Exponential Function

Question

A helium-filled balloon is slowly deflating. During any 24-hour period, it loses 5 percent of the helium it had at the beginning of that period. The balloon held 8000 cc of helium at noon on Monday. How much helium did it contain 3 days later? 4.5 days later? 20 days later? n days later? 12 hours later? k hours later? Approximately how much time is needed for the balloon to lose half its helium? This time is called the half-life. Be as accurate as you can.

Answer 1

8000 is initial volume

In 1 day (24 hr) loses 5% of volume at start of day, so to get volume after 1 day we reduce 8000 by 5%. 100%-5% is 95%. 95% = 0.95, so our decay factor every day is 0.95

After n days we have 8000 (0.95)ⁿ

3 days later n=3

4.5 days later n=4.5

20 days later n=20

12 hours later, n=0.5

k hours later, n=k/2

To lose half helium after h days

Half of 8000 is 4000

4000 = 8000 (0.95)^h

0.5 = 0.95^h

ln (0.5) = h ln (0.95)

h = ln (0.5) / ln (0.95) = 13.513 days

Answer 2

For exponential:

A = Pe^rt

For this case, P=1 and A = 1 - 0.05 = .95, t = 1 day

.95 = e^r(1)

ln(.95) = r

A= Pe^ln(.95)t = P(.95)^t

1) A = 8000(.95)³

2) A = 8000(.95)^4.5

3) A = 8000(.95)²⁰

4) A = 8000(.95)ⁿ

5) A = 8000(.95)^0.5 12 hrs = half a day

6) A = 8000(.95)^k/24

7) 4000 = 8000(.95)^t/24

ln(0.5) = (t/24)ln(.95)

t = 24(ln(.5)/ln(.95)) ≈ 324.32 hrs = 13.51 days