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Table of Contents
I.
Compound Interest Formula
II. Continuous Compound Interest Formula
III. Population Growth Formula
IV. Doubling-Time Growth Formula
V. Exponential Decay
VI. Half-Life Decay Formula
VII. Newton’s Law of Cooling
r
A  P 1  n 
nt
r
A  P 1  n 
nt
r
A  P 1  n 
P
nt
is the Principle invested
The amount of money invested in a Savings Institution
The value of an object:
House
New Car
Piece of Property
Bicycle
r
A  P 1  n 
A
nt
is the
total value of the investment
after t years
r
A  P 1  n 
t
nt
is the
number of years the Principle
has been invested
r
A  P 1  n 
r
nt
is the
Rate of interest
Expressed as a decimal
r
A  P 1  n 
nt
n
is the
Number of times the interest is
Compounded per year
“Compounded annually”
n=1
r
A  P 1  n 
nt
n
is the
Number of times the interest is
Compounded per year
“Compounded semi-annually”
n=2
r
A  P 1  n 
nt
n
is the
Number of times the interest is
Compounded per year
“Compounded quarterly”
n=4
r
A  P 1  n 
nt
n
is the
Number of times the interest is
Compounded per year
“Compounded monthly”
n = 12
r
A  P 1  n 
nt
One thousand dollars is invested at
12% interest
compounded annually.
Determine how much the investment is worth
after 2 years.
r
A  P 1  n 
nt
One thousand dollars is invested at
12% interest
compounded annually.
Determine how much the investment is worth
after 2 years.
r
A  P 1  n 
nt
One thousand dollars is invested at
12% interest
compounded annually.
Determine how much the investment is worth
after 2 years.
P  1,000
r
A  P 1  n 
nt
One thousand dollars is invested at
12% interest
compounded annually.
Determine how much the investment is worth
after 2 years.
P  1,000
r
A  P 1  n 
nt
One thousand dollars is invested at
12% interest
compounded annually.
Determine how much the investment is worth
after 2 years.
P  1,000
r  0.12
r
A  P 1  n 
nt
One thousand dollars is invested at
12% interest
compounded annually.
Determine how much the investment is worth
after 2 years.
P  1,000
r  0.12
r
A  P 1  n 
nt
One thousand dollars is invested at
12% interest
compounded annually.
Determine how much the investment is worth
after 2 years.
P  1,000
r  0.12
n 1
r
A  P 1  n 
nt
One thousand dollars is invested at
12% interest
compounded annually.
Determine how much the investment is worth
after 2 years.
P  1,000
r  0.12
n 1
r
A  P 1  n 
nt
A  the unknown
One thousand dollars is invested at
12% interest
compounded annually.
Determine how much the investment is worth
after 2 years.
P  1,000
r  0.12
n 1
r
A  P 1  n 
nt
A  the unknown
One thousand dollars is invested at
12% interest
compounded annually.
Determine how much the investment is worth
after 2 years.
P  1,000
r  0.12
n 1
r
A  P 1  n 
nt
A  the unknown
One thousand dollars is invested at
12% interest
compounded annually.
Determine how much the investment is worth
after 2 years.
P  1,000
r  0.12
n 1
t2
r
A  P 1  n 
nt
A  the unknown
P  1,000
r  0.12
n 1
t2
r
A  P 1  n 
nt
A  the unknown

