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

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

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

simplify

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

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

Solution steps

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

x ln (e) = x

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

- ln ∣ u + 1 ∣ + ln ∣ u - 1 ∣ + C

ln (\frac{35 + 100}{738 + 100})

10 \cdot lo g_{10} (3.0057 \cdot 1 0^{- 13}) + 30

ln (x^{2}) + ln (z)

f (x) = ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 1

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