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I am currently learning the sigmoid/logistic function and have completely forgotten how the maths behind this equivalence works:

$$ \dfrac{1}{1+ e^{-x}} = \dfrac{e^{x}}{1+e^{x}} $$

By this I mean how the left side equates with the right side. I know that the left side is simply multiplied by (1/exp(x)) but what are the rules governing this? Any help would be much appreciated, especially the rules so that I can revisit them.

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    $\begingroup$ $\dfrac a b =\dfrac ab \times \dfrac cc = \dfrac{ac}{bc}$ for $b,c \not=0$ $\endgroup$
    – Henry
    Jul 6, 2021 at 11:04

2 Answers 2

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It is easy.

$ \dfrac{1}{1+ e^{-x}} = \dfrac{e^{x}}{1+e^{x}} $

Consider lhs

$ \dfrac{1}{1+ \frac{1}{e^{x}}} $

which is equal to

$ \dfrac{e^{x}}{e^{x}+ 1} $

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    $\begingroup$ Multiply by $e^x/e^x$ To get to the final line. $\endgroup$
    – Dave
    Jul 6, 2021 at 4:07
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$$\begin{align} \dfrac{1}{1+ e^{-x}} &=\dfrac{1}{1+ e^{-x}} \\ &= \dfrac{1}{1+ e^{-x}} \cdot \frac{e^x}{ e^x} \\ &= \dfrac{e^{x}}{e^x+1} \end{align} $$

  • Multiplying a number by 1 doesn't change its value, and $\frac{e^x}{ e^x}$ is just one way to write 1.

  • Multiplication adds exponents of a common base: $a^b \cdot a^c = a^{b+c}$. We use this in the final line to get $e^x \cdot e^{-x}=e^0 = 1$.

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