0.12
A  1,000 1 
1
P  1,000
r  0.12
n 1

12
t2

0.12
A  1,000 1 
1

12
A  1,000 1  0.12 
A  1,000 1.12 
2
2

0.12
A  1,000 1 
1

12
A  1,000 1  0.12 
A  1,000 1.12 
A  1254.4
2
2
A  1254.4
One thousand dollars is invested at
12% interest
compounded annually.
Determine how much the investment is worth
after 2 years.
The investment is worth $1,254.40
r
A  P 1  n 
nt
The value of a new $500 television
Decreases 10% per year.
Find its value after
5 years.
r
A  P 1  n 
nt
The value of a new $500 television
Decreases 10% per year.
Find its value after
5 years.
P?
r ?
n?
t ?
A?
r
A  P 1  n 
nt
The value of a new $500 television
Decreases 10% per year.
Find its value after
5 years.
P?
r ?
n?
t ?
A?
r
A  P 1  n 
nt
The value of a new $500 television
Decreases 10% per year.
Find its value after
5 years.
P  500
r ?
r
A  P 1  n 
nt
The value of a new $500 television
Decreases 10% per year.
Find its value after
5 years.
P  500
r   0.10
n?
r
A  P 1  n 
nt
The value of a new $500 television
Decreases 10% per year.
Find its value after
5 years.
P  500
r   0.10
n 1
t ?
r
A  P 1  n 
nt
A  the unknown
The value of a new $500 television
Decreases 10% per year.
Find its value after
5 years.
P  500
r   0.10
n 1
t 5
r
A  P 1  n 
nt
A  the unknown
P  500
r   0.10
n 1
t 5
r
A  P 1  n 
nt
A  the unknown


0.10
A  500 1 
1
P  500
r   0.10
n 1

15

t 5


0.10
A  500 1 
1
A  500 1  0.10 
5
A  500  0.90 
5

15



0.10
A  500 1 
1

15

A  500 1  0.10 
5
A  500  0.90 
5
A  295.25
The television is worth $295.25
r
A  P 1  n 
nt
One hundred dollars is invested at
7.2% interest
compounded quarterly.
How long will it take for the
investment to double?
r
A  P 1  n 
nt
One hundred dollars is invested at
7.2% interest
compounded quarterly.
How long will it take for the
investment to double?
P?
r ?
n?
t ?
A?
r
A  P 1  n 
nt
One hundred dollars is invested at
7.2% interest
compounded quarterly.
How long will it take for the
investment to double?
P?
r ?
n?
t ?
A?
r
A  P 1  n 
nt
One hundred dollars is invested at
7.2% interest
compounded quarterly.
How long will it take for the
investment to double?
P  100
r ?
n?
t ?
A?
r
A  P 1  n 
nt
One hundred dollars is invested at
7.2% interest
compounded quarterly.
How long will it take for the
investment to double?
P  100
r  .072
n?
t ?
A?
r
A  P 1  n 
nt
One hundred dollars is invested at
7.2% interest
compounded quarterly.
How long will it take for the
investment to double?
P  100
r  .072
n4
t ?
A?
r
A  P 1  n 
nt
One hundred dollars is invested at
7.2% interest
compounded quarterly.
How long will it take for the
investment to double?
P  100
r  .072
n4
t  unknown A  ?
r
A  P 1  n 
nt
One hundred dollars is invested at
7.2% interest
compounded quarterly.
How long will it take for the
investment to double?
P  100
r  .072
n4
t  unknown A  200
r
A  P 1  n 
P  100
r  .072
n4
nt
t  unknown A  200
r
A  P 1  n 

nt
.072
200  100 1 
4
P  100
r  .072
n4

4t
t  unknown A  200

.072
200  100 1 
4

200  100 1  .018
200  100 1.018 
4t
4t
4t
200  100 1.018 
4t
200  100 1.018 
4t
200  100 1.018 
100
4t
100
2  1.018
4t
log 2  log 1.018
4t
log 2  4t log 1.018 
log 2  4t log 1.018 
4log 1.018 4log 1.018
log 2
t
4log 1.018 
log 2  4t log 1.018 
4log 1.018 4log 1.018
log 2
t
4log 1.018 
9.7  t
It takes approximately 10 years
Nearest
Year
Link
Back to
Table of
Contents
A  Pe
rt
Where did this formula come from?
r
A  P 1  n 
nt
If Compounded Continually
Then how many times per year?
Then
n is approaching infinity
r
A  P 1  n 
n
Let x 
r
nt
n approaches infinity
r
remains constant
x approaches infinity
r
A  P 1  n 
n
If
r  x  r
r
nt
n  xr
r
A  P 1  n 
n
If
r  x  r
r
xrt
n  xr
r
A  P 1  n 
n
If x 
r
xrt
1 r

