College Algebra (11th Edition)

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

Chapter 4 - Review Exercises - Page 469: 34

Answer

$\approx6.0486$

Work Step by Step

By using the change of base formula we can calculate easily the value of $\log_{3}769$. $\log_{3}769=\frac{\log_{10}769}{\log_{10}3}\approx6.0486$
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