Find the largest value of \( x \) that satisfies: \( \log _{2}\left(x^{2}\right)-\log _{2}(x+4)=2 \) \( x=\square \)

To solve the equation \( \log_{2}(x^{2}) - \log_{2}(x+4) = 2 \), we can use the property of logarithms that states \(\log_{b}(a) - \log_{b}(c) = \log_{b}\left(\frac{a}{c}\right)\). Applying this property: \[ \log_{2}\left(\frac{x^{2}}{x+4}\right) = 2 \] This means we can rewrite the equation as: \[ \frac{x^{2}}{x+4} = 2^{2} \] Calculating \(2^{2}\) gives us 4, so: \[ \frac{x^{2}}{x+4} = 4 \] Next, we can cross-multiply to eliminate the fraction: \[ x^{2} = 4(x + 4) \] Expanding the right side: \[ x^{2} = 4x + 16 \] Rearranging the equation gives us: \[ x^{2} - 4x - 16 = 0 \] To solve this quadratic equation, we can apply the quadratic formula \(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\), where \(a = 1\), \(b = -4\), and \(c = -16\). Calculating the discriminant: \[ b^{2} - 4ac = (-4)^{2} - 4(1)(-16) = 16 + 64 = 80 \] Now we can apply the quadratic formula: \[ x = \frac{-(-4) \pm \sqrt{80}}{2(1)} = \frac{4 \pm 4\sqrt{5}}{2} = 2 \pm 2\sqrt{5} \] This gives us two possible solutions: \[ x = 2 + 2\sqrt{5} \quad \text{and} \quad x = 2 - 2\sqrt{5} \] Calculating the approximate value of \(2 + 2\sqrt{5}\): \[ \sqrt{5} \approx 2.236 \implies 2 + 2 \times 2.236 \approx 2 + 4.472 = 6.472 \] And for \(2 - 2\sqrt{5}\): \[ 2 - 2 \times 2.236 \approx 2 - 4.472 = -2.472 \] Since \(x\) must be positive (as it is inside a logarithm), we discard \(2 - 2\sqrt{5}\). Thus, the largest value of \(x\) that satisfies the equation is: \[ \boxed{2 + 2\sqrt{5}} \]