a) We would apply the formula for calculating exponential decay which is expressed as
y = a(1 - r)^t
where
y is the final value after time t
t is the time
r is the rate of decay
a is teh initail amount
The halving time refers to the time when y = a/2
From the information given,
r = 3% = 3/100 = 0.03
y = a/2
Thus,
a/2 = a(1 - 0.03)^t
0.5 = (1 - 0.03)^t = 0.97^t
Take natural log of both sides,
ln 0.5 = ln0.97^t = tln0.97
t = ln 0.5/ln 0.97
t = 22.76
Approximate halving time is 22.76 weeks
b) We would find the exact halving time by using the formula
t1/2 = ln2/r = ln2 = 0.03
exact havling time = 23.104 weeks
c) Given that a = 3
time = 2 years
Number of weeks in a year = 52
Number of weeks in 2 years = 52 x 2 = 104
using the exact halving time,
y = 3(1/2)^(104/23.104)
y = 0.132
The amount of trash that would be dumped in 2 years = 0.132 tons
Using the approximate halving time,
y = 3(1/2)^(104/22.76)
y = 0.126
The amount of trash that would be dumped in 2 years = 0.126 tons
e) Relative error = (exact - approximate)/exact x 100
Relative error = (0.132 - 0.126)/0.132 x 100
Relative error = 4.54 %