Precalculus (6th Edition) Blitzer

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

Chapter 3 - Section 3.4 - Exponential and Logarithmic Equations - Exercise Set - Page 491: 146

Answer

$$ \{ 1, e^2 \}$$

Work Step by Step

$$(\ln x)^2= \ln x^2 \quad \Rightarrow \quad (\ln x )(\ln x)= 2 \ln x \quad \Rightarrow \quad \ln x =0 \quad \text{ or } \quad \ln x=2 \quad \Rightarrow \quad x=1 \quad \text{ or } \quad x=e^2$$Both values satisfy the original equation:$$(\ln 1)^2 =0 = \ln 1^2, \\ (\ln e^2)^2=4=\ln (e^2)^2$$
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