This is a question that I came across a few days ago，Although it is not particularly like a research problem, the following problem is that I study the zero distribution of a class of elementary transcendental functions.And I don't think the following problems are easy to deal with. I 've been thinking about them for a few days and they 've all failed
if $f(x)=x^2-x-\ln{x}-\ln{a}$,and $f(x_{1})=f(x_{2})=0,0<x_{1}<x_{2}$.I have conjecture the roots $x_{1},x_{2}$ such $$\dfrac{3}{2a+1}<x_{1}x_{2}<\dfrac{\ln{a}}{a-1}$$$$\dfrac{3}{2a+1}<x_{1}x_{2}<\dfrac{\ln{a}}{a-1}，a>1$$
This is my attempt
since $$x^2_{1}-x_{1}-\ln{x_{1}}=x^2_{2}-x_{2}-\ln{x_{2}}=\ln{a}$$ let $x_{2}=tx_{1},t>1$,then we have $$x^2_{1}-x_{1}-\ln{x_{1}}=t^2x^2_{1}-tx_{1}-\ln{t}-\ln{x_{1}}$$ $$x_{1}=\dfrac{t-1+\sqrt{(t-1)^2+4\ln{t}\cdot(t^2-1)}}{2(t^2-1)}=f(t)$$ then $$x_{2}x_{1}=t(x_{1})^2=t\left(\dfrac{t-1+\sqrt{(t-1)^2+4\ln{t}\cdot(t^2-1)}}{2(t^2-1)}\right)^2=t(f(t))^2$$ and $$a=e^{x^2_{1}-x_{1}-\ln{x_{1}}}=e^{f^2(t)-f(t)-\ln{f(t)}}$$ so it must prove $$\dfrac{3}{2e^{f^2(t)-f(t)-\ln{f(t)}}+1}\le t(f(t))^2<\dfrac{f^2(t)-f(t)-\ln{f(t)}}{e^{f^2(t)-f(t)-\ln{f(t)}}-1},t>1$$Next thing I know, it's pretty complicated.