Inverse function on interval where it is one-to-one

Question

I have to find inverse function of

${\displaystyle{f(x)=\frac{\log x}{\log 2}+\frac{\log 2}{\log x}.}}$

${\displaystyle{y=\frac{\log x}{\log 2}+\frac{\log 2}{\log x}}}$

${\displaystyle{y=\frac{\log^2 x+\log^2 2}{\log 2\log x}}}$

${\displaystyle{y\log 2\log x={\log^2 x+\log^2 2}}}$

${\displaystyle{{\log^2 x+\log^2 2}}}-y\log 2\log x=0$

${\displaystyle{(\log x}-{\frac{y\log 2}{2})}^2-\frac{y^2}{4}\log 2\log x+\log 2\log x=0}$

${\displaystyle{(\log x}-{\frac{y\log 2}{2})}^2=\frac{y^2-4}{4}\log 2\log x}$

Could you help me with next step, please? Thanks for help.

Answer 1

HINT

Start with finding the inverse of $g(x)=x+\frac 1 x$ (Notice that $f(x)=g(\frac {\log x} {log2})$ and you can easily find the inverse of $h(x)=\frac {\log x} {log2}$).

You have $f=g\circ h$ so $f^{-1}=h^{-1} \circ g^{-1}$

Answer 2

Hint focused on simplification:

Let $h_{a}\left(x\right)$ denote the inverse of $g_{a}\left(x\right)=\frac{x}{a}+\frac{a}{x}$ where $a>0$.

Then the inverse of $f\left(x\right)=g_{a}\left(\log x\right)$ is $e^{h_{a}\left(x\right)}$.