Algebra and Trigonometry 10th Edition

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

Chapter 5 - 5.4 - Exponential and Logarithmic Equations - 5.4 Exercises - Page 395: 41



Work Step by Step

$e^{2x}-4e^x-5=0$ $(e^x)^2-4e^x-5=0~~$ ($a=1,b=-4,c=-5$): $e^x=\frac{-b±\sqrt {b^2-4ac}}{2a}=\frac{-(-4)±\sqrt {(-4)^2-4(1)(-5)}}{2(1)}=\frac{4±\sqrt {36}}{2}=\frac{4±6}{2}=2±3$ $e^x=5$ $\ln e^x=\ln5$ $x=1.609$ $e^x=-1$. But, there is no $x$ such that $e^x=-1$.
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