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

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Question

(B) We are given ${\log _7}2 = m$.We need to evaluate ${\log _{49}}28$.Using the change of base formula,${\log _{49}}28 = \frac{{\log _7}28}{{\log _7}

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) We are given ${\log _7}2 = m$.We need to evaluate ${\log _{49}}28$.Using the change of base formula,${\log _{49}}28 = \frac{{\log _7}28}{{\log _7}49}$.Since $28 = 7 	imes 4$,we have ${\log _7}28 = {\log _7}(7 	imes 4) = {\log _7}7 + {\log _7}4 = 1 + {\log _7}(2^2) = 1 + 2{\log _7}2$.Since $49 = 7^2$,we have ${\log _7}49 = 2$.Substituting these values,${\log _{49}}28 = \frac{1 + 2{\log _7}2}{2}$.Substituting $m = {\log _7}2$,we get ${\log _{49}}28 = \frac{1 + 2m}{2}$.

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

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