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 \(x\).

(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 \(x \approx -0.481\).

Let \(u = e^x\). Then \(e^{-x} = \frac{1}{u}\).

The equation becomes:

\(\frac{u + \frac{1}{u}}{u + 1} = 4\).

Multiply through by \(u(u + 1)\) to clear the fraction:

\(u^2 + 1 = 4u(u + 1)\).

Expand and simplify:

\(u^2 + 1 = 4u^2 + 4u\).

Rearrange to form a quadratic equation:

\(3u^2 + 4u - 1 = 0\).

Use the quadratic formula \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 3\), \(b = 4\), \(c = -1\):

\(u = \frac{-4 \pm \sqrt{4^2 - 4 \times 3 \times (-1)}}{2 \times 3}\).

\(u = \frac{-4 \pm \sqrt{16 + 12}}{6}\).

\(u = \frac{-4 \pm \sqrt{28}}{6}\).

\(u = \frac{-4 \pm 2\sqrt{7}}{6}\).

\(u = \frac{-2 \pm \sqrt{7}}{3}\).

Choose the positive root: \(u = \frac{\sqrt{7} - 2}{3}\).

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

\(x = \ln\left(\frac{\sqrt{7} - 2}{3}\right)\).

Calculate \(x\) to 3 decimal places:

\(x \approx -1.536\).

Start with the equation:

\(\frac{2e^x + e^{-x}}{e^x - e^{-x}} = 4\)

Multiply both sides by \(e^x - e^{-x}\):

\(2e^x + e^{-x} = 4(e^x - e^{-x})\)

Expand the right side:

\(2e^x + e^{-x} = 4e^x - 4e^{-x}\)

Rearrange terms:

\(2e^x + e^{-x} + 4e^{-x} = 4e^x\)

\(2e^x + 5e^{-x} = 4e^x\)

Rearrange to isolate terms with \(e^x\):

\(2e^x - 4e^x = -5e^{-x}\)

\(-2e^x = -5e^{-x}\)

Divide both sides by \(-2\):

\(e^x = \frac{5}{2}e^{-x}\)

Multiply both sides by \(e^x\):

\(e^{2x} = \frac{5}{2}\)

Take the natural logarithm of both sides:

\(2x = \ln\left(\frac{5}{2}\right)\)

Solve for \(x\):

\(x = \frac{1}{2} \ln\left(\frac{5}{2}\right)\)

Calculate \(x\) to 2 decimal places:

\(x \approx 0.46\)

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:

\(x \approx -0.405\).