Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 6 - Logarithmic Functions - 6.5 Solving Exponential Equations - 6.5 Exercises - Page 528: 70

Answer

$b = -3.5, 1$

Work Step by Step

$-2b^{2} -5b + 10 = 3$ $-2b^{2} -5b + 10 - 3 = 0$ $-2b^{2} -5b + 7 = 0$ $b = \frac{-(-5)±\sqrt {(-5)^{2}-4(-2)(7)}}{2(-2)}$ $b = \frac{5±\sqrt {25-4(-2)(7)}}{-4}$ $b = \frac{5±\sqrt {25+56}}{-4}$ $b = \frac{5±\sqrt {81}}{-4}$ $b = \frac{5±9}{-4}$ $b = -3.500, 1$
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