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

The solution to ln((e^{-y}y^x)/(x!)) is -y+xln(y)-ln(x!)

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

simplify

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

- y + x ln (y) - ln (x!)

Solution steps

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Apply log rule:

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

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

= - y ln (e) + x ln (y) - ln (x!)

y ln (e) = y

= - y + x ln (y) - ln (x!)

lo g_{c} (m) - lo g_{c} (n) - lo g_{c} (z)

2 ln (x) + ln (\frac{x ^{3}}{y ^{2}}) - 4 ln (x + y)

ln (x^{3} + 8) - ln (x + 2) - 2 ln (x^{2} - 2 x + 4)

ln (x + 2) - ln (5 x)

What is ln((e^{-y}y^x)/(x!)) ?
The solution to ln((e^{-y}y^x)/(x!)) is -y+xln(y)-ln(x!)