What is ln(-e^{-x}(1+1/3 e^{-2x})) ?

The solution to ln(-e^{-x}(1+1/3 e^{-2x})) is -x+ln(1+1/3 e^{-2x})

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simplify ln(-e^{-x}(1+1/3 e^{-2x}))

simplify

ln (- e^{- x} (1 + \frac{1}{3} e^{- 2 x}))

- x + ln (1 + \frac{1}{3} e^{- 2 x})

Solution steps

ln (- e^{- x} (1 + \frac{1}{3} e^{- 2 x}))

Apply log rule:

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

ln (- e^{- x} (1 + \frac{1}{3} e^{- 2 x})) = ln (e^{- x}) + ln (1 + \frac{1}{3} e^{- 2 x})

= ln (e^{- x}) + ln (\frac{1}{3} e^{- 2 x} + 1)

Apply log rule:

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

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

= - x ln (e) + ln (1 + \frac{1}{3} e^{- 2 x})

x ln (e) = x

= - x + ln (\frac{1}{3} e^{- 2 x} + 1)

x - 3 + ln (\frac{x}{( x + 2 )})

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ln (\frac{3}{2} a e^{- \frac{3}{2} a x})

ln (2) - 2 ln (6)

2 ln (x) + \frac{1}{2} ln (x - 1) - 3 ln (x + 7)

What is ln(-e^{-x}(1+1/3 e^{-2x})) ?
The solution to ln(-e^{-x}(1+1/3 e^{-2x})) is -x+ln(1+1/3 e^{-2x})