What is ln(1/(2pis^2)) ?

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

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simplify ln(1/(2pis^2))

simplify

ln (\frac{1}{2 π s ^{2}})

- 2 ln (s) - ln (2) - ln (π)

Solution steps

ln (\frac{1}{2 π s ^{2}})

Apply log rule:

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

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

= ln (1) - ln (2 π s^{2})

Apply log rule:

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

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

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

Apply log rule:

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

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

= ln (1) - (ln (2) + ln (π) + 2 ln (s))

Apply log rule:

lo g_{a} (1) = 0

= 0 - (2 ln (s) + ln (2) + ln (π))

0 - (ln (2) + ln (π) + 2 ln (s)) = - (ln (2) + ln (π) + 2 ln (s))

= - (2 ln (s) + ln (2) + ln (π))

Distribute parentheses

= - (2 ln (s)) - (ln (2)) - (ln (π))

Apply minus-plus rules

+ (- a) = - a

= - 2 ln (s) - ln (2) - ln (π)

ln (\frac{x}{3})

lo g_{\frac{1}{4}} (332)

3 lo g_{10} (X) + 2 lo g_{10} (y)

ln (a e^{- k t})

- lo g_{10} (4.1 \cdot 1 0^{- 6})