What is ln(1/(sqrt(2pi))*e^{-((x-y)^2)/2}) ?

The solution to ln(1/(sqrt(2pi))*e^{-((x-y)^2)/2}) is (-x^2+2xy-y^2-ln(2pi))/2

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simplify ln(1/(sqrt(2pi))*e^{-((x-y)^2)/2})

simplify

ln (\frac{1}{2 π} \cdot e^{- \frac{( x - y ) ^{2}}{2}})

\frac{- x ^{2} + 2 x y - y ^{2} - ln ( 2 π )}{2}

Solution steps

ln (\frac{1}{2 π} e^{- \frac{( x - y ) ^{2}}{2}})

Apply log rule:

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

ln (\frac{1}{2 π} e^{- \frac{( x - y ) ^{2}}{2}}) = ln (\frac{1}{2 π}) + ln (e^{- \frac{( x - y ) ^{2}}{2}})

= ln (\frac{1}{2 π}) + ln (e^{- \frac{( x - y ) ^{2}}{2}})

Apply log rule:

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

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

= ln (\frac{1}{2 π}) - \frac{( x - y ) ^{2}}{2} ln (e)

Apply log rule:

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

ln (\frac{1}{2 π}) = ln (1) - ln (2 π)

= ln (1) - ln (2 π) - ln (e) \frac{( x - y ) ^{2}}{2}

ln (1) = 0

ln (2 π) = \frac{1}{2} ln (2 π)

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

= 0 - \frac{1}{2} ln (2 π) - \frac{( x - y ) ^{2}}{2}

0 - \frac{1}{2} ln (2 π) - \frac{( x - y ) ^{2}}{2} = - \frac{1}{2} ln (2 π) - \frac{( x - y ) ^{2}}{2}

= - \frac{( x - y ) ^{2}}{2} - \frac{1}{2} ln (2 π)

Multiply

\frac{1}{2} ln (2 π) : \frac{ln ( 2 π )}{2}

= - \frac{( x - y ) ^{2}}{2} - \frac{ln ( 2 π )}{2}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- ( x - y ) ^{2} - ln ( 2 π )}{2}

Expand

- (x - y)^{2} - ln (2 π) : - x^{2} + 2 x y - y^{2} - ln (2 π)

= \frac{- x ^{2} + 2 x y - y ^{2} - ln ( 2 π )}{2}

ln_{e} (2) 2

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What is ln(1/(sqrt(2pi))*e^{-((x-y)^2)/2}) ?
The solution to ln(1/(sqrt(2pi))*e^{-((x-y)^2)/2}) is (-x^2+2xy-y^2-ln(2pi))/2