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

The solution to ln(1/2 e^{-kt}) is -kt-ln(2)

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simplify ln(1/2 e^{-kt})

simplify

ln (\frac{1}{2} e^{- k t})

- k t - ln (2)

Solution steps

ln (\frac{1}{2} e^{- k t})

Apply log rule:

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

ln (\frac{1}{2} e^{- k t}) = ln (\frac{1}{2}) + ln (e^{- k t})

= ln (\frac{1}{2}) + ln (e^{- k t})

Apply log rule:

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

ln (e^{- k t}) = - k t ln (e)

= ln (\frac{1}{2}) - k t ln (e)

Apply log rule:

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

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

= ln (1) - ln (2) - ln (e) k t

ln (1) = 0

k t ln (e) = k t

= 0 - ln (2) - k t

0 - ln (2) - k t = - ln (2) - k t

= - k t - ln (2)

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