We are given the equation:

\[
\log_9 (x+5) + \log_9 (x+4) = 2
\]

### Step 1: Combine the logarithms
Using the logarithmic property \( \log_b A + \log_b B = \log_b (A \times B) \), we can combine the two logarithms on the left-hand side:

\[
\log_9 \left( (x+5)(x+4) \right) = 2
\]

### Step 2: Convert the logarithmic equation to an exponential equation
Next, we convert the logarithmic equation to an exponential form. Recall that \( \log_b y = z \) is equivalent to \( b^z = y \). Applying this to our equation:

\[
(x+5)(x+4) = 9^2
\]
\[
(x+5)(x+4) = 81
\]

### Step 3: Expand the left-hand side
Now expand the product on the left-hand side:

\[
x^2 + 4x + 5x + 20 = 81
\]
\[
x^2 + 9x + 20 = 81
\]

### Step 4: Rearrange the equation
Now, subtract 81 from both sides to set the equation equal to 0:

\[
x^2 + 9x + 20 - 81 = 0
\]
\[
x^2 + 9x - 61 = 0
\]

### Step 5: Solve the quadratic equation
We now solve the quadratic equation \( x^2 + 9x - 61 = 0 \) using the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \( a = 1 \), \( b = 9 \), and \( c = -61 \). Substituting these into the quadratic formula:

\[
x = \frac{-9 \pm \sqrt{9^2 - 4(1)(-61)}}{2(1)}
\]
\[
x = \frac{-9 \pm \sqrt{81 + 244}}{2}
\]
\[
x = \frac{-9 \pm \sqrt{325}}{2}
\]

Since \( \sqrt{325} \approx 18.03 \), we get two possible solutions:

\[
x = \frac{-9 + 18.03}{2} \quad \text{or} \quad x = \frac{-9 - 18.03}{2}
\]
\[
x = \frac{9.03}{2} \quad \text{or} \quad x = \frac{-27.03}{2}
\]
\[
x \approx 4.52 \quad \text{or} \quad x \approx -13.52
\]

### Step 6: Check for valid solutions
Since the logarithm of a negative number is undefined, we need to check the validity of the solutions by substituting them back into the original logarithmic expressions. 

- For \( x \approx 4.52 \), both \( x + 5 \) and \( x + 4 \) are positive.
- For \( x \approx -13.52 \), \( x + 4 \) is negative, which makes this solution invalid.

### Final Answer:
The valid solution is:

\[
x \approx 4.52
\]

Would you like any more details or clarifications?

### Related questions:
1. How do we handle logarithms with bases other than 10 and \( e \)?
2. What happens if both solutions are invalid in a logarithmic equation?
3. Can logarithmic properties be applied to complex numbers?
4. How does the quadratic formula derive solutions for non-integer values of \( x \)?
5. What other methods can solve logarithmic equations apart from converting to exponential form?

### Tip:
Always check the domain of logarithmic functions to ensure that solutions don't lead to taking the logarithm of a negative number or zero.

Learn how to solve the logarithmic equation log9(x+5) + log9(x+4) = 2 by combining logarithms and using the quadratic formula. This step-by-step guide is perfect for students in Grades 9-12, covering logarithmic properties and algebraic techniques.

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