What is ln(e^{-9θ}(θ^9)/(9!)) ?

The solution to ln(e^{-9θ}(θ^9)/(9!)) is -9θln(e)++9ln(θ)-ln(9!)

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simplify ln(e^{-9θ}(θ^9)/(9!))

simplify

ln (e^{- 9 θ} \frac{θ ^{9}}{9 !})

- 9 θ ln (e) + + 9 ln (θ) - ln (9!)

Solution steps

ln (e^{- 9 θ} \frac{θ ^{9}}{9 !})

Apply log rule:

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

ln (e^{- 9 θ} \frac{θ ^{9}}{9 !}) = ln (e^{- 9 θ}) + ln (\frac{θ ^{9}}{9 !})

= ln (e^{- 9 θ}) + ln (\frac{θ ^{9}}{9 !})

Apply log rule:

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

ln (e^{- 9 θ}) = - 9 θ ln (e)

= - 9 θ ln (e) + ln (\frac{θ ^{9}}{9 !})

Apply log rule:

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

ln (\frac{θ ^{9}}{9 !}) = ln (θ^{9}) - ln (9!)

= - 9 ln (e) θ + + ln (θ^{9}) - ln (9!)

Apply log rule:

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

ln (θ^{9}) = 9 ln (θ)

= - 9 θ ln (e) + + 9 ln (θ) - ln (9!)

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6 ln (x) + \frac{1}{2} ln (y) - 5

What is ln(e^{-9θ}(θ^9)/(9!)) ?
The solution to ln(e^{-9θ}(θ^9)/(9!)) is -9θln(e)++9ln(θ)-ln(9!)