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: 91

Answer

$x = \frac{11}{3}$ $x = 3.67$

Work Step by Step

$\ln (x-2) - \ln(x+3) = \ln(x-1) - \ln(x+7)$ $\ln \frac{x-2}{x+3} = \ln \frac{x-1}{x+7}$ $\frac{x-2}{x+3} = \frac{x-1}{x+7}$ $(x-2)(x+7) = (x-1)(x+3)$ $x(x+7) - 2(x+7) = x(x+3)-1(x+3)$ $x^{2} + 7x - 2x - 14 = x^{2} + 3x - x - 3$ $x^{2} + 5x - 14 = x^{2} + 2x - 3$ $x^{2} - x^{2} + 5x - 2x - 14 + 3 = 0$ $3x - 11 = 0$ $3x = 11$ $x = \frac{11}{3}$ $x = 3.67$
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