If `log_(7) 2 = m`, then find ` log_(49) 28` in terms of m.

Explanation

To solve the problem, we need to find \( \log_{49} 28 \) in terms of \( m \), where \( m = \log_{7} 2 \).

### Step-by-Step Solution:

1. **Apply the Change of Base Formula**:
\[ \log_{49} 28 = \frac{\log 28}{\log 49} \]

**Hint**: Remember that the change of base formula allows us to express logarithms in terms of a common base.

2. **Factorize 28 and 49**:
\[ 28 = 2^2 \times 7 \quad \text{and} \quad 49 = 7^2 \]

**Hint**: Factor the numbers into their prime factors to simplify the logarithmic expressions.

3. **Rewrite the Logarithms**:
\[ \log 28 = \log(2^2 \times 7) = \log(2^2) + \log(7) = 2\log(2) + \log(7) \]
\[ \log 49 = \log(7^2) = 2\log(7) \]

**Hint**: Use the properties of logarithms: \( \log(a \times b) = \log a + \log b \) and \( \log(a^n) = n \log a \).

4. **Substitute Back into the Change of Base Formula**:
\[ \log_{49} 28 = \frac{2\log(2) + \log(7)}{2\log(7)} \]

**Hint**: Substitute the expressions for \( \log 28 \) and \( \log 49 \) that we derived.

5. **Separate the Terms**:
\[ \log_{49} 28 = \frac{2\log(2)}{2\log(7)} + \frac{\log(7)}{2\log(7)} = \frac{\log(2)}{\log(7)} + \frac{1}{2} \]

**Hint**: Simplify the fractions by canceling out common terms.

6. **Substitute \( m \)**:
Since \( m = \log_{7} 2 = \frac{\log(2)}{\log(7)} \), we can substitute \( m \) into the equation:
\[ \log_{49} 28 = m + \frac{1}{2} \]

**Hint**: Use the definition of \( m \) to replace \( \frac{\log(2)}{\log(7)} \) with \( m \).

### Final Answer:
\[ \log_{49} 28 = m + \frac{1}{2} \]