Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 1-8 - Cumulative Review: 10

Answer

$x = -7$

Work Step by Step

$\log_5 (x+8) + \log_5 (x+12) = 1$ $\log_5 (x+8)(x+12) = 1$ $5^{1} = (x+8)(x+12)$ $5 = (x+8)(x+12)$ $x(x+12)+8(x+12) = 5$ $x^{2} + 12x + 8x + 96 = 5$ $x^{2} + 20x + 96 - 5 = 0$ $x^{2} + 20x + 91 = 0$ $x^{2} + 7x + 13x + 91 = 0$ $x(x+7) + 13(x+7) = 0$ $(x+13)(x+7) = 0$ $x = -13, -7$ Check: $\log_5 (-13+8) + \log_5 (-13+12) \overset{?}{=} 1$ $\log_5 (-5) + \log_5 (-1) \overset{?}{=} 1$ Since $\log_5$ is undefined for negative numbers, $x=-13$ is not a solution. $\log_5 (-7+8) + \log_5 (-7+12) \overset{?}{=} 1$ $\log_5 (1) + \log_5 (5) \overset{?}{=} 1$ $\log_5 (1)(5) \overset{?}{=} 1$ $\log_5 (5) \overset{?}{=} 1$ $1 = 1$
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