## Intermediate Algebra for College Students (7th Edition)

a) $2$ b) $\log_3(\dfrac{xy^4}{9})$
a) Since, $\log_39=\log _3(3)^3=2$ b) From part (a), we have $\log_3 x+4\log_3 y-2=\log_3x+4\log_3 y-\log_39$ or, $\log_3 x+4\log_3 y-2=\log_3x+\log_3 y^4-\log_39$ or, $\log_3 x+4\log_3 y-2=\log_3(xy^4)-\log_39$ or, $\log_3 x+4\log_3 y-2=\log_3(\dfrac{xy^4}{9})$ Hence, a) $2$ b) $\log_3(\dfrac{xy^4}{9})$