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(B) We have ${\log _4}18 = {\log _{{2^2}}}({3^2} 	imes 2)$.Using the property ${\log _{{a^n}}}{b^m} = {m \over n}{\log _a}b$,we get:${\log _4}18 = {\

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An irrational number

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(B) We have ${\log _4}18 = {\log _{{2^2}}}({3^2} 	imes 2)$.Using the property ${\log _{{a^n}}}{b^m} = {m \over n}{\log _a}b$,we get:${\log _4}18 = {\log _{{2^2}}}({2 	imes 3^2}) = {1 \over 2}{\log _2}(2 	imes 3^2)$.Applying the product rule ${\log _a}(xy) = {\log _a}x + {\log _a}y$:${\log _4}18 = {1 \over 2}({\log _2}2 + {\log _2}{3^2}) = {1 \over 2}(1 + 2{\log _2}3) = {1 \over 2} + {\log _2}3$.Since ${\log _2}3$ is an irrational number,${1 \over 2} + {\log _2}3$ is also an irrational number.Therefore,the correct option is $B$.

${\log _4}18$ is

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