Trouble finding inverse of y = 10^x / (10^x + 1)

**Grep** · December 22nd 2010, 02:05 PM

This is surely where a teacher is handy to have. Luckily, I also have this forum!

I'm obviously missing something, but I can't figure out the inverse of:

$y = \frac{10^x}{10^x + 1}$

I can isolate one x variable, but can't figure out how to isolate it entirely.

I've tried a bunch of ways, but the main one is:

Take the log base 10 of both sides:
$log_{10}(y) = log_{10}(\frac{10^x}{10^x + 1})$

Split off the division into subtraction:
$log_{10}(y) = log_{10}(10^x) - log_{10}(10^x + 1)$

Then I can isolate one x:
$log_{10}(y) = x - log_{10}(10^x + 1)$

I just can't seem to see the way forward. Please, someone help put me out of my misery. Would be much appreciated!

**e^(i*pi)** · December 22nd 2010, 02:13 PM

$10^xy + y - 10^x = 0$

$10^xy-10^x = -y$

$10^x(y-1) = -y$

$10^x = \dfrac{-y}{y-1}$

$x = \log \left(\dfrac{-y}{y-1}\right) = \log(-y) - \log(y-1)$

$f^{-1}(x) = \log \left(\dfrac{-x}{x-1}\right) = \log(-x) - \log(x-1)$

It would appear that no real number satisfies the domain

**FernandoRevilla** · December 22nd 2010, 02:14 PM

Firstly express:

$10^x=\dfrac{y}{1-y}$

...

Fernando Revilla

Edited: I didn't see the previous post.

**Archie Meade** · December 22nd 2010, 02:18 PM

How about

$\displaystyle\ y=\frac{10^x}{10^x+1}=\frac{10^{-x}}{10^{-x}}\;\frac{10^x}{10^x+1}=\frac{1}{1+10^{-x}}\Rightarrow\ 1+10^{-x}=\frac{1}{y}\Rightarrow\ 10^{-x}=\frac{1-y}{y}$

**Grep** · December 22nd 2010, 02:51 PM

Thanks to you all! I see my mistake. I tried to go too quickly to logs, but I needed to manipulate it quite a bit before I even get there. I love that I got two different ways to arrive at the solution. Going through Archie Meade's solution to the end, I get the textbook answer of $y = log_{10}(\frac{x}{1-x})$ .

And e^(i*pi)'s solution reduces to the same thing if I just do the somewhat obvious $y = log_{10}(\frac{-x}{x -1}) = log_{10}(\frac{-x}{-(-x + 1)}) = log_{10}(\frac{x}{1 - x})$ .

Thanks again. And on to the next problem I go.

**Prove It** · December 22nd 2010, 07:33 PM

$\displaystyle \frac{10^x}{10^x + 1} = \frac{10^x + 1 - 1}{10^x + 1} = \frac{10^x + 1}{10^x + 1} - \frac{1}{10^x + 1} = 1 - \frac{1}{10^x + 1}$ .

So if $\displaystyle y = 1 - \frac{1}{10^x + 1}$ , to find the inverse, the $\displaystyle x,y$ values swap, giving

$\displaystyle x = 1 - \frac{1}{10^y + 1}$

$\displaystyle \frac{1}{10^y+1} = 1 - x$

$\displaystyle 10^y + 1 = \frac{1}{1-x}$

$\displaystyle 10^y = \frac{1}{1-x} - 1$

$\displaystyle 10^y = \frac{1}{1-x}- \frac{1-x}{1-x}$

$\displaystyle 10^y = \frac{x}{1-x}$

$\displaystyle y = \log_{10}{\left(\frac{x}{1-x}\right)}$

$\displaystyle y = \log_{10}{(x)} - \log_{10}{(1-x)}$