If `log_(7)2 = lambda`, then the value of `log_(49) `(28) is

Options

A. `(2 lambda + 1)`
B. `( 2 lambda + 3)`
C. `(1)/(2) ( 2 lambda + 1)`
D. `2 (2 lambda + 1)`

Answer

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

### Step-by-Step Solution:

1. **Express the logarithm in terms of base 7**:
\[ \log_{49}(28) = \frac{\log_{7}(28)}{\log_{7}(49)} \]
Since \( 49 = 7^2 \), we have:
\[ \log_{7}(49) = \log_{7}(7^2) = 2 \]
Thus, we can rewrite:
\[ \log_{49}(28) = \frac{\log_{7}(28)}{2} \]

2. **Break down \( \log_{7}(28) \)**:
We can express \( 28 \) as \( 4 \times 7 \):
\[ \log_{7}(28) = \log_{7}(4 \times 7) = \log_{7}(4) + \log_{7}(7) \]
Since \( \log_{7}(7) = 1 \), we have:
\[ \log_{7}(28) = \log_{7}(4) + 1 \]

3. **Express \( \log_{7}(4) \)**:
We can express \( 4 \) as \( 2^2 \):
\[ \log_{7}(4) = \log_{7}(2^2) = 2 \log_{7}(2) \]
Given that \( \log_{7}(2) = \lambda \), we have:
\[ \log_{7}(4) = 2\lambda \]

4. **Substitute back into the expression for \( \log_{7}(28) \)**:
\[ \log_{7}(28) = 2\lambda + 1 \]

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

6. **Simplify the expression**:
\[ \log_{49}(28) = \lambda + \frac{1}{2} \]

### Final Answer:
Thus, the value of \( \log_{49}(28) \) is:
\[ \lambda + \frac{1}{2} \]