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We are tasked with solving the logarithmic equation:

\[
\log_8(x + 4) + \log_8(x + 6) = \log_8(24)
\]

### Step 1: Apply the properties of logarithms

Use the product rule for logarithms, which states:

\[
\log_b(A) + \log_b(B) = \log_b(A \cdot B)
\]

So, we can rewrite the left-hand side as:

\[
\log_8((x + 4)(x + 6)) = \log_8(24)
\]

### Step 2: Remove the logarithms

Since the bases of the logarithms on both sides are the same, we can drop the logs and set the arguments equal to each other:

\[
(x + 4)(x + 6) = 24
\]

### Step 3: Expand and solve the quadratic equation

Now expand the left-hand side:

\[
(x + 4)(x + 6) = x^2 + 6x + 4x + 24 = x^2 + 10x + 24
\]

Now set this equal to 24:

\[
x^2 + 10x + 24 = 24
\]

Subtract 24 from both sides:

\[
x^2 + 10x = 0
\]

### Step 4: Solve the quadratic equation

Factor the equation:

\[
x(x + 10) = 0
\]

This gives two solutions:

\[
x = 0 \quad \text{or} \quad x = -10
\]

### Step 5: Check for extraneous solutions

Since logarithms are only defined for positive arguments, the expressions inside the logarithms must be positive:

1. \(x + 4 > 0 \Rightarrow x > -4\)
2. \(x + 6 > 0 \Rightarrow x > -6\)

Thus, \(x\) must be greater than \(-4\). This means the solution \(x = -10\) is invalid, leaving only \(x = 0\).

### Final Answer:
The solution set is \(\{0\}\).

Would you like further details or have any questions?

Here are 5 related questions to explore further:
1. What is the product rule for logarithms?
2. How can you factor a quadratic equation like \(x^2 + 10x = 0\)?
3. Why are logarithmic equations undefined for negative values inside the logarithms?
4. How would this problem change if the base of the logarithms was different?
5. Can you always drop the logs when the bases are the same in an equation?

**Tip:** Always check for extraneous solutions when dealing with logarithmic equations to ensure the arguments of the logarithms are valid!

Solve the logarithmic equation: log_8(x + 4) + log_8(x + 6) = log_8(24).

Learn how to solve the logarithmic equation log_8(x + 4) + log_8(x + 6) = log_8(24) using the product rule for logarithms. This problem also involves solving a quadratic equation. Step-by-step explanation provided, suitable for Grades 9-12.

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