Inverse Function for (e^x)/(e^x + 1): Steps and Example Solutions

[physstudent1]

Homework Statement

Find the inverse:

y = (e^x)/(e^x + 1)

Homework Equations

The Attempt at a Solution

I switched x with y and solved for y but I ended up getting lne^y - lnx = lne^y +ln1 and then -lnx= ln1

 

[bob1182006]

could you show your work?

and I think you split up e^x+1 to lne^y+ln1? you can't do that.

 

[physstudent1]

ohhh your right it should be

e^y = x(e^y+1)

then

lne^y = lnx + ln(e^y+1) but I'm still stuck from here

 

[bob1182006]

hm..ok take the ln(e^y+1) to the right side and simplify by combining the ln's

edit: whops meant take lne^y to the right side, lnx to the left and you should be able to simplify it

 

[physstudent1]

what is there to combine

 

[physstudent1]

so get -lnx = ln(e^y + 1) - lne^y

then -lnx = ln((e^ y +1)/e^y) ? this doesn't seem like it helped now I am back to where i started.

 

[bob1182006]

hm..no I multiplied by -1 on the RHS and LHS and got x=-lny but plugging that in I get y=1+y >.<

 

[physstudent1]

could you show the steps I'm not seeing it

 

[bob1182006]

wow did that wrong too >.>

my algebra is really bad right now for some reason...hm..try multiplying/dividing by \frac{e^{-x}}{e^{-x}}

ok yes multiply/divide by that and you will find the inverse.

 

[physstudent1]

so do that by the original equation before i start trying to find the inverse ?

 

[physstudent1]

i got -ln(1/x -1) = y for the inverse would anyone agree ?

 

[bob1182006]

yes and try to get x^-x all by itself on the RHS or LHS so you don't have something like ln(e^x+1)=y

 

[bob1182006]

yep that's what I got, and you can check by plugging it in. also should be x=f(y).

 

[physstudent1]

gotcha :)