Intermediate Algebra (6th Edition)

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

Chapter 9 - Section 9.6 - Properties of Logarithms - Exercise Set: 79


Because $log_{b}1=0$

Work Step by Step

We are given that $log_{b}8=log_{b}(8\times1)=log_{b}8+log_{b}1$. Although this seems counterintuitive, we know that $log_{b}8=log_{b}8+log_{b}1$, because $log_{b}1=0$ according to the properties of logarithms. Therefore, $log_{b}8=log_{b}8+log_{b}1=log_{b}8+0=log_{b}8$.
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