To solve the problem, we need to evaluate the integrals \( I_1 \) and \( I_2 \) and find the ratio \( \frac{I_1}{I_2} \).
### Step-by-Step Solution:
1. **Define the Integrals:**
\[
I_1 = \int_{10}^{20} \frac{\ln x}{\ln x + \ln(30 - x)} \, dx
\]
\[
I_2 = \int_{20}^{30} \frac{\ln x}{\ln x + \ln(50 - x)} \, dx
\]
2. **Using Symmetry in \( I_1 \):**
We can use the property of integrals:
\[
\int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx
\]
For \( I_1 \), we have:
\[
I_1 = \int_{10}^{20} \frac{\ln(30 - x)}{\ln(30 - x) + \ln x} \, dx
\]
This is because \( \ln(30 - x) \) can be substituted for \( \ln x \) and vice versa.
3. **Adding the Two Forms of \( I_1 \):**
Now, we add the two expressions for \( I_1 \):
\[
2I_1 = \int_{10}^{20} \left( \frac{\ln x}{\ln x + \ln(30 - x)} + \frac{\ln(30 - x)}{\ln(30 - x) + \ln x} \right) \, dx
\]
The sum of the fractions simplifies to:
\[
2I_1 = \int_{10}^{20} 1 \, dx
\]
4. **Evaluate the Integral:**
\[
2I_1 = \int_{10}^{20} 1 \, dx = [x]_{10}^{20} = 20 - 10 = 10
\]
Thus,
\[
I_1 = \frac{10}{2} = 5
\]
5. **Using Symmetry in \( I_2 \):**
Similarly, for \( I_2 \):
\[
I_2 = \int_{20}^{30} \frac{\ln(50 - x)}{\ln(50 - x) + \ln x} \, dx
\]
Adding both forms of \( I_2 \):
\[
2I_2 = \int_{20}^{30} \left( \frac{\ln x}{\ln x + \ln(50 - x)} + \frac{\ln(50 - x)}{\ln(50 - x) + \ln x} \right) \, dx
\]
This simplifies to:
\[
2I_2 = \int_{20}^{30} 1 \, dx
\]
6. **Evaluate the Integral:**
\[
2I_2 = \int_{20}^{30} 1 \, dx = [x]_{20}^{30} = 30 - 20 = 10
\]
Thus,
\[
I_2 = \frac{10}{2} = 5
\]
7. **Finding the Ratio:**
Now, we can find the ratio:
\[
\frac{I_1}{I_2} = \frac{5}{5} = 1
\]
### Final Answer:
\[
\frac{I_1}{I_2} = 1
\]
