To solve this optimization problem, we first need to define the symbolic representation of the variables and the objective function, along with the constraints given in the problem description.

Let's denote:
- \(x_1\) as the amount invested in the mining industry.
- \(x_2\) as the amount invested in the logging industry.

The objective is to maximize the return on investment. Given that the mining industry yields a 9% return and the logging industry yields a 5% return, the total return can be represented by the objective function:
\[0.09x_1 + 0.05x_2\]

The constraints are as follows:
1. The total amount invested is $100,000: \(x_1 + x_2 = 100,000\)
2. At least 30% of the investment must be in the mining industry: \(x_1 \geq 0.3 \times 100,000\)
3. At most 55% of the investment can be in the logging industry: \(x_2 \leq 0.55 \times 100,000\)
4. Non-negativity constraints: \(x_1 \geq 0\) and \(x_2 \geq 0\), since the amount invested cannot be negative.

Thus, the symbolic representation of the problem is:
```json
{
    'sym_variables': [('x1', 'amount invested in the mining industry'), ('x2', 'amount invested in the logging industry')],
    'objective_function': '0.09*x1 + 0.05*x2',
    'constraints': ['x1 + x2 = 100000', 'x1 >= 30000', 'x2 <= 55000', 'x1 >= 0', 'x2 >= 0']
}
```

To solve this problem using Gurobi in Python, we can write the following code:
```python
from gurobipy import *

# Create a new model
m = Model("Investment_Optimization")

# Define variables
x1 = m.addVar(name="mining_investment", lb=0)
x2 = m.addVar(name="logging_investment", lb=0)

# Set the objective function
m.setObjective(0.09*x1 + 0.05*x2, GRB.MAXIMIZE)

# Add constraints
m.addConstr(x1 + x2 == 100000, name="total_investment")
m.addConstr(x1 >= 30000, name="min_mining_investment")
m.addConstr(x2 <= 55000, name="max_logging_investment")

# Optimize the model
m.optimize()

# Print the results
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Amount invested in mining: {x1.x}")
    print(f"Amount invested in logging: {x2.x}")
    print(f"Total return: {m.objVal}")
else:
    print("No optimal solution found")
```