(a) Start with the equation \(\ln(1 + e^{-x}) + 2x = 0\).

Remove the logarithm: \(1 + e^{-x} = e^{-2x}\).

Substitute \(u = e^x\), then \(e^{-x} = \frac{1}{u}\) and \(e^{-2x} = \frac{1}{u^2}\).

The equation becomes \(1 + \frac{1}{u} = \frac{1}{u^2}\).

Multiply through by \(u^2\) to clear fractions: \(u^2 + u - 1 = 0\).

(b) Solve the quadratic equation \(u^2 + u - 1 = 0\).

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

Calculate the discriminant: \(b^2 - 4ac = 1 + 4 = 5\).

Find the roots: \(u = \frac{-1 \pm \sqrt{5}}{2}\).

Choose the positive root: \(u = \frac{-1 + \sqrt{5}}{2} \approx 0.618\).

Since \(u = e^x\), solve for \(x\): \(x = \ln(0.618)\).

Calculate \(x \approx -0.481\).