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

$\dfrac{3}{2}$
Using $\log_bx^m=m\log_bx$ or the Power Rule of Logarithms, the given expression, $\log_{9}27 ,$ is equivalent to \begin{array}{l}\require{cancel} \log_{9}\left(\sqrt{9}\right)^3 \\\\= \log_{9}9^{3/2} \\\\= \dfrac{3}{2}\log_{9}9 .\end{array} Since $\log_bb=1,$ the expression, $\log_{9}9 ,$ simplifies to $1$. \begin{array}{l}\require{cancel} \dfrac{3}{2}(1) \\\\= \dfrac{3}{2} .\end{array}