The number of negative integral values of x satisfying the inequality `log_((x+(5)/(2)))((x-5)/(2x-3))^(2)lt 0` is :

Answer

To solve the inequality \( \log_{(x+\frac{5}{2})}\left(\frac{(x-5)}{(2x-3)}\right)^{2} < 0 \), we will follow these steps:

### Step 1: Rewrite the Inequality
We start by rewriting the inequality:
\[ \log_{(x+\frac{5}{2})}\left(\frac{(x-5)}{(2x-3)}\right)^{2} < 0 \]
This implies that:
\[ \frac{(x-5)}{(2x-3)}^{2} < 1 \]
and the base \( x + \frac{5}{2} > 0 \).

### Step 2: Analyze the Base
For the logarithm to be defined, the base must be positive:
\[ x + \frac{5}{2} > 0 \implies x > -\frac{5}{2} \]

### Step 3: Solve the Inequality
Now we need to solve:
\[ \frac{(x-5)^{2}}{(2x-3)^{2}} < 1 \]
This can be rewritten as:
\[ (x-5)^{2} < (2x-3)^{2} \]

### Step 4: Expand Both Sides
Expanding both sides gives:
\[ x^{2} - 10x + 25 < 4x^{2} - 12x + 9 \]

### Step 5: Rearrange the Inequality
Rearranging the inequality leads to:
\[ 0 < 3x^{2} - 2x - 16 \]
or
\[ 3x^{2} - 2x - 16 > 0 \]

### Step 6: Factor the Quadratic
To factor \( 3x^{2} - 2x - 16 \), we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \]
where \( a = 3, b = -2, c = -16 \):
\[ x = \frac{2 \pm \sqrt{(-2)^{2} - 4 \cdot 3 \cdot (-16)}}{2 \cdot 3} \]
\[ x = \frac{2 \pm \sqrt{4 + 192}}{6} = \frac{2 \pm \sqrt{196}}{6} = \frac{2 \pm 14}{6} \]
This gives us the roots:
\[ x = \frac{16}{6} = \frac{8}{3} \quad \text{and} \quad x = \frac{-12}{6} = -2 \]

### Step 7: Test Intervals
The critical points are \( x = -2 \) and \( x = \frac{8}{3} \). We will test the intervals:
1. \( (-\infty, -2) \)
2. \( (-2, \frac{8}{3}) \)
3. \( (\frac{8}{3}, \infty) \)

### Step 8: Determine Signs
- For \( x < -2 \): Choose \( x = -3 \):
\[ 3(-3)^{2} - 2(-3) - 16 = 27 + 6 - 16 = 17 > 0 \]
- For \( -2 < x < \frac{8}{3} \): Choose \( x = 0 \):
\[ 3(0)^{2} - 2(0) - 16 = -16 < 0 \]
- For \( x > \frac{8}{3} \): Choose \( x = 3 \):
\[ 3(3)^{2} - 2(3) - 16 = 27 - 6 - 16 = 5 > 0 \]

### Step 9: Conclusion on Intervals
The inequality \( 3x^{2} - 2x - 16 > 0 \) holds for:
\[ x < -2 \quad \text{or} \quad x > \frac{8}{3} \]

### Step 10: Find Negative Integral Values
The only negative integral values satisfying \( x < -2 \) are \( -3, -4, -5, \ldots \). However, we need to check if these values satisfy the base condition \( x + \frac{5}{2} > 0 \):
- For \( x = -3 \): \( -3 + \frac{5}{2} = -3 + 2.5 = -0.5 \) (not valid)
- For \( x = -4 \): \( -4 + \frac{5}{2} = -4 + 2.5 = -1.5 \) (not valid)
- For \( x = -5 \): \( -5 + \frac{5}{2} = -5 + 2.5 = -2.5 \) (not valid)

Thus, there are **no negative integral values of \( x \)** satisfying the original inequality.

### Final Answer
The number of negative integral values of \( x \) satisfying the inequality is **0**.

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