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Basic of Logarithms
easy
${\log _3}\,4{\log _4}\,5{\log _5}\,6{\log _6}\,7{\log _7}\,8{\log _8}\,9= . .$ . .
A
$1$
B
$2$
C
$3$
D
$4$
(IIT-2000)
Solution
(b) ${\log _3}4.{\log _4}5.{\log _5}6.{\log _6}7.{\log _7}8.{\log _8}9$
$ = {{\log 4} \over {\log 3}}.{{\log 5} \over {\log 4}}.{{\log 6} \over {\log 5}}.{{\log 7} \over {\log 6}}.{{\log 8} \over {\log 7}}.{{\log 9} \over {\log 8}} = {{\log 9} \over {\log 3}}$
$ = {\log _3}9 = {\log _3}{3^2} = 2$.
Standard 11
Mathematics
