College Algebra (11th Edition)

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

Chapter 4 - Section 4.4 - Evaluating Logarithms and the Change-of-Base Theorem - 4.4 Exercises - Page 435: 58

Answer

$\approx9.8386$

Work Step by Step

The value of $ \ln33 $ is approximately $ 3.4965 $. The value of $ \ln568 $ is approximately $ 6.3421 $. Hence, the given expression, $ \ln33+\ln568 $ evaluates to \begin{array}{l}\require{cancel} 3.4965+6.3421 \\\\ \approx9.8386 .\end{array}
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