x n

1
A  P 1 x
n
If x 
r

xrt
1 r

x n

1
A  P 1 x

xrt
 
1
1

 x
1
A  P 1 x
xrt
x
x approaches infinity

1
1 x

x

1
Y1  1  x
nd
2
TABLE
x approaches infinity

x

1
1 x
x

1  1x


x
x approaches infinity
4
8
12
100
365
10,000
100,000
1,000,000
2.4414
2.5658
2.6130
2.7048
2.7146
2.7182
2.7183
2.7183
x

1
1 x
x

1  1x
4

x
2.4414
8

x
12
x approaches infinity
100
365
10,000
100,000
1,000,000

1
1 x
x

1  1x

4
8
2.4414
2.5658
x

x
12
x approaches infinity
100
365
10,000
100,000
1,000,000

1
1 x
x

1  1x


x
4
8
12
2.4414
2.5658
2.6130
x
x approaches infinity
100
365
10,000
100,000
1,000,000

1
1 x
x

1  1x


x
x approaches infinity
4
8
12
100
2.4414
2.5658
2.6130
2.7048
x
365
10,000
100,000
1,000,000

1
1 x
x

1  1x


x
x approaches infinity
4
8
12
100
365
2.4414
2.5658
2.6130
2.7048
2.7146
x
10,000
100,000
1,000,000

1
1 x
x

1  1x


x
x approaches infinity
4
8
12
100
365
10,000
2.4414
2.5658
2.6130
2.7048
2.7146
2.7182
x
100,000
1,000,000

1
1 x
x

1  1x


x
x approaches infinity
4
8
12
100
365
10,000
100,000
2.4414
2.5658
2.6130
2.7048
2.7146
2.7182
2.7183
x
1,000,000

1
1 x
x

1  1x


x
x approaches infinity
4
8
12
100
365
10,000
100,000
1,000,000
2.4414
2.5658
2.6130
2.7048
2.7146
2.7182
2.7183
2.7183
x
e




