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$1 + m$
$ = {{\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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