Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 5 - 5.4 - Exponential and Logarithmic Equations - 5.4 Exercises - Page 396: 79



Work Step by Step

$2x\ln x+x=0$ $x(2\ln x+1)=0$ $x=0$ $2\ln x+1=0$ $2\ln x=-1$ $\ln x=-\frac{1}{2}$ $e^{\ln~x}=e^{-\frac{1}{2}}$ $x=e^{-\frac{1}{2}}=0.607$ The domain of $\ln x$ is $(0,∞)$. That is, it does not include the $0$. So, $x=0$ is not a valid solution.
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