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

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2(1+2m)

1+2m2

21+2m

1+m

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detailed solution

Correct option is B

log4928=log 28log 49=log 7+log 42 log 7=log 72 log 7+log 42 log 7=12+12 log74=12+12.2 log72=12+log72=12+m=1+2m2

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If log72=m, then log4928 is equal to

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