Intermediate Algebra (6th Edition)

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

Chapter 11 - Cumulative Review - Page 670: 30a

Answer

$5$

Work Step by Step

Using the property $\log_b b^x=x$ then the expression $ \log 100,000 $ is equal to \begin{array}{l} \log_{10}10^5 \\= 5(\log_{10}10) \\= 5(1) \\= 5 .\end{array}
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