Substitute \(u = e^x\) into the equation:

\(4e^{-x} = 4 \cdot \frac{1}{e^x} = \frac{4}{u}\).

The equation becomes \(\frac{4}{u} = 3u + 4\).

Multiply through by \(u\) to clear the fraction:

\(4 = 3u^2 + 4u\).

Rearrange to form a quadratic equation:

\(3u^2 + 4u - 4 = 0\).

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

\(u = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 3 \cdot (-4)}}{2 \cdot 3}\).

\(u = \frac{-4 \pm \sqrt{16 + 48}}{6}\).

\(u = \frac{-4 \pm \sqrt{64}}{6}\).

\(u = \frac{-4 \pm 8}{6}\).

Solutions for \(u\) are \(u = \frac{4}{6} = \frac{2}{3}\) or \(u = \frac{-12}{6} = -2\).

Since \(u = e^x\) must be positive, we take \(u = \frac{2}{3}\).

Thus, \(e^x = \frac{2}{3}\).

Take the natural logarithm of both sides:

\(x = \ln\left(\frac{2}{3}\right)\).

Calculate \(x\) to 3 significant figures:

\(x \approx -0.405\).