What is ln(1/(β^n)e^{(-1)/(β^n)(-αn+x)}) ?

The solution to ln(1/(β^n)e^{(-1)/(β^n)(-αn+x)}) is -(x-αn)/(β^n)-nln(β)

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simplify ln(1/(β^n)e^{(-1)/(β^n)(-αn+x)})

simplify

ln (\frac{1}{β ^{n}} e^{\frac{- 1}{β ^{n}} (- α n + x)})

- \frac{x - α n}{β ^{n}} - n ln (β)

Solution steps

ln (\frac{1}{β ^{n}} e^{\frac{- 1}{β ^{n}} (- α n + x)})

Apply log rule:

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

ln (\frac{1}{β ^{n}} e^{\frac{- 1}{β ^{n}} (- α n + x)}) = ln (\frac{1}{β ^{n}}) + ln (e^{\frac{- 1}{β ^{n}} (- α n + x)})

= ln (\frac{1}{β ^{n}}) + ln (e^{\frac{- 1}{β ^{n}} (x - α n)})

Apply log rule:

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

ln (e^{\frac{- 1}{β ^{n}} (- α n + x)}) = \frac{- 1}{β ^{n}} (- α n + x) ln (e)

= ln (\frac{1}{β ^{n}}) + \frac{- 1}{β ^{n}} (- α n + x) ln (e)

Apply log rule:

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

ln (\frac{1}{β ^{n}}) = ln (1) - ln (β^{n})

= ln (1) - ln (β^{n}) + ln (e) \frac{- 1}{β ^{n}} (x - α n)

Apply log rule:

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

ln (β^{n}) = n ln (β)

= ln (1) - n ln (β) + \frac{- 1}{β ^{n}} (- α n + x) ln (e)

ln (1) = 0

\frac{- 1}{β ^{n}} (- α n + x) ln (e) = - \frac{- α n + x}{β ^{n}}

= 0 - n ln (β) - \frac{x - α n}{β ^{n}}

0 - n ln (β) - \frac{- α n + x}{β ^{n}} = - n ln (β) - \frac{- α n + x}{β ^{n}}

= - \frac{x - α n}{β ^{n}} - n ln (β)

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What is ln(1/(β^n)e^{(-1)/(β^n)(-αn+x)}) ?
The solution to ln(1/(β^n)e^{(-1)/(β^n)(-αn+x)}) is -(x-αn)/(β^n)-nln(β)