Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 11 - Cumulative Review - Page 670: 29b

Answer

$\log_{3} 6$

Work Step by Step

Using properties of logarithms, the expression $ \log_{3} \dfrac{1}{2}+\log_{3} 12 $ is equivalent to \begin{array}{l} \log_{3} \left( \dfrac{1}{2}\cdot12 \right) \\= \log_{3} 6 .\end{array}
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