What is ln(1/((r-1)!)θ^{-w}x^{w-1}e^{-xθ^{-1}}) ?

The solution to ln(1/((r-1)!)θ^{-w}x^{w-1}e^{-xθ^{-1}}) is ln(x)(w-1)-x/θ-wln(θ)-ln((r-1)!)

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simplify ln(1/((r-1)!)θ^{-w}x^{w-1}e^{-xθ^{-1}})

simplify

ln (\frac{1}{( r - 1 ) !} θ^{- w} x^{w - 1} e^{- x θ^{- 1}})

ln (x) (w - 1) - \frac{x}{θ} - w ln (θ) - ln ((r - 1)!)

Solution steps

ln (\frac{1}{( r - 1 ) !} θ^{- w} x^{w - 1} e^{- x θ^{- 1}})

Apply log rule:

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

ln (\frac{1}{( r - 1 ) !} θ^{- w} x^{w - 1} e^{- x θ^{- 1}}) = ln (\frac{1}{( r - 1 ) !}) + ln (θ^{- w}) + ln (x^{w - 1}) + ln (e^{- x θ^{- 1}})

= ln (\frac{1}{( r - 1 ) !}) + ln (θ^{- w}) + ln (x^{w - 1}) + ln (e^{- θ^{- 1} x})

Apply log rule:

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

ln (θ^{- w}) = - w ln (θ)

= ln (\frac{1}{( r - 1 ) !}) - w ln (θ) + ln (x^{w - 1}) + ln (e^{- x θ^{- 1}})

Apply log rule:

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

ln (x^{w - 1}) = (w - 1) ln (x)

= ln (\frac{1}{( r - 1 ) !}) - w ln (θ) + (w - 1) ln (x) + ln (e^{- x θ^{- 1}})

Apply log rule:

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

ln (e^{- x θ^{- 1}}) = - x θ^{- 1} ln (e)

= ln (\frac{1}{( r - 1 ) !}) - w ln (θ) + (w - 1) ln (x) - x θ^{- 1} ln (e)

Apply log rule:

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

ln (\frac{1}{( r - 1 ) !}) = ln (1) - ln ((r - 1)!)

= ln (1) - ln ((r - 1)!) - w ln (θ) + ln (x) (w - 1) - ln (e) θ^{- 1} x

ln (1) = 0

x θ^{- 1} ln (e) = \frac{x}{θ}

= 0 - ln ((r - 1)!) - w ln (θ) + ln (x) (w - 1) - \frac{x}{θ}

0 - ln ((r - 1)!) - w ln (θ) + (w - 1) ln (x) - \frac{x}{θ} = - ln ((r - 1)!) - w ln (θ) + (w - 1) ln (x) - \frac{x}{θ}

= ln (x) (w - 1) - \frac{x}{θ} - w ln (θ) - ln ((r - 1)!)

- \frac{1}{2} ln (\frac{1}{2}) + \frac{1}{2}

lo g_{6} (2 x) - lo g_{6} (x - 3) 3

\frac{24}{π} ln (2 + 3)

- (- ln (x + 1) + ln (x - 1))

ln (\frac{0.001 \cdot 0.5}{0.5})

What is ln(1/((r-1)!)θ^{-w}x^{w-1}e^{-xθ^{-1}}) ?
The solution to ln(1/((r-1)!)θ^{-w}x^{w-1}e^{-xθ^{-1}}) is ln(x)(w-1)-x/θ-wln(θ)-ln((r-1)!)