## Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole

# Chapter 6 - Logarithmic Functions - Chapter Review Exercises - Page 543: 52

#### Answer

$x = -3$

#### Work Step by Step

$\log_4 (x+4) + \log_4 (x+7) = 1$ $\log_4 (x+4)(x+7) = 1$ $(x+4)(x+7) = 4^{1}$ $(x+4)(x+7) = 4$ $x(x+7)+4(x+7) = 4$ $x^{2} + 7x + 4x+28 = 4$ $x^{2} + 11x + 28 - 4 = 0$ $x^{2} + 11x + 24 = 0$ $x^{2} + 8x + 3x + 24 = 0$ $x(x+8) + 3(x+8) = 0$ $(x+3)(x+8) = 0$ $x = -3, -8$ $x = -3$ Check: When $x= -3$ $\log_4 (x+4) + \log_4 (x+7) \overset{?}{=} 1$ $\log_4 ((-3)+4) + \log_4 ((-3)+7) \overset{?}{=} 1$ $\log_4 (1) + \log_4 (4) \overset{?}{=} 1$ $0 + 1 \overset{?}{=} 1$ $1 = 1$ When $x = -8$ $\log_4 (x+4) + \log_4 (x+7) \overset{?}{=} 1$ $\log_4 (-8+4) + \log_4 (-8+7) \overset{?}{=} 1$ $\log_4 (-4) + \log_4 (-1) \overset{?}{=} 1$ Does not exist, since you cannot have a negative logarithm.

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