2. Applications of Log and Exponential Formulas
Recall the following compound interest formulas.
Pina (periodic compounding)
and Perta (continuous compounding)
3. Applications of Log and Exponential Formulas
Recall the following compound interest formulas.
Pina (periodic compounding)
P = principal, i = periodic rate,
N = total number of periods, A = accumulated value
and Perta (continuous compounding)
4. Applications of Log and Exponential Formulas
Recall the following compound interest formulas.
Pina (periodic compounding)
P = principal, i = periodic rate,
N = total number of periods, A = accumulated value
then P(1 + i )N = A
and Perta (continuous compounding)
5. Applications of Log and Exponential Formulas
Recall the following compound interest formulas.
Pina (periodic compounding)
P = principal, i = periodic rate,
N = total number of periods, A = accumulated value
then P(1 + i )N = A
and Perta (continuous compounding)
P = principal, r = yearly or annual rate,
t = total number of years, A = accumulated value
6. Applications of Log and Exponential Formulas
Recall the following compound interest formulas.
Pina (periodic compounding)
P = principal, i = periodic rate,
N = total number of periods, A = accumulated value
then P(1 + i )N = A
and Perta (continuous compounding)
P = principal, r = yearly or annual rate,
t = total number of years, A = accumulated value
then Pert = A
7. Applications of Log and Exponential Formulas
Recall the following compound interest formulas.
Pina (periodic compounding)
P = principal, i = periodic rate,
N = total number of periods, A = accumulated value
then P(1 + i )N = A
and Perta (continuous compounding)
P = principal, r = yearly or annual rate,
t = total number of years, A = accumulated value
then Pert = A
The Pina formula has the base (1+ i) which varies
with i.
8. Applications of Log and Exponential Formulas
Recall the following compound interest formulas.
Pina (periodic compounding)
P = principal, i = periodic rate,
N = total number of periods, A = accumulated value
then P(1 + i )N = A
and Perta (continuous compounding)
P = principal, r = yearly or annual rate,
t = total number of years, A = accumulated value
then Pert = A
The Pina formula has the base (1+ i) which varies
with i. All the discussions below are related to
continuous–compounding growth which utilizes the
Perta formula with the fixed natural base e.
9. Applications of Log and Exponential Formulas
Example A. a. An investment gives 4% annual return
compounded continuously. How long would it take
for an initial investment of $500 to reach $1,000?
10. Applications of Log and Exponential Formulas
Example A. a. An investment gives 4% annual return
compounded continuously. How long would it take
for an initial investment of $500 to reach $1,000?
We are to find t using the Perta formula with
P = $500, A= $1,000, and r = 4% = 0.04.
11. Applications of Log and Exponential Formulas
Example A. a. An investment gives 4% annual return
compounded continuously. How long would it take
for an initial investment of $500 to reach $1,000?
We are to find t using the Perta formula with
P = $500, A= $1,000, and r = 4% = 0.04. So
500e0.04t = 1000
12. Applications of Log and Exponential Formulas
To find the r or t in the formula Pert = A, divide both
sides by P.
Example A. a. An investment gives 4% annual return
compounded continuously. How long would it take
for an initial investment of $500 to reach $1,000?
We are to find t using the Perta formula with
P = $500, A= $1,000, and r = 4% = 0.04. So
500e0.04t = 1000
13. Applications of Log and Exponential Formulas
To find the r or t in the formula Pert = A, divide both
sides by P.
Example A. a. An investment gives 4% annual return
compounded continuously. How long would it take
for an initial investment of $500 to reach $1,000?
We are to find t using the Perta formula with
P = $500, A= $1,000, and r = 4% = 0.04. So
500e0.04t = 1000
e0.04t = 2 (=1000/500)
14. Applications of Log and Exponential Formulas
To find the r or t in the formula Pert = A, divide both
sides by P. We get ert = A so that r*t = ln( A ).
P P
Example A. a. An investment gives 4% annual return
compounded continuously. How long would it take
for an initial investment of $500 to reach $1,000?
We are to find t using the Perta formula with
P = $500, A= $1,000, and r = 4% = 0.04. So
500e0.04t = 1000
e0.04t = 2 (=1000/500)
15. Applications of Log and Exponential Formulas
To find the r or t in the formula Pert = A, divide both
sides by P. We get ert = A so that r*t = ln( A ).
