Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 2 - Nonlinear Functions - 2.5 Logarithmic Functions - 2.5 Exercises - Page 98: 51

Answer

$x=3$

Work Step by Step

$\log_{3}(x-2) + \log_{3}(x+6)=2$ $\log_{3}[(x-2)(x+6)]=2$ $x^{2} + 4x -12=3^{2}$ $x^{2}+4x -15=0$ $x=3$ and $x=-7$ Since $x=-7$ is not a valid value for x in the original equation, the only solution is $x=3$.
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