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The problem asks you to:

1. **Find the value of \( a \in \mathbb{R} \)** such that the slope of the tangent line to the graph of \( f(x) = a \cdot \ln(2x + 3) + x \) at the point \( x_0 = -1 \) is equal to 3.
2. For the value of \( a \) found in the previous step, explicitly calculate the equation of the tangent line at \( x_0 = -1 \).

### Step 1: Find \( a \)
The slope of the tangent line to a curve is given by the derivative of the function at the point of interest. Therefore, we need to find \( f'(x) \), the derivative of \( f(x) \), and then evaluate it at \( x_0 = -1 \).

#### Derivative of \( f(x) \):
The function is:
\[
f(x) = a \cdot \ln(2x + 3) + x
\]
Using the chain rule and the fact that the derivative of \( \ln(g(x)) \) is \( \frac{g'(x)}{g(x)} \), the derivative \( f'(x) \) becomes:
\[
f'(x) = a \cdot \frac{2}{2x + 3} + 1
\]

#### Slope at \( x = -1 \):
Substitute \( x_0 = -1 \) into \( f'(x) \):
\[
f'(-1) = a \cdot \frac{2}{2(-1) + 3} + 1 = a \cdot \frac{2}{1} + 1 = 2a + 1
\]
We are told that the slope at \( x_0 = -1 \) is 3, so:
\[
2a + 1 = 3
\]
Solve for \( a \):
\[
2a = 2 \quad \Rightarrow \quad a = 1
\]

### Step 2: Find the equation of the tangent line
Now that we know \( a = 1 \), the function becomes:
\[
f(x) = \ln(2x + 3) + x
\]

#### Find \( f(-1) \):
To find the equation of the tangent line, we need the point \( (x_0, f(x_0)) \) where the tangent touches the curve. Evaluate \( f(x) \) at \( x_0 = -1 \):
\[
f(-1) = \ln(2(-1) + 3) + (-1) = \ln(1) - 1 = 0 - 1 = -1
\]
So the point of tangency is \( (-1, -1) \).

#### Equation of the tangent line:
The equation of a line is given by:
\[
y - y_1 = m(x - x_1)
\]
where \( m \) is the slope, and \( (x_1, y_1) \) is the point of tangency. We know the slope is 3, and the point is \( (-1, -1) \). Substituting into the equation:
\[
y - (-1) = 3(x - (-1)) \quad \Rightarrow \quad y + 1 = 3(x + 1)
\]
Simplifying:
\[
y = 3x + 3 - 1 \quad \Rightarrow \quad y = 3x + 2
\]

### Final Answer:
- The value of \( a \) is \( 1 \).
- The equation of the tangent line is \( y = 3x + 2 \).

Do you have any questions or need more details?

### Related Questions:
1. What is the significance of the chain rule in finding derivatives?
2. How would the result change if \( a \) was a different value?
3. How do you find the derivative of more complex functions like logarithmic expressions?
4. What does the tangent line represent geometrically in this context?
5. How do you verify if the tangent line correctly represents the slope at the given point?

### Tip:
Always double-check your derivative calculations, especially when applying the chain rule to composite functions.

Given the function f(x) = a * ln(2x + 3) + x, find a ∈ ℝ such that the slope of the tangent line to the graph of f(x) at the point x_0 = -1 is equal to 3. For the value of a found, explicitly calculate the equation of the tangent line at x_0 = -1.

Step-by-step solution to finding the value of a such that the slope of the tangent line to the function f(x) = a ln(2x + 3) + x at x = -1 is 3, and calculating the explicit equation of the tangent line. Concepts include calculus, derivatives, and logarithmic functions.

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