If `log_(7)2=m` then `log_(49)28` is equal to

Options

A. `(2)/(1+2m)`
B. `(1+2m)/(2)`
C. `2(1+2m)`
D. `(1+m)`

Answer

To solve the problem, we need to find the value of \( \log_{49} 28 \) given that \( \log_{7} 2 = m \).

### Step-by-Step Solution:

1. **Express \( \log_{49} 28 \) in terms of base 7**:
\[ \log_{49} 28 = \frac{\log_{7} 28}{\log_{7} 49} \]

2. **Simplify \( \log_{7} 49 \)**:
\[ \log_{7} 49 = \log_{7} (7^2) = 2 \log_{7} 7 = 2 \cdot 1 = 2 \]

3. **Express \( 28 \) in terms of its prime factors**:
\[ 28 = 4 \cdot 7 = 2^2 \cdot 7 \]

4. **Use the property of logarithms to expand \( \log_{7} 28 \)**:
\[ \log_{7} 28 = \log_{7} (2^2 \cdot 7) = \log_{7} (2^2) + \log_{7} 7 = 2 \log_{7} 2 + 1 \]

5. **Substitute \( \log_{7} 2 \) with \( m \)**:
\[ \log_{7} 28 = 2m + 1 \]

6. **Substitute back into the expression for \( \log_{49} 28 \)**:
\[ \log_{49} 28 = \frac{\log_{7} 28}{\log_{7} 49} = \frac{2m + 1}{2} \]

7. **Final answer**:
\[ \log_{49} 28 = \frac{2m + 1}{2} \]

### Summary:
Thus, the value of \( \log_{49} 28 \) in terms of \( m \) is:
\[ \frac{2m + 1}{2} \]