e

1
A  P 1 x
x approaches infinity
1
1 x
xrt
x




e

1
A  P 1 ex
x approaches infinity
1
1 x
xrt
x
A  Pe
rt
Continuous Compound
Interest Formula
A  Pe
rt
Mrs. Johnson received a bonus equivalent to
10% of her yearly salary
and has decided to deposit it in a savings account
in which interest is compounded continuously.
Her salary is $38,500 per year and
the account pays 7.5% interest.
How long will it take for her deposit to double in value?
A  Pe
rt
Mrs. Johnson received a bonus equivalent to
10% of her yearly salary
and has decided to deposit it in a savings account
in which interest is compounded continuously.
Her salary is $38,500 per year and
the account pays 7.5% interest.
How long will it take for her deposit to double in value?
P?
r ?
t ?
A?
A  Pe
rt
Mrs. Johnson received a bonus equivalent to
10% of her yearly salary
and has decided to deposit it in a savings account
in which interest is compounded continuously.
Her salary is $38,500 per year and
the account pays 7.5% interest.
How long will it take for her deposit to double in value?
P?
r ?
t ?
A?
A  Pe
rt
Mrs. Johnson received a bonus equivalent to
10% of her yearly salary
and has decided to deposit it in a savings account
in which interest is compounded continuously.
Her salary is $38,500 per year and
the account pays 7.5% interest.
How long will it take for her deposit to double in value?
P  3850
r ?
t ?
A?
A  Pe
rt
Mrs. Johnson received a bonus equivalent to
10% of her yearly salary
and has decided to deposit it in a savings account
in which interest is compounded continuously.
Her salary is $38,500 per year and
the account pays 7.5% interest.
How long will it take for her deposit to double in value?
P  3850 r  0.075
t ?
A?
A  Pe
rt
Mrs. Johnson received a bonus equivalent to
10% of her yearly salary
and has decided to deposit it in a savings account
in which interest is compounded continuously.
Her salary is $38,500 per year and
the account pays 7.5% interest.
How long will it take for her deposit to double in value?
P  3850 r  0.075 t  unknown
A?
A  Pe
rt
Mrs. Johnson received a bonus equivalent to
10% of her yearly salary
and has decided to deposit it in a savings account
in which interest is compounded continuously.
Her salary is $38,500 per year and
the account pays 7.5% interest.
How long will it take for her deposit to double in value?
P  3850 r  0.075 t  unknown A  7700
A  Pe
rt
7700  3850e
0.075t
P  3850 r  0.075 t  unknown A  7700
7700  3850e
0.075t
7700  3850e
0.075t
3850
3850
2e
0.075t
ln 2  ln e
0.075t
ln 2  0.075t ln e
ln 2  0.075t ln e ln e  ?
ln 2  0.075t
ln 2
t
0.075
To the
nearest
year
ln 2  0.075t ln e
ln 2  0.075t
ln 2
t
0.075
9t
To the nearest year
A  Pe
rt
Following the birth of child, a parent wants to make an
initial investment of $2000 and would like it to grow to a
value of $10,000 by the Childs 20th birthday.
What annual interest rate compounded continually
should the parent look for?
A  Pe
rt
Following the birth of child, a parent wants to make an
initial investment of $2000 and would like it to grow to a
value of $10,000 by the Childs 20th birthday.
What annual interest rate compounded continually
should the parent look for?
P?
r ?
t ?
A?
A  Pe
rt
Following the birth of child, a parent wants to make an
