# how to justify the logistic function is the inverse of the natural logit function?

per wiki

The logistic function is the inverse of the natural logit function

The standard logistic function looks like (equation_1)

{\displaystyle {\begin{aligned} f(x)&={\frac {1}{1+e^{-x}}}={\frac {e^{x}}{e^{x}+1}}={\frac {1}{2}}+{\frac {1}{2}}\tanh({\frac {x}{2}})\\ \end{aligned}}}

the natural logit function looks like (equation_2)

$$logit(p) = \log\left(\dfrac{p}{1-p}\right)$$

how to justify equation_1 is the inverse of equation_2?

• maybe you can write $x= \frac{1}{1+e^{-y}}$ and solve for $y$ ... Commented Jun 6, 2019 at 11:58

let $$y = logit(x) = log \dfrac{x}{1-x}$$
$$e^y = \dfrac{x}{1-x}$$
$$1 + e^y = \dfrac{1-x}{1-x} + \dfrac{x}{1-x} = \dfrac{1}{1-x}$$
$$\dfrac{1}{1 + e^y} = 1 - x$$
$$x = 1 - \dfrac{1}{1 + e^y} = \dfrac{e^y}{1 + e^y}$$
just calculate $$f\big(\operatorname{logit}(p)\big)=p$$ and $$\operatorname{logit}\big(f(x)\big)=x$$; $$f\big(\operatorname{logit}(p)\big) = \frac{1}{1+e^{-\log\left(\frac{p}{1-p}\right)}}=\frac{1}{1+\frac{1-p}{p}}=\frac{1}{\frac{1}{p}}=p$$and$$\operatorname{logit}\big(f(x)\big)=\log\left(\frac{\frac{1}{1+e^{-x}}}{1-\frac{1}{1+e^{-x}}}\right)=\log\left(\frac{1}{1+e^{-x}-1}\right)=\log\left(e^x\right)=x.$$