Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 12 - Exponential Functions and Logarithmic Functions - Test: Chapter 12 - Page 845: 23

Answer

$\dfrac{\log 14}{\log 3}\approx2.4022$

Work Step by Step

Using $\log_b x=\dfrac{\log_a x}{\log_a b}$ or the Change-of-Base Formula, the given expression, $ \log_3 14 ,$ is equivalent to \begin{align*} & \dfrac{\log_{10} 14}{\log_{10} 3} \\\\&= \dfrac{\log 14}{\log 3} .\end{align*} With the bases now expressed in base-$10$, the calculator can then be used. Hence, $\dfrac{\log 14}{\log 3}\approx2.4022$.
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