P P
Example A. a. An investment gives 4% annual return
compounded continuously. How long would it take
for an initial investment of $500 to reach $1,000?
We are to find t using the Perta formula with
P = $500, A= $1,000, and r = 4% = 0.04. So
500e0.04t = 1000
e0.04t = 2 (=1000/500) in the log–form
0.04t = ln(2)
16. Applications of Log and Exponential Formulas
To find the r or t in the formula Pert = A, divide both
sides by P. We get ert = A so that r*t = ln( A ).
P P
Example A. a. An investment gives 4% annual return
compounded continuously. How long would it take
for an initial investment of $500 to reach $1,000?
We are to find t using the Perta formula with
P = $500, A= $1,000, and r = 4% = 0.04. So
500e0.04t = 1000
e0.04t = 2 (=1000/500) in the log–form
0.04t = ln(2) or
t = ln(2)/0.04 ≈ 17.5 year
17. Applications of Log and Exponential Formulas
To find the r or t in the formula Pert = A, divide both
sides by P. We get ert = A so that r*t = ln( A ).
P P
Example A. a. An investment gives 4% annual return
compounded continuously. How long would it take
for an initial investment of $500 to reach $1,000?
We are to find t using the Perta formula with
P = $500, A= $1,000, and r = 4% = 0.04. So
500e0.04t = 1000
e0.04t = 2 (=1000/500) in the log–form
0.04t = ln(2) or
t = ln(2)/0.04 ≈ 17.5 year
So it would take roughly than 17½ years for the $500
investment to reach $1,000.
18. Applications of Log and Exponential Formulas
b. As before with r = 4%, does it take more time for
an investment of $500 to become $1,000
than for an investment of $1 to become $2?
19. Applications of Log and Exponential Formulas
b. As before with r = 4%, does it take more time for
an investment of $500 to become $1,000
than for an investment of $1 to become $2?
No. By viewing the $500 investment as 500 separate
$1 accounts, we see that the time needed for
$500 to grow into $1,000 is the same the time needed
for $1 to grow into $2
20. Applications of Log and Exponential Formulas
b. As before with r = 4%, does it take more time for
an investment of $500 to become $1,000
than for an investment of $1 to become $2?
No. By viewing the $500 investment as 500 separate
$1 accounts, we see that the time needed for
$500 to grow into $1,000 is the same the time needed
for $1 to grow into $2 which is ln(2)/0.04 ≈ 17.5 years.
21. Applications of Log and Exponential Formulas
b. As before with r = 4%, does it take more time for
an investment of $500 to become $1,000
than for an investment of $1 to become $2?
No. By viewing the $500 investment as 500 separate
$1 accounts, we see that the time needed for
$500 to grow into $1,000 is the same the time needed
for $1 to grow into $2 which is ln(2)/0.04 ≈ 17.5 years.
Doubling Time and the Rule of 72
22. Applications of Log and Exponential Formulas
b. As before with r = 4%, does it take more time for
an investment of $500 to become $1,000
than for an investment of $1 to become $2?
No. By viewing the $500 investment as 500 separate
$1 accounts, we see that the time needed for
$500 to grow into $1,000 is the same the time needed
for $1 to grow into $2 which is ln(2)/0.04 ≈ 17.5 years.
Doubling Time and the Rule of 72
From the above reasoning we see that given a fixed
growth rate r, the amount of time needed to double in
size is the same regardless of the initial size.
23. Applications of Log and Exponential Formulas
b. As before with r = 4%, does it take more time for
an investment of $500 to become $1,000
than for an investment of $1 to become $2?
No. By viewing the $500 investment as 500 separate
$1 accounts, we see that the time needed for
$500 to grow into $1,000 is the same the time needed
for $1 to grow into $2 which is ln(2)/0.04 ≈ 17.5 years.
Doubling Time and the Rule of 72
From the above reasoning we see that given a fixed
growth rate r, the amount of time needed to double in
size is the same regardless of the initial size.
The amount of time D that’s needed for doubling in
size is called the “doubling time (at rate r)”.
24. Applications of Log and Exponential Formulas
b. As before with r = 4%, does it take more time for
an investment of $500 to become $1,000
than for an investment of $1 to become $2?
No. By viewing the $500 investment as 500 separate
$1 accounts, we see that the time needed for
$500 to grow into $1,000 is the same the time needed
for $1 to grow into $2 which is ln(2)/0.04 ≈ 17.5 years.
