Exponential Decay Models
Models the amount of a radioactive substance remaining at time t
A radioactive substance has a half life of 1 month.
At time t = 0 there are 100 grams of this substance.
Write an equation for the exponential function that gives the amount as a function of time. Use base B = 0 .
5
Write an equation for the exponential function that gives the amount as a function of time
Use base e.
A(t) = 100 (0 .
5 t )
A(t) = 100 e kt e kt
= 0 .
5 t e k
=0 .
5 k = ln 0 .
5 = −0 .
6931
A(t) = 100 e
−0 .
6931 t b = .5
After 1 month, 50% of the initial amount still remains.
The monthly decay rate is 1 – b = 1 − .5 = .5
The substance decays at the rate of 50% per month. k = −0.6931 is the continuous monthly decay rate
A radioactive substance has a half life of 2 weeks.
At time t = 0 there are 6 kg of this substance.
Write an equation for the exponential function that gives the amount as a function of time.
Use base B= 0 .
5
Write an equation for the exponential function that gives the amount as a function of time with an exponent of t.
Find the appropriate base b.
A(t) = 6 (0 .
5 t/2
)
A(t) = 6 (0 .
5
1/2
) b = .5
1/2 b = 0 .
707
A(t) = 6 (0 .
707) t t
Write an equation for the exponential function that gives the amount as a function of time
Use base e.
A(t) = 6 e kt e kt
= 0 .
5 t/2 e k
= 0 .
5
1/2 k = ln (0 .
5
1/2
) = .5 ln 0 .
5 k = − 0
.
3466
A(t) = 6 e
−0 .
3466 t b = .5
1/2
= 0.707
After 1 week, 70.7% of the initial amount still remains
The weekly decay rate is 1 – b =
1 − 0.707 = .293
The substance decays at the rate of 29.3% per week . k = −0.3466 is the continuous weekly decay rate

A radioactive substance has a half life of 14 days.
At time t = 0 there are 6 kg of this substance.
Write an equation for the Write an equation for the exponential function that gives the amount as a function of time.
Use B = 0 .
5 exponential function that gives the amount as a function of time with an exponent of t.
Find the appropriate base b.
Write an equation for the exponential function that gives the amount as a function of time
Use base e.
A(t) = 6 (0 .
5 t/14 )
A(t) = 6 (0 .
5
1/14
)
t b =0 .
5
1/14 b = 0 .
9517
A(t) = 6 (0 .
9517 ) t
A(t) = 6 (0 .
5) t/14 e kt
= 0 .
5 t/14 e k
= 0 .
5
1/14 k = ln 0 .
5
1/14
k = (1/14) ln 0 .
5 k = −0
.
0495
A(t) = 6 e
−0 .
0495 t b = .5 1/14
= 0.9517
After 1day, 95.17% of the initial amount still remains
The daily decay rate is 1 − 0.9517 = .0483
The substance decays at the rate of 4.83% per day. k = −0.0495 is the continuous daily decay rate
Function modeling exponential decay equation in which an initial amount A o
decreases so that
B = fraction of original amount that remains
after specified number of time periods .
A(t) = A o B t/m b = B 1/m = fraction of original amount that remains after one time period annual if years can be hourly, daily, monthly depending on how time is defined in the problem r = 1 − b = 1 − B 1/m = decay rate per unit of time k = ln b = ln B 1/m = continuous decay rate per 1 unit of time b = e k = B
1/m
See also examples in the textbook in section 5.5
Example 5 pages 328 – 329 and Example 7 page 330