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

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

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

simplify

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

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

Solution steps

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

y ln (e) = y

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

ln (\frac{4 x + 1}{3 x - 1})

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

- lo g_{10} (4.8 \cdot 1 0^{- 6})

\frac{9}{14} lo g_{10} (\frac{9}{14}) + \frac{5}{14} lo g_{\frac{5}{14}} (x)

ln (\frac{16}{3})

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