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

The solution to ln((x^ye^{-x})/(y!(1-e^{-x))}) is yln(x)-x-ln(y!)-ln(1-e^{-x})

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

simplify

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

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

Solution steps

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Apply log rule:

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

ln (x^{y}) = y ln (x)

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

Apply log rule:

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

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

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

x ln (e) = x

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

- (ln (y!) + ln (1 - e^{- x})) : - ln (y!) - ln (1 - e^{- x})

= y ln (x) - x - ln (y!) - ln (1 - e^{- x})

lo g_{3} (\frac{x ^{6}}{5 y})

20 lo g_{10} (\frac{2.5 \cdot 1 0 ^{- 3}}{2}) + 4

- ln (x) + 9 ln (y) - 3 ln (z)

ln (\frac{n + 1}{n - 1})

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

What is ln((x^ye^{-x})/(y!(1-e^{-x))}) ?
The solution to ln((x^ye^{-x})/(y!(1-e^{-x))}) is yln(x)-x-ln(y!)-ln(1-e^{-x})