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

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

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

simplify

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

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

Solution steps

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

ln (1) = 0

x ln (e) = x

= 0 - ln (- e^{- x} + 1) - x

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

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

12 (ln ∣ (4) + 2 ∣ - ln ∣ (4) + 3 ∣) + C

y [ln (2 y) - ln (1 - 2 y)] - y [ln (y) - ln (1 - y)]

ln (\frac{60}{120})

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\frac{1}{10} ln (5000)

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