Start with the equation \(\ln(2x + 3) = 2 \ln x + \ln 3\).

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

Substitute: \(\ln(2x + 3) = \ln(x^2) + \ln 3\).

Use the logarithm product rule: \(\ln(x^2) + \ln 3 = \ln(3x^2)\).

Equate the logarithms: \(\ln(2x + 3) = \ln(3x^2)\).

Remove the logarithms: \(2x + 3 = 3x^2\).

Rearrange to form a quadratic equation: \(3x^2 - 2x - 3 = 0\).

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

Calculate the discriminant: \(b^2 - 4ac = (-2)^2 - 4 \times 3 \times (-3) = 4 + 36 = 40\).

Find the roots: \(x = \frac{2 \pm \sqrt{40}}{6}\).

Simplify: \(x = \frac{2 \pm 2\sqrt{10}}{6} = \frac{1 \pm \sqrt{10}}{3}\).

Calculate the positive root: \(x = \frac{1 + \sqrt{10}}{3} \approx 1.39\).

Thus, the solution is \(x = 1.39\) to 3 significant figures.

Start with the equation \(\ln(3x + 4) = 2 \ln(x + 1)\).

Apply the logarithm power rule: \(\ln(3x + 4) = \ln((x + 1)^2)\).

Since the logarithms are equal, set the arguments equal: \(3x + 4 = (x + 1)^2\).

Expand the right side: \(3x + 4 = x^2 + 2x + 1\).

Rearrange to form a quadratic equation: \(x^2 - x - 3 = 0\).

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

Calculate the discriminant: \(b^2 - 4ac = (-1)^2 - 4(1)(-3) = 1 + 12 = 13\).

Find the roots: \(x = \frac{1 \pm \sqrt{13}}{2}\).

Calculate the positive root: \(x = \frac{1 + \sqrt{13}}{2} \approx 2.30\).

Thus, the solution is \(x = 2.30\) to 3 significant figures.

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

Use the properties of logarithms: \(\ln a - \ln b = \ln \left(\frac{a}{b}\right)\) and \(\ln a^b = b \ln a\).

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

This becomes \(\ln \left(\frac{1 + x^2}{x^2}\right) = 1\).

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

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

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

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

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

Calculate \(x\) using \(e \approx 2.718\):

\(x \approx \sqrt{\frac{1}{2.718 - 1}} \approx 0.763\).

Therefore, the solution is \(x = 0.763\) to 3 significant figures.

Start with the equation \(\ln(5-x) = \ln 5 - \ln x\).

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

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

Since the logarithms are equal, their arguments must be equal:

\(5-x = \frac{5}{x}\).

Multiply both sides by \(x\) to eliminate the fraction:

\(x(5-x) = 5\).

Expand and rearrange to form a quadratic equation:

\(x^2 - 5x + 5 = 0\).

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1\), \(b = -5\), \(c = 5\):

\(x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1}\).

\(x = \frac{5 \pm \sqrt{25 - 20}}{2}\).

\(x = \frac{5 \pm \sqrt{5}}{2}\).

Calculate the roots:

\(x_1 = \frac{5 + \sqrt{5}}{2} \approx 3.62\).

\(x_2 = \frac{5 - \sqrt{5}}{2} \approx 1.38\).

Thus, the solutions are \(x = 1.38\) and \(x = 3.62\).

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

Exponentiate both sides to remove the natural logarithm: \(2 + e^{-x} = e^2\).

Rearrange to solve for \(e^{-x}\): \(e^{-x} = e^2 - 2\).

Take the natural logarithm of both sides: \(-x = \ln(e^2 - 2)\).

Multiply by -1 to solve for \(x\): \(x = -\ln(e^2 - 2)\).

Calculate the value: \(x \approx -1.68\).