If {log _7}2 = m, then {log _{49}}28 is equal | vedclass

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Question

(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}

Vedclass · Accepted Answer

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

Vedclass · Answer

(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}$

If ${\log _7}2 = m,$ then ${\log _{49}}28$ is equal to

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