To solve the problem of finding the probability that \( \log_a b \) is an integer when \( a \) and \( b \) are distinct numbers selected from the set \( \{5^1, 5^2, 5^3, \ldots, 5^9\} \), we can follow these steps:
### Step 1: Understanding the logarithmic condition
The expression \( \log_a b \) can be rewritten using the change of base formula:
\[
\log_a b = \frac{\log b}{\log a}
\]
For \( \log_a b \) to be an integer, \( \frac{\log b}{\log a} \) must be an integer. This occurs when \( \log b \) is a multiple of \( \log a \).
### Step 2: Expressing \( a \) and \( b \)
Let \( a = 5^m \) and \( b = 5^n \), where \( m \) and \( n \) are distinct integers chosen from the set \( \{1, 2, 3, \ldots, 9\} \). Then:
\[
\log_a b = \log_{5^m} (5^n) = \frac{n \log 5}{m \log 5} = \frac{n}{m}
\]
Thus, \( \log_a b \) is an integer if and only if \( n \) is divisible by \( m \).
### Step 3: Counting favorable cases
We need to count the pairs \( (m, n) \) such that \( n \) is divisible by \( m \) and \( m \neq n \).
- If \( n = 1 \), \( m \) can be \( 2, 3, 4, 5, 6, 7, 8, 9 \) (8 choices).
- If \( n = 2 \), \( m \) can be \( 4, 6, 8 \) (3 choices).
- If \( n = 3 \), \( m \) can be \( 6, 9 \) (2 choices).
- If \( n = 4 \), \( m \) can be \( 8 \) (1 choice).
- If \( n = 5, 6, 7, 8, 9 \), there are no valid \( m \) values since \( m \) must be less than \( n \).
Now, we sum the favorable cases:
\[
8 + 3 + 2 + 1 = 14
\]
### Step 4: Total number of ways to choose \( a \) and \( b \)
The total number of ways to choose 2 distinct numbers from 9 numbers is given by \( \binom{9}{2} \):
\[
\binom{9}{2} = \frac{9 \times 8}{2} = 36
\]
### Step 5: Calculating the probability
The probability that \( \log_a b \) is an integer is given by the ratio of favorable cases to total cases:
\[
P(\log_a b \text{ is an integer}) = \frac{14}{36} = \frac{7}{18}
\]
### Final Answer
Thus, the probability that \( \log_a b \) is an integer is:
\[
\frac{7}{18}
\]
