Determinate $\lambda\in R$ so that the following equation has 2 real,distinct solutions. Determinate $\lambda\in R$ so that the following equation has 2 real,distinct solutions. $$2x+\ln x-\lambda(x-\ln x)=0$$
I think this should be solved using Rolle property for finding intervals with solutions.So i calculated $f^|(x)=\frac{2x-\lambda x+1+\lambda}{x}$.So $x=\frac{\lambda+1}{\lambda -2}$ so $f(x)=-\lambda-1+(1-\lambda )\ln \frac{2}{\lambda-2}-2$ Here I got stuck.Any help?

Hint: defining $$f(x)=2x+2\ln(x)-\lambda(x-\ln(x))$$ then $$f'(x)=2+\frac{2}{x}-\lambda\left(1-\frac{1}{x}\right)$$ then $$f'(x)=0$$ if $$x_E=\frac{\lambda+2}{\lambda-2}$$ and $$f''(x_E)=-\frac{(-2+\lambda)^2}{\lambda+2}$$ then must hold
$$f''(x_E)<0$$ and $$f(x_E)=(\lambda+2)\left(\ln(x_E)-1\right)>0$$
