How to show $\frac{1}{\lambda}(z-x)+\ln z = 0$ has a positive solution $z$ for any $x$?
It seems no problem $z>0$ because of the $\ln(\cdot)$ function. However, how to show there exists an solution for this equation?
I use the method such as taking exponential; however still cannot find a way to show it.
Hope for hint.
If you consider in the real domain the function $$f(z)=\frac{1}{\lambda}(z-x)+\ln z $$ its derivative is $$f'(z)=\frac{1}{\lambda }+\frac{1}{z}$$ which is always positive. So, the function increases and only one root does exist.
As gammatester commented, the solution is given in terms of Lambert function $$z=\lambda W\left(\frac{e^{x/\lambda }}{\lambda }\right)$$ which is always defined for any positive value of $\lambda$.