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

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