College Algebra (11th Edition)

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

Chapter 4 - Test: 8

Answer

$\log_{e} 2388\approx7.7782$

Work Step by Step

By definition, $\ln x=\log_{e} x$, therefore the given logarithm can be also written as: $\ln 2388 =\log_{e} 2388$ which can be easily calculated: $\log_{e} 2388\approx7.7782$
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