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

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

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

simplify

1 - ln (2) + ln (1 + e^{- 1})

1 + ln (\frac{e + 1}{2 e})

Decimal Notation

0.62011 \dots

Solution steps

1 - ln (2) + ln (1 + e^{- 1})

Apply log rule:

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

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

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

\frac{1 + e ^{- 1}}{2} = \frac{e + 1}{2 e}

= 1 + ln (\frac{e + 1}{2 e})

ln (\frac{27}{16})

ln (\frac{y}{( y - 2 )})

lo g_{10} (\frac{1 0 ^{- 10}}{2})

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

ln (e ln (\frac{1}{e}))

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