Precalculus (6th Edition) Blitzer

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

Chapter 4 - Section 4.5 - Graphs of Sine and Cosine Functions - Exercise Set - Page 599: 122


It does not have any solution.

Work Step by Step

We solve as follows: $$\log_2(2x+1) -\log_2(x-2)=1 \quad \Rightarrow \quad \log_2 \left ( \frac{2x+1}{x-2} \right ) = \log_22 \quad \Rightarrow \quad \frac{2x+1}{x-2}=2 \quad \Rightarrow \quad 2x+1=2(x-2), \quad (x \neq 2) \quad \Rightarrow \quad 1=-4$$Since we reached a wrong result, we conclude that there exists no $x$ satisfying the original equation.
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