Doubling Time and the Rule of 72
From the above reasoning we see that given a fixed
growth rate r, the amount of time needed to double in
size is the same regardless of the initial size.
The amount of time D that’s needed for doubling in
size is called the “doubling time (at rate r)”.
So at r = 4%, the doubling time D ≈ 17.5 yrs.
25. Applications of Log and Exponential Formulas
In example A. the doubling time D at r = 0.04 is
D = In(2) ≈ 17.5 yrs.
0.04
26. Applications of Log and Exponential Formulas
In example A. the doubling time D at r = 0.04 is
D = In(2) ≈ 17.5 yrs.
0.04
Similarly if r = 5% = 0.05, we solve for t in the equation
e0.05t = ln(2)
27. Applications of Log and Exponential Formulas
In example A. the doubling time D at r = 0.04 is
D = In(2) ≈ 17.5 yrs.
0.04
Similarly if r = 5% = 0.05, we solve for t in the equation
In(2)
e0.05t
= ln(2) so the doubling time is
0.05 ≈ 14.0 yrs.
28. Applications of Log and Exponential Formulas
In example A. the doubling time D at r = 0.04 is
D = In(2) ≈ 17.5 yrs.
0.04
Similarly if r = 5% = 0.05, we solve for t in the equation
In(2)
e0.05t
= ln(2) so the doubling time is
0.05 ≈ 14.0 yrs.
Leaving the rate r as a variable, the same algebra
gives us the following.
The Doubling Time Formula
29. Applications of Log and Exponential Formulas
In example A. the doubling time D at r = 0.04 is
D = In(2) ≈ 17.5 yrs.
0.04
Similarly if r = 5% = 0.05, we solve for t in the equation
In(2)
e0.05t
= ln(2) so the doubling time is
0.05 ≈ 14.0 yrs.
Leaving the rate r as a variable, the same algebra
gives us the following.
The Doubling Time Formula
For the exponential growth with rate r, i.e. Pe rt = A,
the doubling time D = r In(2) ( ≈ 0.69897 )
r .
Proof.
30. Applications of Log and Exponential Formulas
In example A. the doubling time D at r = 0.04 is
D = In(2) ≈ 17.5 yrs.
0.04
Similarly if r = 5% = 0.05, we solve for t in the equation
In(2)
e0.05t
= ln(2) so the doubling time is
0.05 ≈ 14.0 yrs.
Leaving the rate r as a variable, the same algebra
gives us the following.
The Doubling Time Formula
For the exponential growth with rate r, i.e. Pe rt = A,
the doubling time D = r In(2) ( ≈ 0.69897 )
r .
Proof. To double means to have an initial amount of
P = 1 unit to grow into A = 2 units.
31. Applications of Log and Exponential Formulas
In example A. the doubling time D at r = 0.04 is
D = In(2) ≈ 17.5 yrs.
0.04
Similarly if r = 5% = 0.05, we solve for t in the equation
In(2)
e0.05t
= ln(2) so the doubling time is
0.05 ≈ 14.0 yrs.
Leaving the rate r as a variable, the same algebra
gives us the following.
The Doubling Time Formula
For the exponential growth with rate r, i.e. Pe rt = A,
the doubling time D = r In(2) ( ≈ 0.69897 )
r .
Proof. To double means to have an initial amount of
P = 1 unit to grow into A = 2 units. Hence 1* er*t = 2.
32. Applications of Log and Exponential Formulas
In example A. the doubling time D at r = 0.04 is
D = In(2) ≈ 17.5 yrs.
0.04
Similarly if r = 5% = 0.05, we solve for t in the equation
In(2)
e0.05t
= ln(2) so the doubling time is
0.05 ≈ 14.0 yrs.
Leaving the rate r as a variable, the same algebra
gives us the following.
The Doubling Time Formula
For the exponential growth with rate r, i.e. Pe rt = A,
the doubling time D = r In(2) ( ≈ 0.69897 )
r .
Proof. To double means to have an initial amount of
P = 1 unit to grow into A = 2 units. Hence 1* er*t = 2.
So r * t = ln(2)
33. Applications of Log and Exponential Formulas
In example A. the doubling time D at r = 0.04 is
D = In(2) ≈ 17.5 yrs.
0.04
Similarly if r = 5% = 0.05, we solve for t in the equation
In(2)
e0.05t
= ln(2) so the doubling time is
0.05 ≈ 14.0 yrs.
