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

Apply the logarithm power rule: \(2 \ln x = \ln(x^2)\).

Substitute: \(\ln(2x + 3) = \ln(x^2) + \ln 3\).

Use the logarithm product rule: \(\ln(x^2) + \ln 3 = \ln(3x^2)\).

Equate the logarithms: \(\ln(2x + 3) = \ln(3x^2)\).

Remove the logarithms: \(2x + 3 = 3x^2\).

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

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

Calculate the discriminant: \(b^2 - 4ac = (-2)^2 - 4 \times 3 \times (-3) = 4 + 36 = 40\).

Find the roots: \(x = \frac{2 \pm \sqrt{40}}{6}\).

Simplify: \(x = \frac{2 \pm 2\sqrt{10}}{6} = \frac{1 \pm \sqrt{10}}{3}\).

Calculate the positive root: \(x = \frac{1 + \sqrt{10}}{3} \approx 1.39\).

Thus, the solution is \(x = 1.39\) to 3 significant figures.