# Inverse of f(x)= e^x/e^x+1

Dramen

## Homework Statement

I'm having problems with finding the inverse equation for this function $$e^x/(e^x-1)$$

## The Attempt at a Solution

Currently, I have done about 2 methods to solve it and hitting dead-ends on each one:

Attempt 1. $$x=e^y/(e^y-1)$$ then multiply both sides by $$(e^y-1)$$ so that I have $$xe^y-x=e^y$$, then subtract $$xe^y$$ from both sides to then get $$x=e^y-xe^y$$ after which I factor out $$e^y$$ and then my final solution to this attempt is: $$f(x)^-1= ln(x/(1-X))$$

Attempt 2. I took the reciprocal of the equation so I have $$1/x= (e^y+1)/e^y$$ then split up the fraction to $$1/x= 1+(1/e^y)$$, then multiplied both sides by $$e^y$$ to get $$e^y/x = 1+1$$ then multiplied it by $$x$$ so that my solution was $$f(x)^-1= ln(2x)$$

Obviously by looking at the final solutions neither of them actually work. I would really appreciate any enlightenment on this problem.

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## Homework Statement

I'm having problems with finding the inverse equation for this function $$e^x/(e^x-1)$$

## The Attempt at a Solution

Currently, I have done about 2 methods to solve it and hitting dead-ends on each one:

Attempt 1. $$x=e^y/(e^y-1)$$ then multiply both sides by $$(e^y-1)$$ so that I have $$xe^y-x=e^y$$, then subtract $$xe^y$$ from both sides to then get $$x=e^y-xe^y$$
You made a sign error here. This attempt, otherwise, is fine.
after which I factor out $$e^y$$ and then my final solution to this attempt is: $$f(x)^-1= ln(x/(1-X))$$
X and x are two different variables. Don't mix cases. Also, you want the superscript -1 on the f, not after (x). f-1 denotes the inverse of f, but f(x)-1 means 1/f(x).

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Attempt 2. I took the reciprocal of the equation so I have $$1/x= (e^y+1)/e^y$$ then split up the fraction to $$1/x= 1+(1/e^y)$$, then multiplied both sides by $$e^y$$ to get $$e^y/x = 1+1$$
You have to multiply everything by ##e^y##:
\begin{align*}
\frac{1}{x} &= 1+\frac{1}{e^y} \\
e^y \times \frac{1}{x} &= e^y \times \left(1+\frac{1}{e^y}\right) \\
\frac{e^y}{x} &= e^y + 1
\end{align*}
then multiplied it by $$x$$ so that my solution was $$f(x)^-1= ln(2x)$$

Obviously by looking at the final solutions neither of them actually work. I would really appreciate any enlightenment on this problem.

Dramen
I see the mathematical errors now because when I was first typing it on my home computer the connection was bad and had problems typing it out and previewing it before posting.

However, it seems that even with your assistance I'm still having problems with finding the inverse function as my solution now:

1) $$f-1(x)=ln(x/1-x)$$ which doesn't work when x≥1

Homework Helper
I get it differently

$$y^{-1} (x) = \ln\left|\frac{x}{x-1}\right|$$

for certain values of x which you have to figure out yourself.

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I see the mathematical errors now because when I was first typing it on my home computer the connection was bad and had problems typing it out and previewing it before posting.

However, it seems that even with your assistance I'm still having problems with finding the inverse function as my solution now:

1) $$f^{-1}(x)=ln(x/1-x)$$ which doesn't work when x≥1
Don't use tags, etc in LaTeX.

You need Parentheses for the denominator.

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Dramen ,

I suggest writing $\displaystyle f(x)=\frac{e^x}{e^x-1}\ \text{ as }\ f(x)=\frac{1}{1-e^{-x}}\ .$

It should be easy from there.

Dramen
I must be an idiot or still suffering from my high school senioritis because even with all your help (which is very appreciated). I still can't get a decent answer as my calculator gets me error values, low decimals (to check against f(x)), or the numbers are negative.

@SammyS this is what I get from your hint $$x= (1)/(1+e^{-1})$$ multiply by $$1+e^{-1}$$ to get
$$x+xe^{-1}=1$$ then subtract x then divide by x to get $$e^{-y}=\frac{1-x}{x}$$ natural log it to finally get $$y= -\ln (\frac {1-x}{x})$$

@dextercioby I'm not sure if I follow you since I have no idea how you switched the denominator x and 1 positions and also when I check the values through $$f(x)*f^{-1}(x)$$ it doesn't come out to 1

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e^x = p. You get nice algebra afterwards: p = (p-1)y => p(1-y)=-y => p=y(y-1)^(-1). Thus x=ln|y/(y-1)| or y^(-1)(x)= what I wrote in post 5.

Bohrok
One more possible route to take is$$\frac{e^x}{e^x-1} = \frac{e^x-1+1}{e^x-1} = 1 + \frac{1}{e^x-1}$$
However,
...when I check the values through $$f(x)*f^{-1}(x)$$ it doesn't come out to 1

That's not true in general for the product of a function and its inverse. What is true is that f(f-1(x)) = f-1(f(x)) = x

Dramen
@dextercioby okay I was able to get my answer though I would like to know if the steps I made to recreate your answer are right $$x=\frac{e^y}{e^y+1}$$ multiply the denominator on both sides to get $$xe^y+x=e^y$$ multiply both sides by -1? $$-e^y+xe^y=-x$$ factor out the ey to then get $$e^y(x-1)=-x$$ divide both sides by (x-1) $$e^y=\frac{-x}{x-1}$$ then natural log it to get your answer of $$f^{-1}(x)=\ln \vert \frac {-x}{x-1}\vert$$

@bohrok seems that I'm starting to lose my grasp of math, that I'm thinking that the check for reciprocals is the same as the check for inverses

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@dextercioby okay I was able to get my answer though I would like to know if the steps I made to recreate your answer are right $$x=\frac{e^y}{e^y+1}$$ multiply the denominator on both sides to get $$xe^y+x=e^y$$ multiply both sides by -1? $$-e^y+xe^y=-x$$ factor out the ey to then get $$e^y(x-1)=-x$$ divide both sides by (x-1)$$e^y=\frac{-x}{x-1}$$ then natural log it to get your answer of $$f^{-1}(x)=\ln \vert \frac {-x}{x-1}\vert$$

@bohrok seems that I'm starting to lose my grasp of math, that I'm thinking that the check for reciprocals is the same as the check for inverses
Why are you using the absolute value in that result?

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Dramen
Yeah I have no idea why I have it either only put it there because dextercioby had one. Also if your talking about the $$e^y+1$$ then its alright because somehow I fudged typing the equation right and accidentally turned it from an addition sign to a subtraction one.

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Yeah I have no idea why I have it either only put it there because dextercioby had one. Also if your talking about the $$e^y+1$$ then its alright because somehow I fudged typing the equation right and accidentally turned it from an addition sign to a subtraction one.
So you should get

$\displaystyle f^{-1} (x) = \ln\left(\frac{x}{x-1}\right)\ .$

I think it's probably easiest to get to this using rewriting f(x) in the manner Bohrok suggested.

Dramen
Okay I finally got it thanks for all the help, however my answer differs from yours in that mine is: $$f^{-1}(x)=\ln \frac {-x}{x-1}$$
Thing is though when I run mine through the checks it still works, but when I used yours I just get errors.

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$$\frac{e^x}{e^x+1}$$ or the one in your original post, which is
$$\frac{e^x}{e^x-1}?$$