How can I find inverse for this functions?

Question

I have this function:

$f(x) = \frac{1}{x\log_e(\frac{b}{a})}$

with domain $a\leqslant x \leqslant b$, where $b>a>0$

I think its range is $(0,1)$. I am not able to figure out what is the rule for $f^{-1}(x)$ for general $a$ and $b$ and its domain and range.

Thanks @Chris Custer, I have arrived at following solution after looking around on the internet:

1. Substitute y for x:

   $y = \frac{1}{x\log_e(\frac{b}{a})}$

2. Let $f^{-1}(x) = y$

3. $\implies f^{-1}(x)=\frac{1}{x\log_e(\frac{b}{a})}$

4. using $g(f(x))$

   $g(\frac{1}{x\log_e(\frac{b}{a})}) = \frac{1}{(\frac{1}{x\log_e\frac{b}{a}})\log_e(\frac{b}{a})}$

5. $g(\frac{1}{x\log_e(\frac{b}{a})}) = \frac{1}{\frac{\log_e(\frac{b}{a})}{x\log_e(\frac{b}{a})}}$

6. $g(\frac{1}{x\log_e(\frac{b}{a})}) = x$

7. since $g(f(x)) =x$

   $f^{-1}(x) = \frac{1}{x\log_e(\frac{b}{a})}$

Question: I'm not sure aobut step 4 though :(

Answer 1

Interchange $x$ and $y$, and solve for $y$. Get $x=\dfrac1{y\log_e(\frac ba)}\implies y=\dfrac1{x\log_e(\frac ba)}$.

Thus $f$ is its own inverse.