Start with the equation:

\(\ln(x^4 - 4) = 4 \ln x - \ln 4\)

Use the logarithm property \(\ln a - \ln b = \ln \left( \frac{a}{b} \right)\):

\(\ln(x^4 - 4) = \ln \left( \frac{x^4}{4} \right)\)

Since the logarithms are equal, set the arguments equal:

\(x^4 - 4 = \frac{x^4}{4}\)

Multiply through by 4 to clear the fraction:

\(4(x^4 - 4) = x^4\)

Expand and simplify:

\(4x^4 - 16 = x^4\)

Rearrange to form a quadratic equation:

\(3x^4 = 16\)

Divide by 3:

\(x^4 = \frac{16}{3}\)

Take the fourth root of both sides:

\(x = \left( \frac{16}{3} \right)^{1/4}\)

Calculate the value:

\(x \approx 1.52\)

Start with the equation \(\ln(x^2 + 1) = 1 + 2 \ln x\).

Use the logarithm power rule: \(2 \ln x = \ln(x^2)\).

Substitute to get: \(\ln(x^2 + 1) = 1 + \ln(x^2)\).

Rearrange to: \(\ln(x^2 + 1) - \ln(x^2) = 1\).

Use the logarithm quotient rule: \(\ln\left(\frac{x^2 + 1}{x^2}\right) = 1\).

Exponentiate both sides: \(\frac{x^2 + 1}{x^2} = e\).

Simplify: \(1 + \frac{1}{x^2} = e\).

Rearrange to find \(x^2\): \(\frac{1}{x^2} = e - 1\).

Thus, \(x^2 = \frac{1}{e - 1}\).

Take the square root: \(x = \frac{1}{\sqrt{e - 1}}\).

Calculate \(x\) to 3 significant figures: \(x \approx 0.763\).

To solve the equation \(\ln(1 + 2^x) = 2\), we first remove the logarithm by exponentiating both sides:

\(1 + 2^x = e^2\).

Next, solve for \(2^x\):

\(2^x = e^2 - 1\).

Now, take the logarithm base 2 of both sides to solve for \(x\):

\(x = \log_2(e^2 - 1)\).

Using a calculator, we find:

\(x \approx 2.676\).

Start with the equation \(\ln(x^2 + 4) = 2 \ln x + \ln 4\).

Apply the logarithm power rule: \(2 \ln x = \ln(x^2)\).

Thus, the equation becomes \(\ln(x^2 + 4) = \ln(x^2) + \ln 4\).

Use the logarithm product rule: \(\ln(x^2) + \ln 4 = \ln(4x^2)\).

Now, the equation is \(\ln(x^2 + 4) = \ln(4x^2)\).

Since the logarithms are equal, set the arguments equal: \(x^2 + 4 = 4x^2\).

Rearrange to get: \(4x^2 - x^2 = 4\).

Simplify: \(3x^2 = 4\).

Divide by 3: \(x^2 = \frac{4}{3}\).

Take the square root: \(x = \frac{2}{\sqrt{3}}\).

Start with the equation \(\ln(x + 4) = 2 \ln x + \ln 4\).

Apply the logarithm power rule: \(2 \ln x = \ln(x^2)\), so the equation becomes \(\ln(x + 4) = \ln(x^2) + \ln 4\).

Use the logarithm product rule: \(\ln(x^2) + \ln 4 = \ln(4x^2)\), so the equation becomes \(\ln(x + 4) = \ln(4x^2)\).

Since the logarithms are equal, set the arguments equal: \(x + 4 = 4x^2\).

Rearrange to form a quadratic equation: \(4x^2 - x - 4 = 0\).

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

Calculate the discriminant: \(b^2 - 4ac = (-1)^2 - 4 \times 4 \times (-4) = 1 + 64 = 65\).

Find the roots: \(x = \frac{-(-1) \pm \sqrt{65}}{8}\).

Calculate: \(x = \frac{1 \pm \sqrt{65}}{8}\).

Only the positive root is valid: \(x = \frac{1 + \sqrt{65}}{8} \approx 1.13\).