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If $\log_{7} 2 = m$ , then $\log_{49} 28$ is equal to

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$2 (1 + 2 m)$

$\frac{1 + 2 m}{2}$

$\frac{2}{1 + 2 m}$

$1 + m$

answer is B.

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Detailed Solution

$\log_{49} 28 = \frac{\log 28}{\log 49} = \frac{\log 7 + \log 4}{2 \log 7}$

$= \frac{\log 7}{2 \log 7} + \frac{\log 4}{2 \log 7} = \frac{1}{2} + \frac{1}{2} \log_{7} 4$

$= \frac{1}{2} + \frac{1}{2} . 2 \log_{7} 2 = \frac{1}{2} + \log_{7} 2 = \frac{1}{2} + m = \frac{1 + 2 m}{2}$

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