Let `f(x)=x+3ln(x-2)&g(x)=x+5ln(x-1),` then the set of `x` satisfying the inequality `f^(prime)(x)

Options

A. `(2, 7//2)`
B. `(1,2)cup(7//2,oo)`
C. `(2,oo)`
D. `7//2,oo`

Answer

To solve the problem, we need to find the set of \( x \) satisfying the inequality \( f'(x) < g'(x) \), where \( f(x) = x + 3 \ln(x - 2) \) and \( g(x) = x + 5 \ln(x - 1) \).

### Step-by-Step Solution:

1. **Find the derivatives \( f'(x) \) and \( g'(x) \)**:
- For \( f(x) = x + 3 \ln(x - 2) \):
\[ f'(x) = 1 + \frac{3}{x - 2} \]
- For \( g(x) = x + 5 \ln(x - 1) \):
\[ g'(x) = 1 + \frac{5}{x - 1} \]

2. **Set up the inequality**:
We need to solve the inequality:
\[ f'(x) < g'(x) \]
This translates to:
\[ 1 + \frac{3}{x - 2} < 1 + \frac{5}{x - 1} \]

3. **Simplify the inequality**:
By subtracting 1 from both sides, we have:
\[ \frac{3}{x - 2} < \frac{5}{x - 1} \]

4. **Cross-multiply to eliminate the fractions**:
We can cross-multiply (keeping in mind the signs of the expressions):
\[ 3(x - 1) < 5(x - 2) \]

5. **Expand both sides**:
\[ 3x - 3 < 5x - 10 \]

6. **Rearrange the inequality**:
\[ -3 + 10 < 5x - 3x \]
\[ 7 < 2x \]
or equivalently,
\[ x > \frac{7}{2} \]

7. **Consider the domain of the functions**:
Since \( f(x) \) and \( g(x) \) are defined for \( x > 2 \) and \( x > 1 \) respectively, the effective domain for our inequality is \( x > 2 \).

8. **Combine the results**:
The solution to the inequality \( x > \frac{7}{2} \) must also respect the domain \( x > 2 \). Therefore, the solution set is:
\[ x \in \left(\frac{7}{2}, \infty\right) \]

### Final Answer:
The set of \( x \) satisfying the inequality \( f'(x) < g'(x) \) is:
\[ \text{Option d: } \left(\frac{7}{2}, \infty\right) \]