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 - 12.6 Solving Exponential Equations and Logarithmic Equations - 12.6 Exercise Set: 39

Answer

$x\approx26.799$

Work Step by Step

In exponential form, the given logarithmic equation, $ \ln (2x+1)=4 ,$ is equivalent to \begin{array}{l}\require{cancel} \log_e (2x+1)=4 \\\\ 2x+1=e^4 .\end{array} Hence, the value of the variable that satisfies the given equation is \begin{array}{l}\require{cancel} 2x=e^4-1 \\\\ x=\dfrac{e^4-1}{2} \\\\ x\approx26.799 .\end{array}
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