How many after 2.5 ?
x = Xi e^-kt
x = 75 e^-2.5 k
ln(x/75) = -2.5 k
k = ln(x/75)/(-2.5) solve for k
so
ln .5 = -k t
solve for t
suppose you start with 75 grams of the substance after 2.5 min you have grams left. how much longer will it take you have half remaining
2 answers
Oh, yes, the diff eq which everyone already knows the solution for
da/dt = -ka
da/a = - k dt
ln a = -kt
a = A e^-kt
da/dt = -ka
da/a = - k dt
ln a = -kt
a = A e^-kt