Basic of Logarithms
medium

The value of ${\log _2}.{\log _3}....{\log _{100}}{100^{{{99}^{{{98}^{{.^{{.^{{{.2}^1}}}}}}}}}}}$ is

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

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