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Basic of Logarithms
medium
${\log _2}.{\log _3}....{\log _{100}}{100^{{{99}^{{{98}^{{.^{{.^{{{.2}^1}}}}}}}}}}}= . . . $.
A
$0$
B
$1$
C
$2$
D
$100!$
Solution
(b) ${\log _2}.{\log _3}…..{\log _{99}}$ ${\log _{100}}{100^{{{99}^{{{98}^{{.^{{.^{{.^{{2^1}}}}}}}}}}}}}$
$ = {\log _2}.{\log _3}….{\log _{98}}^{{{98}^{{{97}^{{.^{{.^{{.^{{2^1}}}}}}}}}}}}$
$ = {\log _2}\,\,2'{\log _3}3 = {\log _2}2 = 1$.
Standard 11
Mathematics