Algebra and Trigonometry 10th Edition

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

Chapter 5 - Review Exercises - Page 413: 90


$x=1.386$ or $x=0.693$

Work Step by Step

Notice that $e^{2x}=(e^x)^2$. First we solve the equation for $e^x$: $e^{2x}-6e^x+8=0$ $(e^x)^2-6(e^x)+8=0~~$ ($a=1,~b=-6,~c=8$) $e^x=\frac{-(-6)±\sqrt {(-6)^2-4(1)(8)}}{2(1)}=\frac{6±\sqrt 4}{2}=\frac{6±2}{2}=3±1$ First solution: $e^x=4$ $\ln e^x=\ln4~~$ (Using the Inverse Property $\ln e^x=x$): $x=\ln4=1.386$ Second solution: $e^x=2$ $\ln e^x=\ln2~~$ (Using the Inverse Property $\ln e^x=x$): $x=\ln2=0.693$
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