What is ln(((λ^{(x_{1})}e^{(-λ)}))/((x_{1)!)}) ?

The solution to ln(((λ^{(x_{1})}e^{(-λ)}))/((x_{1)!)}) is x_{1}ln(λ)-λ-ln(x_{1}!)

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simplify ln(((λ^{(x_{1})}e^{(-λ)}))/((x_{1)!)})

simplify

ln (\frac{( λ ^{(x_{1})} e ^{(- λ)} )}{( x _{1} ! )})

x_{1} ln (λ) - λ - ln (x_{1}!)

Solution steps

ln (\frac{( λ ^{(x_{1})} e ^{(- λ)} )}{( x _{1} ! )})

Apply log rule:

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

ln (\frac{( λ ^{(x_{1})} e ^{(- λ)} )}{( x _{1} ! )}) = ln ((λ^{(x_{1})} e^{(- λ)})) - ln ((x_{1}!))

= ln (λ^{x_{1}} e^{- λ}) - ln (x_{1}!)

Apply log rule:

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

ln (λ^{x_{1}} e^{- λ}) = ln (λ^{x_{1}}) + ln (e^{- λ})

= ln (λ^{x_{1}}) + ln (e^{- λ}) - ln (x_{1}!)

Apply log rule:

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

ln (λ^{x_{1}}) = x_{1} ln (λ)

= x_{1} ln (λ) + ln (e^{- λ}) - ln (x_{1}!)

Apply log rule:

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

ln (e^{- λ}) = - λ ln (e)

= x_{1} ln (λ) - λ ln (e) - ln (x_{1}!)

λ ln (e) = λ

= x_{1} ln (λ) - λ - ln (x_{1}!)

0.026 ln (5 \cdot 1 0^{8})

lo g_{a} (a^{2} b c^{3})

lo g_{10} (2 x - 4) - 0.5 lo g_{10} (x)

ln (\frac{2 ^{0.538 + 1} - 1}{( 0.538 - 1 ) x})

\frac{ln ( x ) - ln ( 4 )}{x - 4}

What is ln(((λ^{(x_{1})}e^{(-λ)}))/((x_{1)!)}) ?
The solution to ln(((λ^{(x_{1})}e^{(-λ)}))/((x_{1)!)}) is x_{1}ln(λ)-λ-ln(x_{1}!)