# How are these two equations equal?

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

I was told to sketch a curve but couldn't figure out the first step. The solution manual rewrote the left hand side of the equation above as the right hand side. I cannot figure out what they did to get this. Could someone explain this to me?

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Multiply top and bottom of the lhs by $e^x$. About your title: the two functions are equal, which yields one equation. – 1015 May 24 '13 at 0:12
Try multiplying the top and bottom by $e^{x}$. – Andy Bromberg May 24 '13 at 0:12

$$\frac{1}{1+e^{-x}}=\frac{e^x}{e^x}\frac{1}{1+e^{-x}}=\frac{e^x}{e^x+e^x\cdot e^{-x}}=\frac{e^x}{e^x+1}$$

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They just multiplied the fraction by $1$, in the form $\dfrac{e^x}{e^x}$.

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Multiply the left hand side by $1 = e^x/e^x$. So, you have

$$\frac{1}{1+e^{-x}}\cdot\frac{e^x}{e^x} = \frac{e^x}{1+e^{x}}$$ $$\frac{e^x}{e^x + e^{x-x}} = \frac{e^x}{1+e^{x}}$$ $$\frac{e^x}{e^x + 1} = \frac{e^x}{1+e^{x}}$$

($e^0 = 1$)

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Left-hand side of the equation is

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

which is equal to the right-hand side $\dfrac{e^x}{1+e^x}$.

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