If `3 leale2015,3leble2015` such that `log_ab+6log_ba=5`, the number of ordered pairs (a,b) of integers is

Options

A. 48
B. 50
C. 52
D. 54

Answer

To solve the problem, we need to find the number of ordered pairs \((a, b)\) of integers such that \( \log_b a + 6 \log_a b = 5 \).

### Step-by-Step Solution:

1. **Rewrite the logarithmic expression**:
We start with the equation:
\[ \log_b a + 6 \log_a b = 5 \]
We can use the property of logarithms that states \( \log_a b = \frac{1}{\log_b a} \). Let's denote \( \log_b a = t \). Therefore, we can rewrite the equation as:
\[ t + 6 \cdot \frac{1}{t} = 5 \]

2. **Multiply through by \( t \)**:
To eliminate the fraction, multiply the entire equation by \( t \):
\[ t^2 + 6 = 5t \]

3. **Rearrange into a standard quadratic equation**:
Rearranging gives us:
\[ t^2 - 5t + 6 = 0 \]

4. **Factor the quadratic**:
We can factor this quadratic equation:
\[ (t - 2)(t - 3) = 0 \]
Thus, the solutions for \( t \) are:
\[ t = 2 \quad \text{or} \quad t = 3 \]

5. **Convert back to logarithmic form**:
Recall that \( t = \log_b a \). Therefore, we have two cases:
- Case 1: \( \log_b a = 2 \) implies \( b = a^2 \)
- Case 2: \( \log_b a = 3 \) implies \( b = a^3 \)

6. **Determine the range for \( a \) and \( b \)**:
We know from the problem that \( a \) and \( b \) must be integers such that \( 3 \leq a \leq 2015 \) and \( 3 \leq b \leq 2015 \).

### Case 1: \( b = a^2 \)

- We need \( a^2 \leq 2015 \).
- The maximum integer \( a \) can take is \( \lfloor \sqrt{2015} \rfloor = 44 \).
- The possible integer values for \( a \) are from 3 to 44.
- The total number of values for \( a \) is \( 44 - 3 + 1 = 42 \).

### Case 2: \( b = a^3 \)

- We need \( a^3 \leq 2015 \).
- The maximum integer \( a \) can take is \( \lfloor \sqrt[3]{2015} \rfloor = 12 \).
- The possible integer values for \( a \) are from 3 to 12.
- The total number of values for \( a \) is \( 12 - 3 + 1 = 10 \).

### Final Calculation:
Now, we sum the number of ordered pairs from both cases:
- From Case 1: 42 pairs
- From Case 2: 10 pairs

Thus, the total number of ordered pairs \((a, b)\) is:
\[ 42 + 10 = 52 \]

### Conclusion:
The number of ordered pairs \((a, b)\) of integers is \( \boxed{52} \).