College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 4 - Review Exercises: 43

Answer

$x\approx2.102$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ 10e^{3x-7}=5 ,$ use the properties of equality to isolate the $e$ expression. Then take the natural logarithm of both sides. Use the laws of logarithms and the properties of equality to isolate the variable. Finally, Express the answer with $3$ decimal places. $\bf{\text{Solution Details:}}$ Using the properties of equality to isolate the $e$ expression results to \begin{array}{l}\require{cancel} \dfrac{10e^{3x-7}}{10}=\dfrac{5}{10} \\\\ e^{3x-7}=\dfrac{1}{2} .\end{array} Taking the natural logarithm of both sides, the equation above is equivalent to \begin{array}{l}\require{cancel} \ln e^{3x-7}=\ln \dfrac{1}{2} .\end{array} Using the Power Rule of Logarithms, which is given by $\log_b x^y=y\log_bx,$ the equation above is equivalent \begin{array}{l}\require{cancel} (3x-7)\ln e=\ln \dfrac{1}{2} .\end{array} Since $\ln e =1,$ the equation above is equivalent to \begin{array}{l}\require{cancel} (3x-7)(1)=\ln \dfrac{1}{2} \\\\ 3x-7=\ln \dfrac{1}{2} .\end{array} Using the Quotient Rule of Logarithms, which is given by $\log_b \dfrac{x}{y}=\log_bx-\log_by,$ the expression above is equivalent \begin{array}{l}\require{cancel} 3x-7=\ln 1-\ln2 .\end{array} Since $\ln1=0,$ the equation above is equivalent to \begin{array}{l}\require{cancel} 3x-7=0-\ln2 \\\\ 3x-7=-\ln2 .\end{array} Using the properties of equality to isolate the variable results to \begin{array}{l}\require{cancel} 3x=7-\ln2 \\\\ x=\dfrac{7-\ln2}{3} \\\\ x\approx2.102 .\end{array}
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