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 - Page 825: 26



Work Step by Step

Using the properties of equality, the given equation, $ 29=3e^{2x} ,$ is equivalent to \begin{array}{l}\require{cancel} \dfrac{29}{3}=e^{2x} .\end{array} Taking the natural logarithm of both sides and using the properties of logarithms, the value of the variable that satisfies the equation, $ \dfrac{29}{3}=e^{2x} ,$ is \begin{array}{l}\require{cancel} \ln\dfrac{29}{3}=\ln e^{2x} \\\\ \ln\dfrac{29}{3}=2x(\ln e) \\\\ \ln\dfrac{29}{3}=2x(1) \\\\ \ln\dfrac{29}{3}=2x \\\\ \dfrac{\ln\dfrac{29}{3}}{2}=x \\\\ x\approx1.134 .\end{array}
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