Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 4 - Inverse, Exponential, and Logarithmic Functions - 4.4 Evaluation Logarithms and the Change-of-Base Theorem - 4.4 Exercises - Page 456: 6

Answer

$\dfrac{\log 12}{\log 3}$

Work Step by Step

Use logarithmic property $\log_{n} m=\dfrac{\log m}{\log n}$ $\log_{3} (12) =\dfrac{\log 12}{\log 3}$ Therefore, our answer is: $\dfrac{\log 12}{\log 3}$
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