problem 15.23B

[Amy Zheng 2l]

For part B,how come we use In(Ainitial/Afinal)=kt? shouldn't it be -kt?

[Anna Goldberg 2I]

ln(At)/(Ainitial)=-kt

When you do ln(A initial)/(At), the swap of the numerator and denominator makes ln(A initial)/(At)=kt

[Hena Sihota 1L]

ln[A]t-ln[A]0=-kt If you divide both sides of this equation by -1 then you get -[A]t+[A]0=kt which can be rewritten as [A]0-[A]t=kt. According to log rules, this equation can be rewritten as ln[A]0/ln[A]t=kt

[Rohan Chaudhari- 1K]

You can also just use [A]=[A]initial * e^-kt, which is just a simplified verision of the ln[A]=-kt+ln[A]initial equation