How do I find $x$ if I know $y$ I have the equation for a sigmoid with the following

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

How do I find what the value of $x$ is if I know $y$?
For example:
if $y = 0.5$ then what is $x$?
(The answer for the example, I believe, is $0$)
Edit: If you are please going to downvote, please explain why, otherwise I cannot improve my questions in the future
 A: All you need to do is solve for $x$:
$$y=\frac{1}{1+e^{-x}}$$
$$\frac{1}{y}=1+e^{-x}$$
$$\frac{1}{y}-1=e^{-x}$$
$$\ln\bigg(\frac{1}{y}-1\bigg)=-x$$
$$x=-\ln\bigg(\frac{1}{y}-1\bigg)$$
As for the case $y=0.5$, we have
$$x=-\ln\bigg(\frac{1}{0.5}-1\bigg)$$
$$x=-\ln(2-1)$$
$$x=-\ln(1)$$
$$x=0$$
So your guess was correct.
A: $$y=\frac{1}{1+e^{-x}} \implies 1+e^{-x}=\frac 1y \implies e^{-x}=\frac 1y-1$$
$$e^{-x}=\frac{1-y}{y}\implies e^x=\frac{y}{y-1}$$$$ \implies \boxed{x=\ln \left(\frac{y}{1-y}\right)}$$
Now plug-in $y=0.5$.
$$\implies x=\ln \left(\frac{0.5}{1-0.5}\right)=\ln (1)=0$$
A: solving $$y=\frac{1}{1+e^{-x}}$$ for $$x$$:
$$1+e^{-x}=\frac{1}{y}$$ and we get
$$-x=\ln\left(\frac{1}{y}-1\right)$$
and now you can insert $$y=0.5$$ in this equation
A: $0.5=\frac{1}{1+e^{-x}} \implies 1=e^{-x}\implies -x\ln e=\ln 1 $ which yields $x=0$. 
I let you conclude what is happening there, which is easy enough if you know a bit about logarithms.
A: Your hunch is right.
$$0.5 = \frac{1}{1+e^{-x}}$$
$$\frac{1}{0.5} = 1 + e^{-x}$$
$$ln\ [\frac{1}{0.5} - 1] = ln\ [e^{-x}]$$
$$ln\ [1] = -x$$
$$x = 0$$
A: We wish to solve
$$ 0.5 = \frac{1}{1+e^{-x}} $$
Okay, sure. Cross-multiply/butterfly/whatever you want to call it first.
$$ 0.5(1 + e^{-x} ) = 1 $$
Divide by $0.5$.
$$ 1 + e^{-x} = 2 $$
Subtract $1$, take logs.
$$ -x = 0 $$
Hence $ x = 0$.
