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Basic of Logarithms
medium
જો ${\log _7}2 = m$ તો ${\log _{49}}28 = . . . .$
A
$2\,(1 + 2m)$
B
${{1 + 2m} \over 2}$
C
${2 \over {1 + 2m}}$
D
$1 + m$
Solution
(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}$
Standard 11
Mathematics
