# Inverse Functions

1. Oct 7, 2007

### physstudent1

1. The problem statement, all variables and given/known data
Find the inverse:

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

2. Relevant equations

3. 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

2. Oct 7, 2007

### bob1182006

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

3. Oct 7, 2007

### 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

4. Oct 7, 2007

### 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

5. Oct 7, 2007

### physstudent1

what is there to combine

6. Oct 7, 2007

### 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 im back to where i started.

7. Oct 7, 2007

### 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 >.<

8. Oct 7, 2007

### physstudent1

could you show the steps I'm not seeing it

9. Oct 7, 2007

### 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.

Last edited: Oct 7, 2007
10. Oct 7, 2007

### physstudent1

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

11. Oct 7, 2007

### physstudent1

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

12. Oct 7, 2007

### 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

13. Oct 7, 2007

### bob1182006

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

14. Oct 7, 2007

gotcha :)