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?

We have
$$
(2-\lambda)x+(\lambda+1)\ln x = 0
$$
now making $z = \ln x$
$$
\frac{2-\lambda}{\lambda+1} = (-z)e^{-z}
$$
Now using the Lambert function $W$ we have
$$
-z = W\left(\frac{2-\lambda}{\lambda+1}\right)\Rightarrow x = e^{- W\left(\frac{2-\lambda}{\lambda+1}\right)}
$$
Now you can choose a suitable $\lambda$ which gives you two solutions.
https://en.wikipedia.org/wiki/Lambert_W_function
