ \log_{3}\left(\frac{x^{2}}{x + 3}\right) = 5 \] Next, we can rewrite the equation in exponential form: \[ \frac{x^{2}}{x + 3} = 3^{5} \] Calculating \( 3^{5} \), \[ 3^{5} = 243 \] Now, substituting this back into the equation gives us: \[ \frac{x^{2}}{x + 3} = 243 \] Next, we cross-multiply to eliminate the fraction: \[ x^{2} = 243(x + 3) \] This expands and simplifies to: \[ x^{2} = 243x + 729 \] Rearranging the equation yields: \[ x^{2} - 243x - 729 = 0 \] Now, we can apply the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -243, c = -729 \). Calculating the discriminant: \[ b^2 - 4ac = (-243)^2 - 4 \times 1 \times (-729) \] \[ = 59049 + 2916 = 61965 \] Next, we take the square root of the discriminant: \[ \sqrt{61965} = 249.93 \text{ (approximately)} \] Now, substituting back into the quadratic formula: \[ x = \frac{243 \pm 249.93}{2} \] Calculating the two solutions: 1. \( x = \frac{243 + 249.93}{2} \approx \frac{492.93}{2} \approx 246.465 \) 2. \( x = \frac{243 - 249.93}{2} \approx \frac{-6.93}{2} \approx -3.465 \) Since \( x \) must be positive (as logarithms of non-positive numbers are undefined), we select the largest value: \[ x \approx 246.465 \] Thus, the largest value of \( x \) that satisfies the original equation is \[ \boxed{246} \] (rounded to the nearest whole number).