Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 9 - Section 9.5 - Common and Natural Logarithms - 9.5 Exercises - Page 620: 4


B. 1 and 2

Work Step by Step

We know that logarithms with base $e$ are natural logarithms, and $ln(x)$ is equivalent to $log_{e}x$. Therefore, $ln(6.3)=log_{e}6.3$. We know that for all positive numbers $a$ (where $a\ne1$), and all positive numbers $x$, $y=log_{a}x$ means the same as $x=a^{y}$. Therefore, $log_{e}6.3=x$ is equivalent to $e^{x}=6.3$. So, we know that $log_{e}6.3$ must be between 1 and 2, because $e^{1}\approx2.718$ and $e^{2}\approx7.389$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.