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

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

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

simplify

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

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

Solution steps

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

Apply log rule:

lo g_{c} (ab) = lo g_{c} (a) + lo g_{c} (b)

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

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

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

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

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ln (\frac{x ^{2} - 4 + x}{2})

ln (2 x) \cdot 2

2 ln (- 2 + 2)

- 8 ln (4) + 8 ln (8)

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