Question

Practice

Answer 1

To solve the equation \( \ln(x - 2) + \ln(x + 4) = 2\ln 4 \), we can use properties of logarithms.

1. **Combine the logarithms on the left side**:
\[
\ln((x - 2)(x + 4)) = 2\ln 4
\]
Using the property \( \ln a + \ln b = \ln(ab) \).

2. **Simplify the right side**:
\[
2\ln 4 = \ln(4^2) = \ln 16
\]
Thus, the equation now is:
\[
\ln((x - 2)(x + 4)) = \ln 16
\]

3. **Remove the logarithms** (assuming both sides are defined and positive):
\[
(x - 2)(x + 4) = 16
\]

4. **Expand and rearrange the equation**:
\[
x^2 + 4x - 2x - 8 = 16
\]
\[
x^2 + 2x - 8 - 16 = 0
\]
\[
x^2 + 2x - 24 = 0
\]

5. **Factor the quadratic**:
\[
(x + 6)(x - 4) = 0
\]
Therefore, the solutions are:
\[
x + 6 = 0 \quad \Rightarrow \quad x = -6
\]
\[
x - 4 = 0 \quad \Rightarrow \quad x = 4
\]

6. **Check for extraneous solutions**:
- For \( x = -6 \):
\[
\ln(-6 - 2) + \ln(-6 + 4) = \ln(-8) + \ln(-2) \quad \text{(undefined)}
\]
So, \( x = -6 \) is extraneous.

- For \( x = 4 \):
\[
\ln(4 - 2) + \ln(4 + 4) = \ln(2) + \ln(8)
\]
\[
\ln(2 \cdot 8) = \ln(16) \quad \text{(valid)}
\]
And checking the original equation:
\[
2 \ln 4 = \ln(16),
\]
which holds true.

Thus, the only valid solution is \( x = 4 \).

**Final Answer**:
\[
\boxed{4}
\]