initial investment of $2000 and would like it to grow to a
value of $10,000 by the Childs 20th birthday.
What annual interest rate compounded continually
should the parent look for?
P  2000
A  Pe
rt
Following the birth of child, a parent wants to make an
initial investment of $2000 and would like it to grow to a
value of $10,000 by the Childs 20th birthday.
What annual interest rate compounded continually
should the parent look for?
P  2000
r  unknown
A  Pe
rt
Following the birth of child, a parent wants to make an
initial investment of $2000 and would like it to grow to a
value of $10,000 by the Childs 20th birthday.
What annual interest rate compounded continually
should the parent look for?
P  2000
r  unknown t  20
A  Pe
rt
Following the birth of child, a parent wants to make an
initial investment of $2000 and would like it to grow to a
value of $10,000 by the Childs 20th birthday.
What annual interest rate compounded continually
should the parent look for?
P  2000
r  unknown t  20
A  10,000
A  Pe
rt
Following the birth of child, a parent wants to make an
initial investment of $2000 and would like it to grow to a
value of $10,000 by the Childs 20th birthday.
What annual interest rate compounded continually
should the parent look for?
P  2000
r  unknown t  20
A  10,000
A  Pe
rt
10,000  2000 e
P  2000
r  unknown t  20
r  20
A  10,000
10,000  2000 e
20 r
10,000  2000 e
2000
2000
5e
20 r
ln 5  ln e
20 r
20 r
ln5  ln e
20 r
ln5  20r lne ln e  ?
ln5  20r
ln5  20r
ln 5
r
20
Nearest hundredth
of a percent
ln5  20r
ln 5
 r = 0.0805
20
8.05%
How long will it take to double
and investment at 4.5% interest
How long will it take to double
and investment at 4.5% interest
A  Pe
P?
r ?
rt
t ?
A?
How long will it take to double
and investment at 4.5% interest
A  Pe
P?
rt
How long will it take to double
and investment at 4.5% interest
A  Pe
PP
r ?
rt
How long will it take to double
and investment at 4.5% interest
A  Pe
PP
r  0.045
rt
t ?
How long will it take to double
and investment at 4.5% interest
A  Pe
PP
r  0.045
rt
t  unknown
A?
How long will it take to double
and investment at 4.5% interest
A  Pe
PP
r  0.045
rt
t  unknown A  2P
How long will it take to double
and investment at 4.5% interest
A  Pe
rt
2P  Pe
0.045t
PP
r  0.045
t  unknown A  2P
How long will it take to double
and investment at 4.5% interest
2P  Pe
0.045t
Solve for t to the nearest hundredth
2P  Pe
0.045t
P
P
2e
0.045t
ln 2  ln e
0.045t
ln 2  ln e
0.045t
ln e  ?
ln 2  0.045t ln e
ln 2  0.045t
ln 2  0.045t
ln 2
0.045
t
How long will it take to double
and investment at 4.5% interest
ln 2
t
0.045
Solve for t to the nearest hundredth
t  15.40 years
What interest rate should an
investor seek to double his money
in 20 years?
You solve for r to the nearest tenth of a percent
A  Pe
rt
What interest rate should an
investor seek to double his money
in 20 years?
You solve for r to the nearest tenth of a percent
ln 2
r
20
r  0.035
3.5%
Will invests $2000 in a bond trust that pays 9%
interest compounded semiannually.
His friend Henry invests $2000 in a Certificate of
Deposit that pays 8 ½ % compounded
continuously.
Who will have more money after 20 years,
Will or Henry? How much more money?
Will invests $2000 in a bond trust that pays 9%
interest compounded semiannually.

