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: 25

Answer

4.1506

Work Step by Step

We know that logarithms with base $e$ are natural logarithms, and $ln(x)$ can be written as $log_{e}x$. Therefore, $ln(8.59\times e^{2})=log_{e}(8.59\times e^{2})$. 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, $log_{e}(8.59\times e^{2})\approx4.1506$, because $e^{4.1506}\approx e^{(8.59\times e^{2})}$.
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