Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 4 - Inverse, Exponential, and Logarithmic Functions - 4.5 Exponential and Logarithmic Equations - 4.5 Exercises - Page 469: 77

Answer

$\dfrac {1\pm \sqrt {41}}{4}$

Work Step by Step

$\log _{2}\left( 2x-3\right) +\log _{2}\left( x+1\right) =1\Rightarrow \log _{2}\left( \left( 2x-3\right) \left( x+1\right) \right) =\log _{2}2$ $\Rightarrow \left( 2x-3\right) \left( x+1\right) =2\Rightarrow 2x^{2}-x-3=2\Rightarrow 2x^{2}-x-5=0$ $x_{1,2}=\dfrac {-b\pm \sqrt {b^{2}-4ac}}{2a}=\dfrac {1\pm \sqrt {1-4\times 2\times \left( -5\right) }}{2\times 2}=\dfrac {1\pm \sqrt {41}}{4}$
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