Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 5 - Cumulative Test for Chapters 3-5 - Page 417: 33


$x=1.946$ $x=1.792$

Work Step by Step

$e^{2x}-13e^x+42=0$ $(e^x)^2-13(e^x)+42=0~~$ ($a=1,b=-13,c=42$): $e^x=\frac{-b±\sqrt {b^2-4ac}}{2a}=\frac{-(-13)±\sqrt {(-13)^2-4(1)(42)}}{2(1)}=\frac{13±\sqrt {1}}{2}=\frac{13±1}{2}$ $e^x=7$ $\ln e^x=\ln7$ $x=\ln7=1.946$ $e^x=6$ $\ln e^x=\ln6$ $x=\ln6=1.792$
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