Exponential Models
The general exponential model is: y = aekt.
a = the parameter that represents the initial amount (amount
at t = 0)
k = the parameter that represents the rate of growth (k > 0)
or decay (k < 0)
t = the variable that represents time
y = the variable that represents the amount at time t
e = constant
We can use this model to represent numerous situations of
exponential growth or decay. Examples are radioactive
decay and population growth.
Radioactive Decay
Any radioactive element has a half-life. Half-life is the
amount of time it takes for a substance to go from a given
amount to half of that amount.
Ex: A Given substance has a half-life of 1200 years. If
you start with 100g of the substance, how much will there
be after:
a) 1200 years? 50g
b) 2400 years? 25g
c) 3600 years? 12.5g
Radium is a radioactive isotope. 226Ra (read as Radium
226) is the most stable isotope of Radium. The 226 is its
atomic mass, which means it has a total of 226 protons and
neutrons. This isotope decays into radon gas.
1) 226Ra has a half-life of 1620 years. If there is an initial
quantity of 24g, how much will there be in 1000
years?
Model: y = 24ekt
The first step of every problem involving radioactivity is
to use the half-life to find the value of k.
12 = 24ek(1620)
½ = ek(1620)
ln(½) = ln(e1620k)
1620k = ln(½)
1
ln  
2
k   ≈
1620
-.0004 (Store this value!)
Now our model is complete, so we can find the amount
after 1000 years.
y = 24e -.0004(1000) ≈ 15.646 g
2) 226Ra has a half-life of 1620 years. If there will be 50g
in 700 years, what is the initial amount?
Since we already know the value of k, we can use this to
answer the question. If we didn’t already know it, we
would find k first.
Model: y = ae -.0004t
50 = ae -.0004(700)
50
a
≈ 67.460 g
e .0004( 700)
3) Let t = 0 be the year 1990. If there were 700 students
in AHS in 1990, and 825 students in the year 2000,
predict the number of students in AHS in 2010.
Model: y = 700ekt
We will use the given information to find k, and then use
the completed model to answer the question.
825 = 700ek(10)
t = 10)
(Since the year 2000 corresponds to
825
 e10k
700
 825 
10 k
ln 
  ln e
 700 
 
 825 
10k  ln 

 700 
 825 
ln 

700 

k
 .016
10
Since the year 2010 corresponds to t = 20,
y = 700e .016(20) ≈ 963.989 students