College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 4 - Exponential and Logarithmic Functions - Exercise Set 4.4: 87

Answer

Since $x \ne -5$, then $x = 2$

Work Step by Step

$\log x + \log (x+3) = \log 10$ $\log [x(x+3)] = \log 10$ $\log (x^{2} + 3x) = \log 10$ $x^{2} + 3x = 10$ $x^{2} + 3x - 10 = 0$ $(x+5)(x-2) = 0$ $x = -5, 2$ Since $x \ne -5$, then $x = 2$
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