Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Cumulative Review Exercises - Page 647: 3


$ x=4$

Work Step by Step

$\log_{2} x+\log_{2} (x-2)=0 \implies 3=\log_{2} 2^3$ Re-arrange as: $ x(x-2)=2^3$ $\implies x^2-2x-8=0$ $\implies (x^2+2x)-(4x+8)=0$ or, $(x+2)(x-4)=0$ So, $ x=-2,4$ Neglect the negative term of $ x $, because it does not belong to the domain of the variable; so we have $ x=4$
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