What is ln((λ^x)/(x!)e^{-λ}) ?

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

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simplify ln((λ^x)/(x!)e^{-λ})

simplify

ln (\frac{λ ^{x}}{x !} e^{- λ})

x ln (λ) - ln (x!) - λ

Solution steps

ln (\frac{λ ^{x}}{x !} e^{- λ})

Apply log rule:

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

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

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

Apply log rule:

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

ln (e^{- λ}) = - λ ln (e)

= ln (\frac{λ ^{x}}{x !}) - λ ln (e)

Apply log rule:

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

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

= ln (λ^{x}) - ln (x!) - ln (e) λ

Apply log rule:

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

ln (λ^{x}) = x ln (λ)

= x ln (λ) - ln (x!) - λ ln (e)

λ ln (e) = λ

= x ln (λ) - ln (x!) - λ

ln (a) + 12 ln (3)

ln (\frac{1}{1 - x}) - x

3 ln (x - 3) + 2 ln (x + 1) - ln (2 x - 3)

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

3.9 + lo g_{10} (\frac{7.25}{3})

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