I am confused as to how the inverse of $f(x)=\frac{3+\ln x}{3-\ln x}$ is found. The problem within my textbook is $f(x)=$$\frac{3+\ln{x}}{3-\ln{x}}$ I checked the answer and it is $y=$$e^\frac{3x-3}{x+1}$
I've tried many times to simplify the equation after switching the variables but I don't know how to separate the $y$ from the natural logarithm. If someone could show me the steps involved I would appreciate it.
 A: Steps:
Start with:$$y=\frac{3+\ln x}{3-\ln x}$$
Rewrite the equality as: $$\ln x=g(y)$$ and conclude that: $$x=e^{\ln x}=e^{g(y)}$$
Now switch $x$ and $y$ to get the final result.
A: Well, as always you want to solve for $y$ in
$x = \frac {3+\ln y}{3-\ln y}$
$x(3-\ln y) = (3 + \ln y)$
$3x - 3 = \ln y + x\ln y$
$(x+1) \ln y = 3x-3$
$\ln y = \frac {3x-3}{x+1}$
$e^{\ln y} = e^{\frac {3x-3}{x+1}}$
So $y = e^{\frac {3x-3}{x+1}}$

I don't know how to seperate the y from the natural logarithm

Then inverse of $\ln y = K$ is $e^{\ln y} = y = e^K$.  $\ln x$ and $e^x$ are inverses of each other.  You "cancel" $\ln y=k$ by raising $e$ to the value to get $y = e^k$, and you "cancel" $e^y = M$ by taking the natural log to get $y = \ln M$.
A: $$y=\frac{3+\ln x}{3-\ln x}$$
Swap $(x,y)$ to find inverse. This is the procedure!
$$x=\frac{3+\ln y}{3-\ln y}$$
Componendo/Dividendo
$$\frac{x-1}{x+1}= \dfrac{2 \ln y}{6}$$
$$ y \rightarrow  y_{\,(inv-fn)} = e^{\frac{3(x-1)}{(x+1)} } $$
Also the graphs reflect each other with respect to mirror line $ y=x.$
The basic rule or definition of log /exp is the same
$$ A = ln_{base}C \rightarrow base^{A}= C. $$
