Number of integers satisfying the inequality `log_((x+3))(x^2-x) lt 1` is

Answer

To solve the inequality \( \log_{(x+3)}(x^2 - x) < 1 \), we will follow these steps:

### Step 1: Understand the logarithmic inequality
The inequality \( \log_{(x+3)}(x^2 - x) < 1 \) can be rewritten using the property of logarithms:
\[ x^2 - x < x + 3 \]
This is valid when the base \( x + 3 > 1 \), which implies \( x > -2 \).

### Step 2: Solve the inequality \( x^2 - x < x + 3 \)
Rearranging the inequality gives:
\[ x^2 - x - x - 3 < 0 \implies x^2 - 2x - 3 < 0 \]
Now we can factor the quadratic:
\[ (x - 3)(x + 1) < 0 \]

### Step 3: Determine the intervals
To find the intervals where this inequality holds, we will analyze the sign of the product \( (x - 3)(x + 1) \):
- The roots are \( x = -1 \) and \( x = 3 \).
- The critical points divide the number line into intervals: \( (-\infty, -1) \), \( (-1, 3) \), and \( (3, \infty) \).

Testing a point from each interval:
- For \( x = -2 \) (in \( (-\infty, -1) \)): \( (-2 - 3)(-2 + 1) = (-5)(-1) > 0 \)
- For \( x = 0 \) (in \( (-1, 3) \)): \( (0 - 3)(0 + 1) = (-3)(1) < 0 \)
- For \( x = 4 \) (in \( (3, \infty) \)): \( (4 - 3)(4 + 1) = (1)(5) > 0 \)

Thus, the solution to \( (x - 3)(x + 1) < 0 \) is:
\[ -1 < x < 3 \]

### Step 4: Combine with the base condition
From Step 1, we have the condition \( x > -2 \). Therefore, we need to combine this with our previous result:
\[ -1 < x < 3 \quad \text{and} \quad x > -2 \]
The more restrictive condition is \( -1 < x < 3 \).

### Step 5: Identify integer solutions
The integers that satisfy \( -1 < x < 3 \) are:
- \( 0, 1, 2 \)

Thus, the number of integers satisfying the inequality is:
\[ \text{Number of integers} = 3 \]

### Final Answer
The number of integers satisfying the inequality \( \log_{(x+3)}(x^2 - x) < 1 \) is \( \boxed{3} \).