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 - Page 490: 73

Answer

$x = 5, -13$ $x \ne -13$, therefore $x = 5$

Work Step by Step

$2\log_3(x+4) = \log_3 9 + 2$ $\log_3(x+4)^{2} = \log_3 3^{2} + 2$ $\log_3(x+4)^{2} = 2 + 2$ $\log_3(x+4)^{2} = 4$ $3^{4} =(x+4)^{2}$ $81 = x(x+4)+4(x+4)$ $81 = x^{2} + 8x + 16$ $x^{2} + 8x + 16 - 81 = 0$ $x^{2} + 8x - 65 = 0$ $x^{2} + 13x - 5x - 65 = 0$ $x(x + 13) - 5(x+13) = 0$ $(x -5)(x+13) = 0$ $x = 5, -13$ $x \ne -13$, therefore $x = 5$
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