We are given the equation:

$\ln(3x + 4) = 2\ln(x + 1)$

Step-by-step solution:

1. Use the logarithmic property: $\ln(a^b) = b\ln(a)$ on the right-hand side: $\ln(3x + 4) = \ln((x + 1)^2)$

2. Since both sides of the equation have logarithms, we can equate the arguments: $3x + 4 = (x + 1)^2$

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

4. Rearrange the equation to form a quadratic equation: $3x + 4 = x^2 + 2x + 1$ $0 = x^2 + 2x + 1 - 3x - 4$ $0 = x^2 - x - 3$

5. Solve the quadratic equation $x^2 - x - 3 = 0$ using the quadratic formula: $x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-3)}}{2(1)}$ $x = \frac{1 \pm \sqrt{1 + 12}}{2}$ $x = \frac{1 \pm \sqrt{13}}{2}$

6. Calculate the solutions: $x = \frac{1 + \sqrt{13}}{2} \quad \text{or} \quad x = \frac{1 - \sqrt{13}}{2}$

7. Approximate the solutions: $x_1 = \frac{1 + 3.605}{2} = \frac{4.605}{2} \approx 2.30$ $x_2 = \frac{1 - 3.605}{2} = \frac{-2.605}{2} \approx -1.30$

8. Check the domain of the original logarithmic functions. For $\ln(3x + 4)$, we need $3x + 4 > 0 \Rightarrow x > -\frac{4}{3}$, and for $\ln(x + 1)$, we need $x + 1 > 0 \Rightarrow x > -1$.

   Therefore, $x \approx -1.30$ is not valid because it does not satisfy the domain of $\ln(x + 1)$.

Thus, the only valid solution is:

$x \approx 2.30 \, (\text{to 3 significant figures})$

Would you like further details or have any questions?

Here are some related questions to expand on the topic:

1. What are the properties of logarithms used in solving equations like this one?
2. How can the quadratic formula be applied in other types of logarithmic equations?
3. What are the restrictions on the domain of logarithmic functions?
4. Can logarithmic equations always be solved algebraically?
5. How would the solution change if the logarithmic base was not $e$ (natural logarithms)?

Tip: Always check the domain of the logarithmic functions before solving logarithmic equations to avoid extraneous solutions.