To solve the given optimization problem, we first need to convert the natural language description into a symbolic representation. This involves defining variables and formulating the objective function and constraints using these variables.

Let's denote:
- $x_1$ as the amount of indoor soil purchased,
- $x_2$ as the amount of outdoor soil purchased.

The objective is to minimize the total cost, which can be represented as $2x_1 + 3x_2$, since the indoor soil costs $2 per unit and the outdoor soil costs $3 per unit.

The constraints are:
1. The gardener requires at least 80 units of compost: Given that indoor soil contains 2 units of compost and outdoor soil contains 4 units of compost, this constraint can be represented as $2x_1 + 4x_2 \geq 80$.
2. The gardener requires at least 70 units of loam: With indoor soil containing 3 units of loam and outdoor soil containing 6 units of loam, this constraint is $3x_1 + 6x_2 \geq 70$.

Non-negativity constraints also apply since the amount of soil purchased cannot be negative: $x_1 \geq 0$ and $x_2 \geq 0$.

Thus, the symbolic representation of the problem is:
```json
{
    'sym_variables': [('x1', 'amount of indoor soil'), ('x2', 'amount of outdoor soil')],
    'objective_function': '2*x1 + 3*x2',
    'constraints': ['2*x1 + 4*x2 >= 80', '3*x1 + 6*x2 >= 70', 'x1 >= 0', 'x2 >= 0']
}
```

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

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

# Define variables
x1 = m.addVar(vtype=GRB.CONTINUOUS, name="indoor_soil", lb=0)
x2 = m.addVar(vtype=GRB.CONTINUOUS, name="outdoor_soil", lb=0)

# Set the objective function
m.setObjective(2*x1 + 3*x2, GRB.MINIMIZE)

# Add constraints
m.addConstr(2*x1 + 4*x2 >= 80, "compost_requirement")
m.addConstr(3*x1 + 6*x2 >= 70, "loam_requirement")

# Optimize the model
m.optimize()

# Print the results
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Amount of indoor soil: {x1.x}")
    print(f"Amount of outdoor soil: {x2.x}")
    print(f"Total cost: {m.objVal}")
else:
    print("No optimal solution found")
```