Question

Exp growth decay

Answer 1

i know e is e^((-ln(2)/111)t)

Answer 2

but im not sure what the initial value could be

Answer 3

Why do you need an initial value? P drops to 0.85P.

Answer 4

(.85P)=(P)e^((-ln(2)/111)t)

Answer 5

Given that the half-life is 111 days, you have a relative decay factor \(k\):
\[\frac{1}{2}=e^{111k}~~\iff~~k=-\frac{\ln2}{111}\approx-0.0062446\]
You want to find the time it takes for the sample to decay to 85% its initial amount:
\[0.85=e^{kt}\]
Solve for \(t\) using the \(k\) you found earlier.

Answer 6

the minimum amount remaining is 0.15

so
0.15 = (1/2)^(t/111)
t = 303.803 days

Answer 7

"(.85P)=(P)e^((-ln(2)/111)t)"

Perfect!! Now, divide by P and see if you still need it.

Answer 8

sourwing you da man!

yeah i was confused because its .15 not .85

Answer 9

85% has decayed away, I think sour has the right idea =O
I misread that at first also.

Answer 10

kinda tricky but, thanks everyone!

Answer 11

Yup. sourwing seems to have read the correct problem.