.09
A  2000 1 
2

220
 11,632.73
His friend Henry invests $2000 in a Certificate of
Deposit that pays 8 ½ % compounded
continuously.
.08520
A  2000e
 10,947.89
Who will have more money after 20 years,
Will or Henry? How much more money?
$684.84
At the age of 25 Coris invests $2000
in an Individual Retirement Account (IRA)
that is allowed to accumulate interest
tax-free until she retires.
Pat, who is 35, also invests $2000 in an IRA.
Pat and Coris each earn 8% annually
compounded continuously,
and each withdraws the funds from the account
at age 65.
To the nearest hundred, how much more money
does Coris collect than Pat?
At the age of 25 Coris invests $2000
in an Individual Retirement Account (IRA)
that is allowed to accumulate interest
tax-free until she retires.
Pat, who is 35, also invests $2000 in an IRA.
Pat and Coris each earn 8% annually
compounded continuously,
and each withdraws the funds from the account
at age 65.
To the nearest hundred, how much more money
does Coris collect than Pat?
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$27,000
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P  P0e
rt
P  P0e
P0
rt
is the size of the original
population
P  P0e
rt
P is the size of the population
after t years
P  P0e
rt
r annual growth rate
of the population
P  P0e
rt
t is the number of years
The population in Raleigh-Durham, North Carolina,
grew from 560,774 in 1980 to 665,400 in 1987
a) Using these statistics, find the average growth rate of
the population in Raleigh-Durham (to the nearest tenth
of a percent) and determine an equation of the
population growth curve for this region.
b) Assuming a constant growth rate, predict the
population for Raleigh-Durham
in 1995 to the nearest unit.
The population in Raleigh-Durham, North Carolina,
grew from 560,774 in 1980 to 665,400 in 1987
a) Using these statistics, find the average growth rate of
the population in Raleigh-Durham (to the nearest tenth
of a percent) and determine an equation of the
population growth curve for this region.
b) Assuming a constant growth rate, predict the
population for Raleigh-Durham
in 1995 to the nearest unit.
P  P0e
rt
Find
r
P  P0e
rt
The population in Raleigh-Durham, North Carolina,
grew from 560,774 in 1980 to 665,400 in 1987
P0 
r
P
t
P  P0e
rt
The population in Raleigh-Durham, North Carolina,
grew from 560,774 in 1980 to 665,400 in 1987
P0 
r
P
t
P  P0e
rt
The population in Raleigh-Durham, North Carolina,
grew from 560,774 in 1980 to 665,400 in 1987
P0 
P
560,774
r
t
P  P0e
rt
The population in Raleigh-Durham, North Carolina,
grew from 560,774 in 1980 to 665,400 in 1987
P0 
560,774
r
P
665,400
t
P  P0e
rt
The population in Raleigh-Durham, North Carolina,
grew from 560,774 in 1980 to 665,400 in 1987
P0 
560,774
r
P
665,400
t
unknown
P  P0e
rt
The population in Raleigh-Durham, North Carolina,
grew from 560,774 in 1980 to 665,400 in 1987
P0 
560,774
r
unknown
P
665,400
t
7
P  P0e
rt
Find the population growth rate for Raleigh-Durham,
North Carolina to the nearest tenth of a percent
P0 
560,774
r
unknown
P
665,400
t
7
P  P0e
rt
r 7
665,400  560, 774e
r  unknown
P0  560,774
P
665,400
t
7
P  P0e
rt
665,400  560, 774e
665, 400
7r
e
560,774
r 7
665, 400
7r
e
560,774
ln
ln
665, 400
560, 774
665, 400
560, 774
 ln e
7r
 7r ln e
ln
ln
665, 400
560, 774
665, 400
560, 774
665, 400
ln
560, 774
7
 7r ln e
 7r
r
665, 400
ln
560, 774
7
r
M
log b
 log b M  log b N
N
ln 665,400  ln 560,774
r
7
ln 665,400  ln 560,774
r
7
ln 665,400  ln 560,774
r
7
ln 665,400  ln 560,774
r
7
r  0.024
r  2.4%
Nearest tenth
of a percent
P  P0e
rt
We need to find an equation that MODELS
population growth for
Raleigh-Durham, North Carolina
r  0.024
