What is ln((e^{-(x-θ)})/((1+e^{-(x-θ)))^2}) ?

The solution to ln((e^{-(x-θ)})/((1+e^{-(x-θ)))^2}) is -x+θ-2ln(1+e^{-x+θ})

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simplify ln((e^{-(x-θ)})/((1+e^{-(x-θ)))^2})

simplify

ln (\frac{e ^{- (x - θ)}}{( 1 + e ^{- (x - θ)} ) ^{2}})

- x + θ - 2 ln (1 + e^{- x + θ})

Solution steps

ln (\frac{e ^{- (x - θ)}}{( 1 + e ^{- (x - θ)} ) ^{2}})

Apply log rule:

lo g_{c} (\frac{a}{b}) = lo g_{c} (a) - lo g_{c} (b)

ln (\frac{e ^{- (x - θ)}}{( 1 + e ^{- (x - θ)} ) ^{2}}) = ln (e^{- (x - θ)}) - ln ((1 + e^{- (x - θ)})^{2})

= ln (e^{- (x - θ)}) - ln ((e^{- (x - θ)} + 1)^{2})

Apply log rule:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x)

ln (e^{- (x - θ)}) = - (x - θ) ln (e)

= - (x - θ) ln (e) - ln ((1 + e^{- (x - θ)})^{2})

Apply log rule:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x)

ln ((1 + e^{- (x - θ)})^{2}) = 2 ln (1 + e^{- (x - θ)})

= - (x - θ) ln (e) - 2 ln (1 + e^{- (x - θ)})

(x - θ) ln (e) = x - θ

2 ln (1 + e^{- (x - θ)}) = 2 ln (1 + e^{- x + θ})

= - (x - θ) - 2 ln (e^{θ - x} + 1)

- (x - θ) : - x + θ

= - x + θ - 2 ln (1 + e^{- x + θ})

ln (\frac{e ^{\frac{9}{2}} 2 ( x )}{x ^{3}}) - ln ((y - 2)^{3} h (x))

\frac{1}{2} (lo g_{8} (s) + lo g_{8} (t))

lo g_{8} (x (x - 9))

lo g_{10} (x (x^{2} + 2)^{- \frac{1}{2}})

ln (\frac{x}{x})

What is ln((e^{-(x-θ)})/((1+e^{-(x-θ)))^2}) ?
The solution to ln((e^{-(x-θ)})/((1+e^{-(x-θ)))^2}) is -x+θ-2ln(1+e^{-x+θ})