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I am confused as to how the inverse of $f(x)=frac{3+ln x}{3-ln x}$ is found.

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2 $begingroup$ 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. algebra-precalculus share | cite | improve this question edited Mar 17 at 3:58 YuiTo Cheng 2,136 2 8 37 asked Mar 16 at 16:06 S