(B) We have $\log _4 18 = \log _{2^2} (2 \times 3^2) = \frac{1}{2} \log _2 (2 \times 3^2) = \frac{1}{2} (\log _2 2 + \log _2 3^2) = \frac{1}{2} (1 + 2

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(B) We have $\log _4 18 = \log _{2^2} (2 \times 3^2) = \frac{1}{2} \log _2 (2 \times 3^2) = \frac{1}{2} (\log _2 2 + \log _2 3^2) = \frac{1}{2} (1 + 2 \log _2 3) = \frac{1}{2} + \log _2 3$. Since $\log _2 3$ is an irrational number,$\frac{1}{2} + \log _2 3$ is also an irrational number. Therefore,$\log _4 18$ is an irrational number.