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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.

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    I think this only holds when $b=0$ or case 2 both $a=0,b=0$ – 2017-01-30
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    Please, refrain from editting the question after it has already been answered. – 2017-01-30
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    @user122358 Why change the question when answers had already been done? I think its not good to do that. – 2017-01-30
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    Yeah, I know. I should have written the whole thing in the first place. My apologies. – 2017-01-30

2 Answers 2

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Hint: at $x=0$, we get

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

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

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$$e^{-(ax + b)} = e^{-a(x+b)}$$ 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 $a=1$.