Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 3 - Section 3.3 - Properties of Logarithms - Exercise Set - Page 477: 130



Work Step by Step

The change-of-base property states that$$\log_bM= \frac{\log_aM}{\log_ab}$$So by introducing the new base $10$, we have$$\log_79=\frac{\log 9}{\log 7}=\frac{\log 3^2}{\log 7}= \frac{2 \log 3}{\log 7}= \frac{2A}{B}$$
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