What is ln((e^{(xβ)^2}(xβ)^{2yi})/(yi!)) ?

The solution to ln((e^{(xβ)^2}(xβ)^{2yi})/(yi!)) is β^2x^2+2iyln(βx)-ln(y)-ln(i!)

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simplify ln((e^{(xβ)^2}(xβ)^{2yi})/(yi!))

simplify

ln (\frac{e ^{(x β)^{2}} ( x β ) ^{2 y i}}{y i !})

β^{2} x^{2} + 2 i y ln (β x) - ln (y) - ln (i!)

Solution steps

ln (\frac{e ^{(x β)^{2}} ( x β ) ^{2 y i}}{y i !})

Apply log rule:

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

ln (\frac{e ^{(x β)^{2}} ( x β ) ^{2 y i}}{y i !}) = ln (e^{(x β)^{2}} (x β)^{2 y i}) - ln (y i!)

= ln (e^{(β x)^{2}} (β x)^{2 i y}) - ln (i! y)

Apply log rule:

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

ln (e^{(x β)^{2}} (x β)^{2 y i}) = ln (e^{(x β)^{2}}) + ln ((x β)^{2 y i}), ln (y i!) = ln (y) + ln (i!)

= ln (e^{(β x)^{2}}) + ln ((β x)^{2 i y}) - (ln (y) + ln (i!))

Apply log rule:

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

ln (e^{(x β)^{2}}) = (x β)^{2} ln (e)

= (x β)^{2} ln (e) + ln ((x β)^{2 y i}) - (ln (y) + ln (i!))

Apply log rule:

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

ln ((x β)^{2 y i}) = 2 y i ln (x β)

= (x β)^{2} ln (e) + 2 y i ln (x β) - (ln (y) + ln (i!))

(x β)^{2} ln (e) = β^{2} x^{2}

= β^{2} x^{2} + 2 i y ln (β x) - (ln (y) + ln (i!))

= β^{2} x^{2} + 2 i y ln (β x) - (ln (y) + ln (i!))

- (ln (y) + ln (i!)) : - ln (y) - ln (i!)

= x^{2} β^{2} + 2 y i ln (x β) - ln (y) - ln (i!)

= β^{2} x^{2} + 2 i y ln (β x) - ln (y) - ln (i!)

ln (2 π \cdot 501.24 \cdot 0.98 \cdot 111.2 \cdot 1 0^{- 6})

\frac{1}{4} ln (\frac{1}{4})

ln (\frac{0.34}{0.32})

lo g_{e} (12 x)

4.74 - lo g_{10} (\frac{1}{1.859})

What is ln((e^{(xβ)^2}(xβ)^{2yi})/(yi!)) ?
The solution to ln((e^{(xβ)^2}(xβ)^{2yi})/(yi!)) is β^2x^2+2iyln(βx)-ln(y)-ln(i!)