Determinate $\lambda\in R$ so that the following equation has 2 real,distinct solutions.

Question

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?

Answer 1

The function to study is $$ f(x)=(2-\lambda)x+(1+\lambda)\ln x $$ If the function must have exactly two real zeros, its derivative must vanish exactly once. Since $$ f'(x)=2-\lambda+\frac{1+\lambda}{x}=\frac{(2-\lambda)x+(1+\lambda)}{x} $$ we need $$ x_0=\frac{\lambda+1}{\lambda-2}>0 $$ that is, $\lambda<-1$ or $\lambda>2$. Note that for $\lambda<-1$ the point $x_0$ is a point of minimum and for $\lambda>2$ it is a point of maximum, because $$ f''(x)=-\frac{1+\lambda}{x^2} $$ which is positive for $\lambda<-1$ and negative for $\lambda>2$

We have $$ f(x_0)=(\lambda+1)(-1+\ln x_0) $$ For $\lambda<-1$ we need $f(x_0)<0$, which translates to $-1+\ln x_0>0$, that is, $$ \frac{\lambda+1}{\lambda-2}>e $$ which is easily seen to be false for all $\lambda<-1$.

For $\lambda>2$ we need $f(x_0)>0$, which translates again to $-1+\ln x_0>0$, that is, $$ \frac{\lambda+1}{\lambda-2}>e $$ or $$ \lambda<\frac{2e+1}{e-1}=\frac{2e-2+3}{e-1}=2+\frac{3}{e-1} $$ Final result $$ 2<\lambda<\frac{2e+1}{e-1}\approx 3.74593 $$

Answer 2

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 $$f(x_E)=(\lambda+2)\left(\ln(x_E)-1\right)>0$$

Answer 3

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