Chapter 5 - 5.4 - Exponential and Logarithmic Equations - 5.4 Exercises - Page 396: 80

$x=0.607$

$2x\ln(\frac{1}{x})-x=0$ $2x\ln x^{-1}-x=0$ $-2x\ln x-x=0$ $-x(2\ln x+1)=0$ $-x=0$ $x=0$. Not a valid solution, because $\frac{1}{x}=\frac{1}{0}$ $2\ln x+1=0$ $2\ln x=-1$ $\ln x=-\frac{1}{2}$ $e^{\ln~x}=e^{-\frac{1}{2}}$ $x=e^{-\frac{1}{2}}=0.607$