the
P  560,774e
0.024 t
This is the equation that MODELS the population
growth in Raleigh-Durham, North Carolina
r  0.024
P  560,774e
0.024 t
Use this equation to project the population in
Raleigh-Durham, North Carolina in 1995
to the nearest unit
P  560,774e
0.024 t
Use this equation to project the population in
Raleigh-Durham, North Carolina in 1995
to the nearest unit
To find P we need to know
the value of
? t
P  560,774e
0.024 t
Use this equation to project the population in
Raleigh-Durham, North Carolina in 1995
to the nearest unit
The population in Raleigh-Durham, North Carolina,
grew from 560,774 in 1980 to 665,400 in 1987
t = 15
P  560,774e
0.024 t
Use this equation to project the population in
Raleigh-Durham, North Carolina in 1995
to the nearest unit
0.02415
P  560,774e
P  560,774e
0.024 t
Use this equation to project the population in
Raleigh-Durham, North Carolina in 1995
to the nearest unit
P  803,774
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A  A0  2
t
d
A  A0  2
A0
t
d
is the size of the
Original Population
Bacteria
Culture of Yeast
ETC.
A  A0  2
t
d
A
is the size of the
Population after t time
Years
Weeks
Minutes
Months
Hours
ETC.
Seconds
A  A0  2
t
d
d
is the amount of
time it takes for the population
to double
Years
Weeks
Minutes
Months
Hours
ETC.
Seconds
A  A0  2
t
t
d
is the time period
Years
Weeks
Minutes
Months
Hours
ETC.
Seconds
A  A0  2
t
d
A certain bacteria population
doubles in size every 12 hours.
By how much will it grow
in 2 days?
A  A0  2
t
d
A certain bacteria population
doubles in size every 12 hours.
By how much will it grow
in 2 days?
A?
A0  ?
d ?
t ?
A  A0  2
t
d
A certain bacteria population
doubles in size every 12 hours.
By how much will it grow
in 2 days?
A A
A0  ?
d ?
t ?
A  A0  2
t
d
A certain bacteria population
doubles in size every 12 hours.
By how much will it grow
in 2 days?
A A
A0  A0
d ?
t ?
A  A0  2
t
d
A certain bacteria population
doubles in size every 12 hours.
By how much will it grow
in 2 days?
A A
A0  A0 d  12 hrs
t ?
A  A0  2
t
d
A certain bacteria population
doubles in size every 12 hours.
By how much will it grow
in 2 days?
A A
A0  A0 d  12 hrs t  2 days
A  A0  2
t
d
A certain bacteria population
doubles in size every 12 hours.
By how much will it grow
in 2 days?
A A
A0  A0 d  12 hrs t  48 hrs
A  A0  2
A A
t
d
A0  A0 d  12 hrs t  48 hrs
A  A0  2
A  A0  2
A A
t
d
48
12
A0  A0 d  12 hrs t  48 hrs
A  A0  2
A  A0  2
A A
t
d
48
12
A0  A0 d  12 hrs t  48 hrs
A  A0  2
48
12
A  A0  2
4
A  A0  16
A  A0  16
A  16  A0
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The population is 16 times greater
than the size of the original population
The population grows by a factor of 16
in 2 days
y= b
x
y= 2
x
y= b
x
x
æ1 ö
y = çç ÷
÷
÷
çè 2 ø
y= 2
x
y= b
x
x
æ1 ö
y = çç ÷
÷
÷
çè 2 ø
x
y= 3
x
æ1 ÷
ö
y = çç ÷
çè3÷
ø
y= 2
x
y= b
x
x
æ1 ö
y = çç ÷
÷
÷
çè 2 ø
x
y= 3
x
b> 1
æ1 ÷
ö
y = çç ÷
çè3÷
ø
y= 2
x
y= b
x
x
æ1 ö
y = çç ÷
÷
÷
çè 2 ø
x
y= 3
x
b> 1
æ1 ÷
ö
y = çç ÷
çè3÷
ø
0< b< 1
y= 2
x
y= b
x
x
æ1 ö
x
÷
ç
yy ==ççè22ø÷÷
x
x
æ1 ÷
ö- x
yy ==çççè33÷÷ø
x
- x
y= 3
y= b
b> 1
y= b
0< b< 1
y = ab
x
a is a constant
y = ab
x
- x
y = ab
Determine whether each equation represents an
exponential growth or an exponential decay curve.
You may want check your answers with your graphing calculator
x
1) y = 5(2)
- 0.2 t
2) y = 5(2)
3) y = 100e
0.2t
- 0.31t
4) y = 100 - 100e
Determine whether each equation represents an
exponential growth or an exponential decay curve.
You may want check your answers with your graphing calculator
x
1) y = 5(2)
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- 0.2 t
2) y = 5(2)
3) y = 100e
0.2t
- 0.31t
4) y = 100 - 100e
A
 
