Solve the following inequation .
(iv) `log_(x^2)(x+2)lt1`

Answer

To solve the inequation \( \log_{x^2}(x+2) < 1 \), we will follow these steps:

### Step 1: Rewrite the Inequation
The given inequation can be rewritten using the change of base formula for logarithms:
\[ \log_{x^2}(x+2) < 1 \implies \frac{\log_{10}(x+2)}{\log_{10}(x^2)} < 1 \]

### Step 2: Simplify the Logarithm
Since \( \log_{10}(x^2) = 2\log_{10}(x) \), we can substitute this into our inequation:
\[ \frac{\log_{10}(x+2)}{2\log_{10}(x)} < 1 \]

### Step 3: Multiply Both Sides by \( 2\log_{10}(x) \)
Assuming \( \log_{10}(x) > 0 \) (which means \( x > 1 \)), we can multiply both sides by \( 2\log_{10}(x) \):
\[ \log_{10}(x+2) < 2\log_{10}(x) \]

### Step 4: Rewrite the Inequation
This can be rewritten as:
\[ \log_{10}(x+2) < \log_{10}(x^2) \]

### Step 5: Exponentiate Both Sides
Now, we exponentiate both sides to eliminate the logarithm:
\[ x + 2 < x^2 \]

### Step 6: Rearrange the Inequation
Rearranging gives us:
\[ 0 < x^2 - x - 2 \]

### Step 7: Factor the Quadratic
Next, we factor the quadratic:
\[ x^2 - x - 2 = (x - 2)(x + 1) \]

### Step 8: Set Up the Inequality
Now we set up the inequality:
\[ (x - 2)(x + 1) > 0 \]

### Step 9: Determine the Intervals
We find the critical points by setting each factor to zero:
- \( x - 2 = 0 \) gives \( x = 2 \)
- \( x + 1 = 0 \) gives \( x = -1 \)

### Step 10: Test the Intervals
We test the intervals:
1. \( (-\infty, -1) \)
2. \( (-1, 2) \)
3. \( (2, \infty) \)

- For \( x < -1 \) (e.g., \( x = -2 \)): \( (-)(-) > 0 \) (True)
- For \( -1 < x < 2 \) (e.g., \( x = 0 \)): \( (-)(+) < 0 \) (False)
- For \( x > 2 \) (e.g., \( x = 3 \)): \( (+)(+) > 0 \) (True)

### Step 11: Combine the Results
Thus, the solution to the inequality is:
\[ x \in (-\infty, -1) \cup (2, \infty) \]

### Final Answer
The solution set is:
\[ (-\infty, -1) \cup (2, \infty) \]