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: 63

Answer

$9;-2$

Work Step by Step

$\ln \left( 7-x\right) +\ln \left( 1-x\right) =\ln \left( 25-x\right) \Rightarrow \ln \left[ \left( 7-x\right) \left( 1-x\right) \right] =\ln \left( 25-x\right) $ $\Rightarrow \ln \left( 7-8x+x^{2}\right) =\ln \left( 25-x\right) \Rightarrow x^{2}-8x+7=25-x\Rightarrow x^{2}-7x-18=0$ $x_{1,2}=\dfrac {-b\pm \sqrt {b^{2}-4ac}}{2a}=\dfrac {7\pm \sqrt {49-4\times 1\times \left( -18\right) }}{2}=\dfrac {7\pm 11}{2}=9;-2$
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