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Question

Practice

Aratox · Answer

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**:
$$
\boxed{4}
$$