Warm Up: Find the final amount
:
Invest $4000 at 6% compounded quarterly for 20 years.
 r
A  P 1  
 n
nt
 0.06 
A  $40001 

4 

A  $13,162.65
420
Invest $5600 at 3.7% compounded continuously for 12 years.
A  Pe
rt
0.03712 
A  5600e
A  $8,730.01
Application
Problems
Growth and Decay
Growth  P 1  r 
t
Decay  P 1  r 
t
The population of Durango is 50,000. If it grows at
a rate of 2%, what will the population be in one
year?
Growth  P 1  r 
t
Growth  50,0001  0.02 
 50,0001.02 
 $51,000
1
The population of Cedar Oak in 1920 was 6500. If the
population is growing at a rate of 5% per year, what will the
population be in 1970?
Growth  P 1  r 
t
A  65001  0.05
A  74,538
50
The population of Walden in 1930 was 350,000. If the
population is decreasing at a rate of 2.5% per year, what will
the population be in 1990?
Decay  P 1  r 
t
A  350,0001  0.025
A  76,621
60
The population of Greenwood in 1940 was 7400. If the
population is growing at a rate of 4.5% per decade, what will
the population be in 2010?
Growth  P 1  r 
t
A  74001  0.045
70
10
A  74001.045
A  10,070
7
The population of Glenview in 1935 was 8200. If the
population is growing at a rate of 3% per decade, what will it
be in 2020?
Growth  P 1  r 
t
A  82001  0.03
85
10
A  82001.03
A  10,542
8.5
If you have 30 bacteria that double every hour, predict the
population after 4 hours.
Growth  P Amout of Change 
t
where t is how many times it changes.
A  302
4
A  3016 
A  480
If you have 65 bacteria that triple every hour, predict the
population after 4 hours.
Growth P  Amout of Change 
t
A  653
4
A  6581
A  5265
If you have 120 bacteria that double every 15 minutes,
predict the population after 3 hours.
Growth P  Amout of Change 
t
Note : It is doubling every 15
MINUTES. So, in 3 hours
it doubles 4  3 times.
A  1202 
34 
A  1202 
12
A  1204096 
A  491,520
Suppose the half-life of a certain radioactive material is 20
days and there are 10 grams initially. How much is left after
60 days?
Decay  P 1  r 
t
A  101  0.5
60
A  100.5
A  100.125
A  1.25
3
20
Suppose the half-life of a certain radioactive material is 15
days and there are 60 grams initially. How much is left after
80 days? Give answer to nearest thousandth.
Decay  P 1  r 
t
A  601  0.5
80
15
A  600.5
A  1.488
80
15
One Last Thing from
Algebra……
Difference of Two
Percent Increase / Decrease 
x100
Original
Example: The original population was 100.
The new population is 60. What is the
percent decrease?
Percent

100  60
Decrease 
x100
100
 40%
Homework

Worksheet