First, let \(u = e^x\). Then the equation becomes \(\frac{2u + \frac{1}{u}}{2 + u} = 3\).

Multiply through by \(u(2 + u)\) to clear the fraction: \(2u^2 + 1 = 3u(2 + u)\).

Expand and simplify: \(2u^2 + 1 = 6u + 3u^2\).

Rearrange to form a quadratic equation: \(u^2 + 6u - 1 = 0\).

Use the quadratic formula \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 1, b = 6, c = -1\).

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

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

Simplify: \(u = -3 \pm \sqrt{10}\).

Since \(u = e^x > 0\), we reject \(u = -3 - \sqrt{10}\) as it is negative.

Thus, \(u = -3 + \sqrt{10}\).

Find \(x\) by taking the natural logarithm: \(x = \ln(-3 + \sqrt{10})\).

Calculate \(x\) to 3 decimal places: \(x = -1.818\).

To show there is only one real root, note that the quadratic equation \(u^2 + 6u - 1 = 0\) has only one positive solution for \(u\), which corresponds to one real solution for \(x\).