College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 4 - Test - Page 472: 7

Answer

$\log_{10} 2388\approx3.3780$

Work Step by Step

By definition, $\log x=\log_{10} x$, therefore this given logarithm can be written also as: $\log 2388=\log_{10} 2388$ which can be easily calculated. $\log_{10} 2388\approx3.3780$
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