How to rearrange $p =e^x/(1+e^x)$ to $x = \ln(p/(1-p))$

Question

I am somehow not able to get my head around this:

I want to rearrange this to x:

$$p =\frac{e^x}{1+e^x}$$

The solution is $x = \ln(p/(1-p))$ But i am not able to rearrange it by myself, because i struggle with the constant 1.

[1] https://i.stack.imgur.com/ojJ5M.png

First find $e^x$ by multiplying the denominator of $RHS$ with the $LHS$ and rearranging the terms. Then Take $\ln$ on both sides. — 19aksh, Oct 27 '20 at 12:07
Similar question: https://math.stackexchange.com/questions/3877726/logistic-regression-converting-equations/3877732#3877732 — player3236, Oct 27 '20 at 12:11

score 3 · Answer 1 · answered Oct 27 '20 at 12:10

3

$p=\frac{e^x}{1+e^x} \implies e^x=\frac{p}{1-p} \implies x=\ln p-\ln(1-p)$.

answered Oct 27 '20 at 12:10

Z Ahmed

43,235

score 2 · Answer 2 · answered Oct 27 '20 at 12:09

2

Hint

First of all, isolate $e^x$.

$$p=\frac{e^x}{1+e^x}\to p(1+e^x)=e^x\to p+pe^x=e^x$$

$$p=e^x-pe^x\to e^x(1-p)=p\to e^x=\frac{p}{1-p}$$

Can you finish?

answered Oct 27 '20 at 12:09

Arnaldo

21,342

2

Thx, with this step to step solution i now get it! – 4sens Oct 27 '20 at 12:16

score 1 · Answer 3 · answered Oct 27 '20 at 12:13

You could divide the numerator and denominator of the fraction by $e^x$, giving us $$p=\frac{1}{1+e^{-x}}\implies1+e^{-x}=\frac{1}{p}\implies e^{-x}=\frac{1-p}{p}$$ $$\implies-x=\ln\frac{1-p}{p}\implies x=\ln\frac{p}{1-p}$$ using the laws of logarithms.

How to rearrange $p =e^x/(1+e^x)$ to $x = \ln(p/(1-p))$

3 Answers3