Mark M. answered 05/31/24
Let A(t) = amount remaining after t days.
A(t) = 100ekt
50 = 100e28k
So, e28k = 0.5
28k = ln(0.5)
k = -0.024755
A(t) = 100e-0.024755t
Find t so that A(t) = 5:
5 = 100e-0.024755t
0.05 = e-0.024755t
ln(0,05) = -0.024755t
So, t ≈ 121 days.
Becca M.
thank you!!05/31/24