Algebra and Trigonometry 10th Edition

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

Chapter 5 - 5.1 - Exponential Functions and Their Graphs - 5.1 Exercises - Page 369: 47


$x_1=3$ and $x_2=-1$

Work Step by Step

$e^{x^2-3}=e^{2x}$ Applying the One-to-One Property: $x^2-3=2x$ $x^2-2x-3=0$ ($a=1,~b=-2,~c=-3$) $x=\frac{-(-2)±\sqrt {(-2)^2-4(1)(-3)}}{2(1)}$ $x=\frac{2±\sqrt {16}}{2}=\frac{2±4}{2}$ $x_1=\frac{6}{2}=3$ and $x_2=\frac{-2}{2}=-1$
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