Start with the equation \(\ln(x + 4) = 2 \ln x + \ln 4\).

Apply the logarithm power rule: \(2 \ln x = \ln(x^2)\), so the equation becomes \(\ln(x + 4) = \ln(x^2) + \ln 4\).

Use the logarithm product rule: \(\ln(x^2) + \ln 4 = \ln(4x^2)\), so the equation becomes \(\ln(x + 4) = \ln(4x^2)\).

Since the logarithms are equal, set the arguments equal: \(x + 4 = 4x^2\).

Rearrange to form a quadratic equation: \(4x^2 - x - 4 = 0\).

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 4\), \(b = -1\), \(c = -4\).

Calculate the discriminant: \(b^2 - 4ac = (-1)^2 - 4 \times 4 \times (-4) = 1 + 64 = 65\).

Find the roots: \(x = \frac{-(-1) \pm \sqrt{65}}{8}\).

Calculate: \(x = \frac{1 \pm \sqrt{65}}{8}\).

Only the positive root is valid: \(x = \frac{1 + \sqrt{65}}{8} \approx 1.13\).