Start by using the properties of logarithms to combine and simplify the equation:

\(\ln 3 + \ln(2x + 5) = \ln(3(2x + 5))\)

\(2 \ln(x + 2) = \ln((x + 2)^2)\)

Equating the two expressions gives:

\(\ln(3(2x + 5)) = \ln((x + 2)^2)\)

Remove the logarithms by setting the arguments equal:

\(3(2x + 5) = (x + 2)^2\)

Expand both sides:

\(6x + 15 = x^2 + 4x + 4\)

Rearrange to form a quadratic equation:

\(x^2 - 2x - 11 = 0\)

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1, b = -2, c = -11\):

\(x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-11)}}{2 \cdot 1}\)

\(x = \frac{2 \pm \sqrt{4 + 44}}{2}\)

\(x = \frac{2 \pm \sqrt{48}}{2}\)

\(x = \frac{2 \pm 4\sqrt{3}}{2}\)

\(x = 1 \pm 2\sqrt{3}\)

Thus, the solutions are \(x = 1 + 2\sqrt{3}\) or \(x = 1 - 2\sqrt{3}\). Since \(x = 1 - 2\sqrt{3}\) is not valid in the original logarithmic equation (as it results in a negative argument), the valid solution is \(x = 1 + 2\sqrt{3}\).