# Inverse of a rational equation with exponential components

• Aug 19th 2010, 05:24 PM
VectorRun
Inverse of a rational equation with exponential components
I'm having trouble coming up with the inverse function of this:

$\displaystyle \frac{1 + e^{-x}} {1 - e^{-x}}=y=f(x)$

I might be overcomplicating how I'm going about doing this; essentially, I ended up using exponent and logarithm laws to arrive at this:

$\displaystyle \ln{(1 + e^{-y})} - \ln{(1 - e^{-y})}=\ln{x}$

Any tips? I'm thinking that I probably did overcomplicate something...
• Aug 19th 2010, 05:55 PM
VectorRun
Wow, I did overcomplicate things, well, it was more of an oversight actually...I was too hesitant to multiply both sides by the denominator since it looked like that might get messy...But yeah, that's the right way to do it; multiply by the denominator, isolate y terms from x and factor out the exp, isolate some more and then use exp/log rules.

The inverse is this then: $\displaystyle \ln{(x + 1)} - \ln{(x - 1)}=f^{-1}{(x)}$