Find the largest value of $x$ that satisfies: \[ \begin{array}{l} \log _{3}\left(x^{2}\right)-\log _{3}(x+5)=1 \\ x= \end{array} \] You may enter the exact value or round to 4 decimal places. Question Help: $\square$ Video Calculator Submit Question

Step 1 ：Rewrite the equation using the properties of logarithms: \(\log_3{\frac{x^2}{x+5}} = 1\)

Step 2 ：This implies that \(\frac{x^2}{x+5} = 3\)

Step 3 ：Cross-multiply to get rid of the fraction: \(x^2 = 3(x+5)\)

Step 4 ：Expand the right side: \(x^2 = 3x + 15\)

Step 5 ：Rearrange the equation to set it equal to zero: \(x^2 - 3x - 15 = 0\)

Step 6 ：Factor the quadratic equation: \((x - 5)(x + 3) = 0\)

Step 7 ：Setting each factor equal to zero gives the solutions \(x = 5\) and \(x = -3\)

Step 8 ：Check these solutions in the original equation. Substituting \(x = 5\) into the original equation gives: \(\log_3{(5^2)} - \log_3{(5+5)} = 1\), \(\log_3{25} - \log_3{10} = 1\), \(\log_3{\frac{25}{10}} = 1\), \(\log_3{2.5} = 1\). This is not true, so \(x = 5\) is not a solution.

Step 9 ：Substituting \(x = -3\) into the original equation gives: \(\log_3{((-3)^2)} - \log_3{(-3+5)} = 1\), \(\log_3{9} - \log_3{2} = 1\), \(\log_3{\frac{9}{2}} = 1\), \(\log_3{4.5} = 1\). This is also not true, so \(x = -3\) is not a solution.

Step 10 ：\(\boxed{\text{Therefore, there are no solutions to the equation.}}\)