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Can anyone, please, tell me what rules and intermediate steps have been used to simplify the equation below? $$\frac{1}{1+\exp(-x\Theta) }=\frac{1}{2}$$ $$= \exp(-x\Theta) = 1 $$ $$= x\Theta = 0 $$

Sebastiano
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thom
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2 Answers2

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$$\frac{1}{1+e^{-x\Phi}}=\frac{1}{2}$$ implies, $$\frac{1}{1+\frac{1}{e^{-x\Phi}}}=\frac{1}{2}$$ implies, $$\frac{e^{-x\Phi}}{1+e^{-x\Phi}}=\frac{1}{2}$$ implies, $$2e^{-x\Phi}=1+e^{-x\Phi}$$ implies, $$e^{-x\Phi}=1$$ implies, $$x\Phi=0.$$

pmun
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This much you might have understood. $\frac{1}{1+e^{-x\Phi}}=\frac{1}{2}$. Now see $a^{-b}=\frac{1}{a^b}$. This means that the expression is reduced to $\frac{1}{1+\frac{1}{e^{-x\Phi}}}=\frac{1}{2}$. Now take the lcm of the denomination of previous expression and take that denominator to numerator.(I know that is confusing do tell me in comment if you didn't understand the pervious line). After that cross multiply the expression. You will get $2e^{-x\Phi}=1+e^{-x\Phi}$. Hope you can continue forward.