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

$x=0.607$

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