Find number of integral solution`log_(x+3)(x^(2)-x)lt1`

Answer

To find the number of integral solutions for the inequality \( \log_{(x+3)}(x^2 - x) < 1 \), we will follow these steps:

### Step 1: Rewrite the logarithmic inequality
Using the property of logarithms, we can rewrite the inequality:
\[ \log_{(x+3)}(x^2 - x) < 1 \implies x + 3^1 > x^2 - x \]
This simplifies to:
\[ x + 3 > x^2 - x \]

### Step 2: Rearrange the inequality
Rearranging the inequality gives:
\[ 0 > x^2 - 2x - 3 \]
or equivalently:
\[ x^2 - 2x - 3 < 0 \]

### Step 3: Factor the quadratic expression
Next, we factor the quadratic:
\[ x^2 - 2x - 3 = (x - 3)(x + 1) \]
Thus, we have:
\[ (x - 3)(x + 1) < 0 \]

### Step 4: Find critical points
The critical points from the factors are \( x = 3 \) and \( x = -1 \). These points divide the number line into intervals.

### Step 5: Test the intervals
We will test the sign of \( (x - 3)(x + 1) \) in the intervals:
1. \( (-\infty, -1) \)
2. \( (-1, 3) \)
3. \( (3, \infty) \)

- For \( x < -1 \) (e.g., \( x = -2 \)):
\[ (-2 - 3)(-2 + 1) = (-5)(-1) > 0 \]
- For \( -1 < x < 3 \) (e.g., \( x = 0 \)):
\[ (0 - 3)(0 + 1) = (-3)(1) < 0 \]
- For \( x > 3 \) (e.g., \( x = 4 \)):
\[ (4 - 3)(4 + 1) = (1)(5) > 0 \]

Thus, the inequality \( (x - 3)(x + 1) < 0 \) holds for the interval \( (-1, 3) \).

### Step 6: Determine integral solutions
The integral solutions in the interval \( (-1, 3) \) are:
- \( 0, 1, 2 \)

### Step 7: Check for validity of each solution
We need to check if these values make \( \log_{(x+3)}(x^2 - x) \) defined:
- For \( x = 0 \): \( \log_{3}(0) \) is undefined.
- For \( x = 1 \): \( \log_{4}(0) \) is undefined.
- For \( x = 2 \): \( \log_{5}(2) \) is defined.

Thus, the only valid integral solution is \( x = 2 \).

### Final Answer
The number of integral solutions is **1**.