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: 76

Answer

$x = 8, -1$ $x \ne -1$, therefore $x = 8$

Work Step by Step

$\log_2 (x-3) + \log_2 x - \log_2 (x+2) = 2$ $\log_2 (x-3)(x) - \log_2(x+2) = 2$ $\log_2 \frac{x(x-3)}{x+2} = 2$ $2^{2} = \frac{x(x-3)}{x+2}$ $\frac{x(x-3)}{x+2} = 4$ $x(x-3) = 4(x+2)$ $x^{2} - 3x = 4x + 8$ $x^{2} - 3x - 4x - 8 = 0$ $x^{2} - 7x - 8 = 0$ $(x-8)(x+1) = 0$ $x = 8, -1$ $x \ne -1$, therefore $x = 8$
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