Applications of Exponential Function - Compound Interest
• One of the applications of the exponential function is
compound interest. The formula is given by:
 r
A ( t )  P 1  
 n
nt
t = time in years
A(t) = amount after t years
P = Principal
r = rate (as a decimal)
n = number of compoundings in a year
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Applications of Exponential Function - Compound Interest
• Example 1:
Find the amount in an account after 4 years, if the initial
investment was $5000, at a rate of 4.5%, compounded
monthly.
t = 4, P = 5000, r = .045, and n =12
12  4
 .045 
A(4)  50001 

12 

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Applications of Exponential Function - Compound Interest
• Entering the expression into a calculator ...
yields an amount of $5984.07 in the account after
4 years.
12  4
 .045 
A(4)  50001 

12 

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Applications of Exponential Function - Compound Interest
• Example 2:
Find the rate of an account with an initial investment of
$10,000 that will yield approximately $22,477.37 after
12 years, compounded daily.
t = 12, A(12) = 22477.37, P = 10,000, and n = 360.
(Banking uses 360 rather than 365.)
r 

A(12)  100001 

 360 
360  12
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 22477.37
Applications of Exponential Function - Compound Interest
• Divide by 10000 ...
r 

1 

 360 
360  12
22477.37

 2.247737
10000
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Applications of Exponential Function - Compound Interest
• Raise both sides to a power of the reciprocal of the
exponent ...
r 

1 
  2.247737
 360 
1
36012
• Note: by pressing the 
symbol, the calculator
automatically supplies
ANS^.
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 1.0001875
Applications of Exponential Function - Compound Interest
• Subtract 1 and multiply by 360 to find ...
r  360(0.0001875)  .0675  6.75%
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Applications of Exponential Function - Compound Interest
• In the formula for compound interest, the letter n
represents the number of compounding periods in one
year. As the number of compounding periods increase
without bound, it is said that the compounding is
“continuous.” The formula for continuous compounding
is given by ...
A( t )  Pe
rt
where e = 2.71828 ...
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Applications of Exponential Function - Compound Interest
• Example 3:
Find the principal that must be invested now at 5%
with continuous compounding in order to have $10,000
in 7 years.
A(7)  Pe
0.05 7
 10000
10000
P  0.057  7046.88
e
• The amount to be invested
is $7046.88
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Applications of Exponential Function - Compound Interest
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