MCQ
If ${\log _7}2 = m,$ then ${\log _{49}}28$ is equal to
  • A
    $2\,(1 + 2m)$
  • ${{1 + 2m} \over 2}$
  • C
    ${2 \over {1 + 2m}}$
  • D
    $1 + m$

Answer

Correct option: B.
${{1 + 2m} \over 2}$
b
(b) ${\log _{49}}28 = {{\log 28} \over {\log 49}} = {{\log 7 + \log 4} \over {2\log 7}}$

$ = {{\log 7} \over {2\log 7}} + {{\log 4} \over {2\log 7}} = {1 \over 2} + {1 \over 2}{\log _7}4$

$ = {1 \over 2} + {1 \over 2}.2{\log _7}2 = {1 \over 2} + {\log _7}2 = {1 \over 2} + m = {{1 + 2m} \over 2}$

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