Let `g(x)=f(logx)+f(2-logx) and f^(prime prime)(x)<0AAx in (0,3)dot` Then find the interval in which `g(x)` increases.

Explanation

To find the interval in which the function \( g(x) = f(\log x) + f(2 - \log x) \) increases, we need to analyze its derivative \( g'(x) \) and determine where it is greater than zero.

### Step-by-step Solution:

1. **Differentiate \( g(x) \)**:
\[ g'(x) = \frac{d}{dx}[f(\log x)] + \frac{d}{dx}[f(2 - \log x)] \]
Using the chain rule, we differentiate each term:
\[ g'(x) = f'(\log x) \cdot \frac{1}{x} + f'(2 - \log x) \cdot \left(-\frac{1}{x}\right) \]
Simplifying this gives:
\[ g'(x) = \frac{1}{x} \left( f'(\log x) - f'(2 - \log x) \right) \]

2. **Set \( g'(x) > 0 \)**:
For \( g(x) \) to be increasing, we need:
\[ g'(x) > 0 \implies f'(\log x) - f'(2 - \log x) > 0 \]
This simplifies to:
\[ f'(\log x) > f'(2 - \log x) \]

3. **Analyze \( f'(x) \)**:
Given that \( f''(x) < 0 \) for \( x \in (0, 3) \), we know that \( f'(x) \) is a decreasing function. This means that if \( a < b \), then \( f'(a) > f'(b) \).

4. **Set up the inequality**:
From \( f'(\log x) > f'(2 - \log x) \), we can infer that:
\[ \log x < 2 - \log x \]
Rearranging this gives:
\[ 2\log x < 2 \implies \log x < 1 \]

5. **Solve for \( x \)**:
The inequality \( \log x < 1 \) implies:
\[ x < e \]
Therefore, \( g(x) \) is increasing for \( x \) in the interval \( (0, e) \).

### Conclusion:
The function \( g(x) \) increases in the interval \( (0, e) \).