The Natural Base, e
Applications
The Formula for continuously
compounded interest is:
A  Pe
rt
A
Total amount
P
Principal (amount invested)
r
Interest rate
t
time
Example 1
What is the total amount for an investment of $500
invested at 5.25% for 40 years and compounded
continuously?
A  Pe
.0525
A  500e
rt
A = $4083.08
40 
Example 2
What is the total amount for an investment of $100
invested at 3.5% for 8 years and compounded
continuously ?
A  Pe
.035
A  100e
rt
A = $132.31
8
The half-life of a substance is the time it takes
for half of the substance to breakdown or
convert to another substance during the
process of decay.
Natural decay is modeled by the function
below.
N (t )  N 0 e
 kt
N(t)
The amount remaining
N0
Initial amount (at t=(0)
k
Decay constant
t
time
Example 3
Pluonium-239 has a half-life of 24,110 years. How long
does it take for a 1 g sample of Pu-239 to decay to .1 g ?
Step 1: Find decay constant
‘k’
 kt
N (t )  N 0 e
1
 1e  k 24110
2
1
24110 k
ln 2  ln e
 ln 2  24110k
 ln 2
k
24110
k  .0000287
Example 3--continued
Pluonium-239 has a half-life of 24,110 years. How long
does it take for a 1 g sample of Pu-239 to decay to .1 g ?
Step 2: Solve for
N (t )  N0e
‘t’
 kt
.0000287 t
.10  1e
.0000287 t
ln .1  ln e
ln .1  .0000287t
 ln .1
t
.0000287
t  80, 229
Pluonium takes
approximately 80,229
years to decay.
Example 4
The half-life of carbon-14 is 5730 years. What is the age
of a fossil that only has 8% of its original carbon-14?
Step 2: Solve for
Step 1: Find decay constant
ln 2
k
5730
k  .00012
The age of the fossil is
approximately 21,048
years.
‘k’
N (t )  N0e
 kt
‘t’
.00012 t
.00012 t
8  100e
.08  e .00012t
ln .08  ln e
ln .08  .00012t
 ln .08
t
.00012
t  21, 048