${\log _4}18 = . . . .$
Easy
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b
(b) \({\log _4}18 = {1 \over 2}{\log _2}({3^2}.2) = {1 \over 2}(2{\log _2}3 + {\log _2}2)\)
(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.
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