Answer
$log_3(\frac{1}{32p^5})$
Work Step by Step
$-\frac{3}{4}log_3(16p^4)-\frac{2}{3}log_3(8p^3)=-[\frac{3}{4}log_3(2^4p^4)+\frac{2}{3}log_3(2^3p^3)]=-[log_3(2^3p^3)+log_3(2^2p^2)]=-[log_3(2^5p^5)]=log_3(\frac{1}{32p^5})$
Precalculus (6th Edition)
by Lial, Margaret L.; Hornsby, John; Schneider, David I.; Daniels, Callie
Published by
Pearson
ISBN 10:
013421742X
ISBN 13:
978-0-13421-742-0
Chapter 4 - Inverse, Exponential, and Logarithmic Functions - 4.3 Logarithmic Functions - 4.3 Exercises - Page 446: 92
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