1
$\begingroup$

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.

  • 0
    @David $\lambda > 0$ Sorry for this. – 2017-02-01
  • 1
    Have you leared the intermediate value theorem? – 2017-02-01
  • 0
    @FrankLu Thanks. I see your point. From $z>>0$ to $z \rightarrow 0$, discuss the function value and we will see it intersect $0$. – 2017-02-01
  • 1
    Using http://en.wikipedia.org/wiki/Lambert_W_function the solution is $$\lambda W\left(\frac{e^{x/\lambda}}{\lambda}\right)$$ It has real positive values for all values $\lambda > 0$. – 2017-02-01

1 Answers 1

1

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 $\lambda$.