1
A0 2
t
h
A
 
1
A0 2
A  A0  2
t
h

t
1 h
A  A0  2 
t
h
A  A0  2 
t
h
A0 is the original quantity of the
radioactive substance
(isotope)
A  A0  2 
t
h
A is the amount of radioactive
substance
after t years
A  A0  2 
t
h
h is the half-life
of the radioactive substance
A  A0  2 
t
h
t is the number of years
A  A0  2 
t
h
The half life of
radioactive radium (226Ra)
is 1599 years.
Given an initial quantity of
10 grams, how much
will remain after 1000 years.
A  A0  2 
t
h
The half life of radioactive radium (226Ra)
is 1599 years. Given an initial quantity of
10 grams, how much will remain after 1000 years.
A0 
t
A
h
A  A0  2 
t
h
The half life of radioactive radium (226Ra)
is 1599 years. Given an initial quantity of
10 grams, how much will remain after 1000 years.
A0  10
t  1000
A
h  1599
Unknown
A  10  2 
1000
 1599
A  10  2 
A  6.48
1000
 1599
grams
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T f  Tr  To  Tr  e
T0
 rt
is the Original temperature
of the object
T f  Tr  To  Tr  e
Tr
 rt
is the temperature of the
surrounding air
T f  Tr  To  Tr  e
Tf
 rt
is the final temperature of
the object after t minutes
T f  Tr  To  Tr  e
r
is the rate at which
the object is cooling
 rt
T f  Tr  To  Tr  e
 rt
A chef wants the soup to be served to the customer
at a temperature of no less than 160º F.
It has been determined that the cooling rate r for this
soup is 0.21ºF per minute.
When the soup is removed from the pot, it is 212ºF.
If the room temperature in the restaurant is 68ºF,
determine in how many minutes the soup must be
served to meet the chef’s request.
To the nearest minute
T f  Tr  To  Tr  e
 rt
160  68   212  68  e
0.21t
A chef wants the soup to be served to the customer at a
temperature of no less than 160º F.
It has been determined that the cooling rate r for this soup is
0.21ºF per minute.
When the soup is removed from the pot, it is 212ºF.
If the room temperature in the restaurant is 68ºF, determine
in how many minutes the soup must be served to meet the
chef’s request.
T f  Tr  To  Tr  e
 rt
160  68   212  68  e
0.21t
A chef wants the soup to be served to the customer at a
temperature of no less than 160º F.
It has been determined that the cooling rate r for this soup is
0.21ºF per minute.
When the soup is removed from the pot, it is 212ºF.
If the room temperature in the restaurant is 68ºF, determine
in how many minutes the soup must be served to meet the
chef’s request.
T f  Tr  To  Tr  e
 rt
160  68   212  68  e
0.21t
A chef wants the soup to be served to the customer at a
temperature of no less than 160º F.
It has been determined that the cooling rate r for this soup is
0.21ºF per minute.
When the soup is removed from the pot, it is 212ºF.
If the room temperature in the restaurant is 68ºF, determine
in how many minutes the soup must be served to meet the
chef’s request.
T f  Tr  To  Tr  e
 rt
160  68   212  68  e
0.21t
A chef wants the soup to be served to the customer at a
temperature of no less than 160º F.
It has been determined that the cooling rate r for this soup is
0.21ºF per minute.
When the soup is removed from the pot, it is 212ºF.
If the room temperature in the restaurant is 68ºF, determine
in how many minutes the soup must be served to meet the
chef’s request.
T f  Tr  To  Tr  e
 rt
160  68   212  68  e
0.21t
A chef wants the soup to be served to the customer at a
temperature of no less than 160º F.
It has been determined that the cooling rate r for this soup is
0.21ºF per minute.
When the soup is removed from the pot, it is 212ºF.
If the room temperature in the restaurant is 68ºF, determine
in how many minutes the soup must be served to meet the
chef’s request.
160  68   212  68  e
160  68  144  e
68
68
92  144 e
144
0.21t
144
92
0.21t
e
144
0.21t
0.21t
92
0.21t
e
144
92
0.21t
ln
 ln e
144
Law #2
Law #3
ln92  ln144   0.21t ln e
Law #2
Law #3
ln92  ln144   0.21t ln e
ln 92  ln144
 t ln e
0.21
ln 92  ln144
t
0.21
To the nearest minute
Law #2
Law #3
ln92  ln144   0.21t ln e
ln 92  ln144
 t ln e
0.21
ln 92  ln144
t
0.21
To the nearest minute
2 min
T f  Tr  To  Tr  e
Given that the original temperature
of the coffee is 155ºF
and the room temperature is 75ºF,
determine after how many minutes
the coffee will be 110ºF
To the nearest hundredth of a minute
 rt
T f  Tr  To  Tr  e
 rt
Given that the original temperature
of the coffee is 155ºF
and the room temperature is 75ºF,
determine after how many minutes
the coffee will be 110ºF
To the nearest hundredth of a minute
2.75 min
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