Start with the equation \(\ln(3x + 4) = 2 \ln(x + 1)\).

Apply the logarithm power rule: \(\ln(3x + 4) = \ln((x + 1)^2)\).

Since the logarithms are equal, set the arguments equal: \(3x + 4 = (x + 1)^2\).

Expand the right side: \(3x + 4 = x^2 + 2x + 1\).

Rearrange to form a quadratic equation: \(x^2 - x - 3 = 0\).

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 1, b = -1, c = -3\).

Calculate the discriminant: \(b^2 - 4ac = (-1)^2 - 4(1)(-3) = 1 + 12 = 13\).

Find the roots: \(x = \frac{1 \pm \sqrt{13}}{2}\).

Calculate the positive root: \(x = \frac{1 + \sqrt{13}}{2} \approx 2.30\).

Thus, the solution is \(x = 2.30\) to 3 significant figures.