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Basic of Logarithms
easy
${\log _4}18$ is
A
A rational number
B
An irrational number
C
A prime number
D
None of these
Solution
(b) ${\log _4}18 = {1 \over 2}{\log _2}({3^2}.2) = {1 \over 2}(2{\log _2}3 + {\log _2}2)$
$ = {\log _2}3 + {1 \over 2},$ which is irrational.
Standard 11
Mathematics