Start with the equation:

\(\ln(1 + e^{-3x}) = 2\)

Exponentiate both sides to remove the natural logarithm:

\(1 + e^{-3x} = e^2\)

Rearrange to solve for \(e^{-3x}\):

\(e^{-3x} = e^2 - 1\)

Calculate \(e^2 - 1\):

\(e^2 \approx 7.389\), so \(e^2 - 1 \approx 6.389\)

Now solve for \(x\):

\(e^{-3x} = 6.389\)

Take the natural logarithm of both sides:

\(-3x = \ln(6.389)\)

Calculate \(\ln(6.389) \approx 1.855\)

So:

\(-3x = 1.855\)

Divide by -3:

\(x = -\frac{1.855}{3} \approx -0.618\)

Thus, the solution is \(x = -0.618\) to 3 decimal places.