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

$x=1086.5129$
Apply $\log_{a}(MN)=\log_{a}M+\log_{a}N$ $\log 492+\log x=5.728\qquad$ ... subtract $\log 492$ $\log x=5.728-\log 492\qquad$ ... apply inverse function $10^{(...)}$ $x=10^{5.728-\log 492}\approx 1086.5129256$ To four decimal places, $x=1086.5129$