Find the sum of all integers satisfying the inequalities
`log_(5)(x-3)+1/2log_(5)3lt1/2log_(5)(2x^(2)-6x+7) and log_(3)x+log_(sqrt3)x+log_(1/3)x lt 6`

Answer

To solve the given inequalities step by step, we will break down each part of the problem.

### Step 1: Solve the first inequality
The first inequality is:
\[ \log_{5}(x-3) + \frac{1}{2}\log_{5}(3) < \frac{1}{2}\log_{5}(2x^{2}-6x+7) \]

#### Step 1.1: Simplify the inequality
We can rewrite the inequality using properties of logarithms:
\[ \log_{5}(x-3) + \log_{5}(3^{1/2}) < \log_{5}((2x^{2}-6x+7)^{1/2}) \]

This can be combined as:
\[ \log_{5}((x-3) \cdot \sqrt{3}) < \log_{5}(\sqrt{2x^{2}-6x+7}) \]

#### Step 1.2: Remove the logarithm
Since the logarithm is a monotonically increasing function, we can remove the logarithm (assuming the arguments are positive):
\[ (x-3) \cdot \sqrt{3} < \sqrt{2x^{2}-6x+7} \]

#### Step 1.3: Square both sides
Squaring both sides gives:
\[ 3(x-3)^{2} < 2x^{2}-6x+7 \]

#### Step 1.4: Expand and simplify
Expanding the left side:
\[ 3(x^{2} - 6x + 9) < 2x^{2} - 6x + 7 \]
\[ 3x^{2} - 18x + 27 < 2x^{2} - 6x + 7 \]

Now, rearranging gives:
\[ 3x^{2} - 2x^{2} - 18x + 6x + 27 - 7 < 0 \]
\[ x^{2} - 12x + 20 < 0 \]

#### Step 1.5: Factor the quadratic
Factoring the quadratic:
\[ (x-10)(x-2) < 0 \]

#### Step 1.6: Determine the intervals
The critical points are \( x = 2 \) and \( x = 10 \). Testing intervals:
- For \( x < 2 \): positive
- For \( 2 < x < 10 \): negative
- For \( x > 10 \): positive

Thus, the solution for the first inequality is:
\[ 2 < x < 10 \]

### Step 2: Solve the second inequality
The second inequality is:
\[ \log_{3}x + \log_{\sqrt{3}}x + \log_{\frac{1}{3}}x < 6 \]

#### Step 2.1: Rewrite the logarithms
Using properties of logarithms:
\[ \log_{3}x + \frac{1}{2}\log_{3}x - \log_{3}x < 6 \]
\[ \frac{1}{2}\log_{3}x < 6 \]

#### Step 2.2: Multiply by 2
Multiplying both sides by 2:
\[ \log_{3}x < 12 \]

#### Step 2.3: Remove the logarithm
Removing the logarithm gives:
\[ x < 3^{12} \]

### Step 3: Combine the results
From the first inequality, we have:
\[ 2 < x < 10 \]

From the second inequality, we have:
\[ x < 3^{12} \]

Since \( 3^{12} \) is much larger than 10, we can combine the results:
\[ 2 < x < 10 \]

### Step 4: Find the integers in the range
The integers satisfying \( 2 < x < 10 \) are:
\[ 3, 4, 5, 6, 7, 8, 9 \]

### Step 5: Calculate the sum of these integers
Calculating the sum:
\[ 3 + 4 + 5 + 6 + 7 + 8 + 9 = 42 \]

### Final Answer
The sum of all integers satisfying the inequalities is:
\[ \boxed{42} \]