Intermediate Algebra: Connecting Concepts through Application

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

Chapter 6 - Logarithmic Functions - 6.6 Solving Logarithmic Equations - 6.6 Exercises: 54

Answer

$c = -7$

Work Step by Step

$\log_5 (c + 12) + \log_5 (c+8) - 2 = -1$ $\log_5 (c+12)(c+8) = -1 + 2$ $\log_5 (c+12)(c+8) = 1$ $5^{1} = (c+12)(c+8)$ $5= (c+12)(c+8)$ $5= c(c+8)+12(c+8)$ $5 = c^{2} + 8c+12c+96$ $c^{2} + 20c + 96 - 5 = 0$ $c^{2} + 20c + 91 = 0$ $c = \frac{-(20)±\sqrt {(20)^{2}-4(1)(91)}}{2(1)}$ $c = \frac{-20±\sqrt {400-4(1)(91)}}{2}$ $c = \frac{-20±\sqrt {400-364}}{2}$ $c = \frac{-20±\sqrt {36}}{2}$ $c = \frac{-20±6}{2}$ $c = -7, -13$ Since we can't take the log of a negative number, the only possible solution is $c = -7$. Check: $\log_5 (-7+ 12) + \log_5 ((-7)+8) - 2 \overset{?}{=} -1$ $\log_5 (5) + \log_5 (1) - 2 \overset{?}{=} -1$ $1(1) + 0 - 2 \overset{?}{=} -1$ $1 - 2 \overset{?}{=} -1$ $-1 = -1$
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