What is ln(e/((2pis^2)^{1/2)}) ?

The solution to ln(e/((2pis^2)^{1/2)}) is 1-1/2 ln(2)-1/2 ln(pi)-ln(s)

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simplify ln(e/((2pis^2)^{1/2)})

simplify

ln (\frac{e}{( 2 π s ^{2} ) ^{\frac{1}{2}}})

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

Solution steps

ln (\frac{e}{( 2 π s ^{2} ) ^{\frac{1}{2}}})

Apply log rule:

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

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

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

Apply log rule:

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

ln ((2 π s^{2})^{\frac{1}{2}}) = \frac{1}{2} ln (2 π s^{2})

= ln (e) - \frac{1}{2} ln (2 π s^{2})

Apply log rule:

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

\frac{1}{2} ln (2 π s^{2}) = \frac{1}{2} ln (2) + \frac{1}{2} ln (π) + \frac{1}{2} ln (s^{2})

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

Apply log rule:

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

ln (s^{2}) = 2 ln (s)

= ln (e) - (\frac{1}{2} ln (2) + \frac{1}{2} ln (π) + \frac{1}{2} \cdot 2 ln (s))

Apply log rule:

lo g_{a} (a) = 1

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

\frac{1}{2} \cdot 2 ln (s) = ln (s)

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

- (\frac{1}{2} ln (2) + \frac{1}{2} ln (π) + ln (s)) : - \frac{1}{2} ln (2) - \frac{1}{2} ln (π) - ln (s)

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

- lo g_{10} (2 \cdot 8.9 \cdot 1 0^{- 2})

ln (\frac{e ^{8}}{14})

lo g_{10} (2 (- 1 + i))

lo g_{8} (11) + lo g_{8} (5)

What is ln(e/((2pis^2)^{1/2)}) ?
The solution to ln(e/((2pis^2)^{1/2)}) is 1-1/2 ln(2)-1/2 ln(pi)-ln(s)