What is ln(θ^y((1+αy)^{y-1})/(y!)e^{-θ(1+αy)}) ?

The solution to ln(θ^y((1+αy)^{y-1})/(y!)e^{-θ(1+αy)}) is yln(θ)++(y-1)ln(1+αy)-ln(y!)-θ(1+αy)ln(e)

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simplify ln(θ^y((1+αy)^{y-1})/(y!)e^{-θ(1+αy)})

simplify

ln (θ^{y} \frac{( 1 + α y ) ^{y - 1}}{y !} e^{- θ (1 + α y)})

y ln (θ) + + (y - 1) ln (1 + α y) - ln (y!) - θ (1 + α y) ln (e)

Solution steps

ln (θ^{y} \frac{( 1 + α y ) ^{y - 1}}{y !} e^{- θ (1 + α y)})

Apply log rule:

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

ln (θ^{y} \frac{( 1 + α y ) ^{y - 1}}{y !} e^{- θ (1 + α y)}) = ln (θ^{y}) + ln (\frac{( 1 + α y ) ^{y - 1}}{y !}) + ln (e^{- θ (1 + α y)})

= ln (θ^{y}) + ln (\frac{( α y + 1 ) ^{y - 1}}{y !}) + ln (e^{- θ (α y + 1)})

Apply log rule:

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

ln (θ^{y}) = y ln (θ)

= y ln (θ) + ln (\frac{( 1 + α y ) ^{y - 1}}{y !}) + ln (e^{- θ (1 + α y)})

Apply log rule:

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

ln (e^{- θ (1 + α y)}) = - θ (1 + α y) ln (e)

= y ln (θ) + ln (\frac{( 1 + α y ) ^{y - 1}}{y !}) - θ (1 + α y) ln (e)

Apply log rule:

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

ln (\frac{( 1 + α y ) ^{y - 1}}{y !}) = ln ((1 + α y)^{y - 1}) - ln (y!)

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

Apply log rule:

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

ln ((1 + α y)^{y - 1}) = (y - 1) ln (1 + α y)

= y ln (θ) + + (y - 1) ln (1 + α y) - ln (y!) - θ (1 + α y) ln (e)

\frac{1}{2} ln (8) - \frac{1}{2} ln (3)

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3 lo g_{7} (u) + 9 lo g_{7} (v)

ln (\frac{2 a ^{3}}{b ^{4}})

lo g_{5} (4 x^{2})

What is ln(θ^y((1+αy)^{y-1})/(y!)e^{-θ(1+αy)}) ?
The solution to ln(θ^y((1+αy)^{y-1})/(y!)e^{-θ(1+αy)}) is yln(θ)++(y-1)ln(1+αy)-ln(y!)-θ(1+αy)ln(e)