Exponential decay

• Mar 31st 2008, 08:30 AM
nick2price
Exponential decay
i have the table
t Caffine
0 125
1 111
2 98
3 87
4 77
5 69
6 61
7 54
8 47
9 42
10 38
11 34
12 30
13 26
14 23
15 21
16 18
17 16
18 14
19 13
20 11
21 10
22 9
23 8
24 7

The time is in hours and the caffine is the decrease in milligrams. Doing this in excel, i got the exponential decay formula of y = 124.96e-0.1202t which seems to be correct because if i do it to like 2 hours i get
124.96e^((-0.1202)*2)=98.26, which corresponds to our table showing 98.
Now i wanted to see how long it would take for the caffine to reach 0 milligrams. I am trying to use the example

Problem 2: According to our model, when will the population reach 300 thousand?

To solve this problem we set 100 e0.08t equal to 300 and solve for t.

100 e0.08t = 300

e0.08t = 3 Take the natural logarithm of both sides.

ln e0.08t = ln 3

0.08t = ln 3

t = (ln 3)/0.08 = 13.73, approximately.

Therefore, the population is expected to reach 300 thousand about three fourths of the way through the year 1993.

Obviously this doesnt relate to my problem, but it shows a way to do it. So if i set my formula to 124.96e^((-0.1202)t)=0 this means i would first have to divide both sides by 124.96 which doesnt really work on the side which is 0, cos when i come to take the natural logarithm, i get ln0 which returns an error on my calculator. Any advice on how i can find out when the caffine miligrams will be 0?
• Mar 31st 2008, 08:33 AM
Mathnasium
In exponential decay, the y-value (your amount of caffeine) never reaches zero. A percentage of it is always decaying (disappearing) but a percentage always remains. It's like if you start with 1 and keep dividing it by 2. No matter how many times you divide it by 2, you still have some fraction remaining - it's never zero.