## Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole

# Chapter 6 - Logarithmic Functions - Chapter Review Exercises - Page 543: 51

#### Answer

$x \approx 7.266$

#### Work Step by Step

$\log_3 (x-4) + \log_3 (x+1) = 3$ $\log_3 (x-4)(x+1) = 3$ $(x-4)(x+1) = 3^{3}$ $x(x+1)-4(x+1) = 27$ $x^{2} + x - 4x - 4 = 27$ $x^{2} - 3x - 31 = 0$ $x = \frac{-b±\sqrt {b^{2}-4ac}}{2a}$ $x = \frac{-(-3)±\sqrt {(-3)^{2}-4(1)(-31)}}{2(1)}$ $x = \frac{3±\sqrt {9-4(1)(-31)}}{2}$ $x = \frac{3±\sqrt {9+124}}{2}$ $x = \frac{3±\sqrt {133}}{2}$ $x = -4.266..., 7.266...$ ($x$ must be greater than 4.) $x \approx 7.266$ Check: $\log_3 (x-4) + \log_3 (x+1) \overset{?}{=} 3$ $\log_3 ((7.266...)-4) + \log_3 ((7.266...)+1) \overset{?}{=} 3$ $\log_3 (3.266...) + \log_3 (8.266...) \overset{?}{=} 3$ $\log_3 (3.266...)(8.266...) \overset{?}{=} 3$ $\log_3 (27)\overset{?}{=} 3$ $\log_3 (3^{3})\overset{?}{=} 3$ $3\log_3 (3)\overset{?}{=} 3$ $3(1)\overset{?}{=} 3$ $3 = 3$

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