Can $e^{\frac{1}{\ln(x)}}$ be simplified?

Question

I would like to find the inverse function of $$f(x) = \ln(x) + \frac{1}{\ln(x)}$$ but I got stuck when trying to remove the exponents: $$e^x = y \cdot e^{\frac{1}{\ln(x)}}$$

Answer 1

$$y=\ln(x)+\frac1{\ln(x)}$$

$$\ln^2(x)-y\ln(x)+1=0$$ $$\ln(x)=\frac{y\pm\sqrt{y^2-4}}{2}$$

$$x=\sqrt{e^{y\pm\sqrt{y^2-4}}}.$$

Answer 2

1. Let $u=\ln x$, to simplify the algebra. Then we have $$y=u+\frac{1}{u}$$
2. Multiply this by $u$ to get a quadratic equation in $u$.
3. Solve for $u$ in terms of $y$.
4. Now put $x=e^u$.