## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$\dfrac{1}{2}$
Using $\log_bx^m=m\log_bx$ or the Power Rule of Logarithms, the given expression, $\log_{16}4 ,$ is equivalent to \begin{array}{l}\require{cancel} \log_{16}\sqrt{16} \\\\= \log_{16}(16)^{1/2} \\\\= \dfrac{1}{2}\log_{16}16 .\end{array} Since $\log_bb=1,$ the expression, $\log_{16}16 ,$ simplifies to $1$. \begin{array}{l}\require{cancel} \dfrac{1}{2}(1) \\\\= \dfrac{1}{2} .\end{array}