Why $\dfrac{1}{1+e^{-(ax + b)}}$ become $\dfrac{1}{1+e^{-a(x+b)}}$

Question

Why does $\dfrac{1}{1+e^{-(ax + b)}}$ become $\dfrac{1}{1+e^{-a(x+b)}}$? Hope someone can help me with the steps for why it becomes like that?

I have modified my question from $e^{-(ax + b)}$ to $\dfrac{1}{1+e^{-(ax + b)}}$. And $\dfrac{1}{1+e^{-(ax + b)}}$ is what was in the lecture in youtube.

Answer 1

Hint: at $x=0$, we get

$$e^{-b}\stackrel?=e^{-ab}$$

Hopefully you'll see this is clearly false, at least in general.

Answer 2

$$e^{-(ax + b)} = e^{-a(x+b)}$$ Is not true. For example, take $x=0, a=3, b=1$. Then $$e^{-1} \neq e^{-3}$$ The equality holds at only certain values. For example, it is only true for all $x$ when $ax+b=a(x+b)$ for all $x$, i.e. when $a=1$.