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

$2\log_a x+3\log_a y-\dfrac{3}{4}\log_a a-\dfrac{5}{4}\log_az$
Using the properties of logarithms, the given expression, $\log_a \sqrt[4]{\dfrac{x^8y^{12}}{a^3z^5}} ,$ is equivalent to \begin{array}{l}\require{cancel} \log_a \left(\dfrac{x^8y^{12}}{a^3z^5} \right)^{1/4} \\\\= \dfrac{1}{4}\log_a \left(\dfrac{x^8y^{12}}{a^3z^5} \right) \\\\= \dfrac{1}{4}\left[ \log_a \left(x^8y^{12}\right)-\log_a \left(a^3z^5 \right) \right] \\\\= \dfrac{1}{4}\left[ \log_a x^8+ \log_a y^{12}-\left(\log_a a^3+\log_az^5 \right) \right] \\\\= \dfrac{1}{4}\left[ \log_a x^8+ \log_a y^{12}-\log_a a^3-\log_az^5 \right] \\\\= \dfrac{1}{4}\left[ 8\log_a x+ 12\log_a y-3\log_a a-5\log_az \right] \\\\= \dfrac{1}{4}(8\log_a x)+\dfrac{1}{4}( 12\log_a y)-\dfrac{1}{4}(3\log_a a)-\dfrac{1}{4}(5\log_az) \\\\= \dfrac{8}{4}\log_a x+\dfrac{12}{4}\log_a y-\dfrac{3}{4}\log_a a-\dfrac{5}{4}\log_az \\\\= 2\log_a x+3\log_a y-\dfrac{3}{4}-\dfrac{5}{4}\log_az .\end{array}