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

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

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

simplify

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

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

Solution steps

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

Apply log rule:

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

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

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

ln (\frac{729}{441})

ln (4) - ln (8)

lo g_{10} (x^{7} z)

ln (10) + ln (20)

ln (\frac{∣ a ∣}{2}) + ln \frac{a}{2} + h

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