What is ln((λ^y)/(y!(e^λ-1))) ?

The solution to ln((λ^y)/(y!(e^λ-1))) is yln(λ)-ln(y!)-ln(e^λ-1)

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simplify ln((λ^y)/(y!(e^λ-1)))

simplify

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

y ln (λ) - ln (y!) - ln (e^{λ} - 1)

Solution steps

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

Apply log rule:

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

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

= ln (λ^{y}) - ln (y! (e^{λ} - 1))

Apply log rule:

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

ln (λ^{y}) = y ln (λ)

= y ln (λ) - ln (y! (e^{λ} - 1))

Apply log rule:

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

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

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

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

= y ln (λ) - ln (y!) - ln (e^{λ} - 1)

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ln (10) - ln (1 0^{- 11} e^{40 x}) - ln (y)

7 lo g_{e} (x) - 0.5 lo g_{e} (y)

What is ln((λ^y)/(y!(e^λ-1))) ?
The solution to ln((λ^y)/(y!(e^λ-1))) is yln(λ)-ln(y!)-ln(e^λ-1)