Leaving the rate r as a variable, the same algebra
gives us the following.
The Doubling Time Formula
For the exponential growth with rate r, i.e. Pe rt = A,
the doubling time D = r In(2) ( ≈ 0.69897 )
r .
Proof. To double means to have an initial amount of
P = 1 unit to grow into A = 2 units. Hence 1* er*t = 2.
So r * t = ln(2) or t = ln(2)/r = D is the doubling time.
34. Applications of Log and Exponential Formulas
The constant ln(2) is rounded up to 0.72 for estimation.
35. Applications of Log and Exponential Formulas
The constant ln(2) is rounded up to 0.72 for estimation.
The Rule of 72 for the Doubling Time
The doubling time D ≈ 0.72 where r is the growth rate.
r
36. Applications of Log and Exponential Formulas
The constant ln(2) is rounded up to 0.72 for estimation.
The Rule of 72 for the Doubling Time
The doubling time D ≈ 0.72 where r is the growth rate.
r
We use 0.72 because 72 can be divided by many
different values, hence convenient to use.
37. Applications of Log and Exponential Formulas
The constant ln(2) is rounded up to 0.72 for estimation.
The Rule of 72 for the Doubling Time
The doubling time D ≈ 0.72 where r is the growth rate.
r
We use 0.72 because 72 can be divided by many
different values, hence convenient to use.
So at r = 4%, the doubling time ≈ 0.72 = 72 = 18
0.04 4
yrs,
38. Applications of Log and Exponential Formulas
The constant ln(2) is rounded up to 0.72 for estimation.
The Rule of 72 for the Doubling Time
The doubling time D ≈ 0.72 where r is the growth rate.
r
We use 0.72 because 72 can be divided by many
different values, hence convenient to use.
So at r = 4%, the doubling time ≈ 0.72 = 72 = 18
0.04 4
if r = 8%, the doubling time ≈ 72 = 9 yrs,
8 yrs,
39. Applications of Log and Exponential Formulas
The constant ln(2) is rounded up to 0.72 for estimation.
The Rule of 72 for the Doubling Time
The doubling time D ≈ 0.72 where r is the growth rate.
r
We use 0.72 because 72 can be divided by many
different values, hence convenient to use.
So at r = 4%, the doubling time ≈ 0.72 = 72 = 18
0.04 4
if r = 8%, the doubling time ≈ 72 = 9 yrs,
8 yrs,
72
if r = 12%, the doubling time ≈ 12 = 6 yrs.
40. Applications of Log and Exponential Formulas
The constant ln(2) is rounded up to 0.72 for estimation.
The Rule of 72 for the Doubling Time
The doubling time D ≈ 0.72 where r is the growth rate.
r
We use 0.72 because 72 can be divided by many
different values, hence convenient to use.
So at r = 4%, the doubling time ≈ 0.72 = 72 = 18
0.04 4
if r = 8%, the doubling time ≈ 72 = 9 yrs,
8 yrs,
72
if r = 12%, the doubling time ≈ 12 = 6 yrs.
The equation D = In(2) sayrs that “the doubling time
r
D is inversely proportional to the growth rate r”.
41. Applications of Log and Exponential Formulas
The constant ln(2) is rounded up to 0.72 for estimation.
The Rule of 72 for the Doubling Time
The doubling time D ≈ 0.72 where r is the growth rate.
r
We use 0.72 because 72 can be divided by many
different values, hence convenient to use.
So at r = 4%, the doubling time ≈ 0.72 = 72 = 18
0.04 4
if r = 8%, the doubling time ≈ 72 = 9 yrs,
8 yrs,
72
if r = 12%, the doubling time ≈ 12 = 6 yrs.
The equation D = In(2) sayrs that “the doubling time
r
D is inversely proportional to the growth rate r”.
So if we double the growth rate r then D is shorten to
1/2 of the time before,
42. Applications of Log and Exponential Formulas
The constant ln(2) is rounded up to 0.72 for estimation.
The Rule of 72 for the Doubling Time
The doubling time D ≈ 0.72 where r is the growth rate.
r
We use 0.72 because 72 can be divided by many
different values, hence convenient to use.
So at r = 4%, the doubling time ≈ 0.72 = 72 = 18
0.04 4
if r = 8%, the doubling time ≈ 72 = 9 yrs,
8 yrs,
72
if r = 12%, the doubling time ≈ 12 = 6 yrs.
The equation D = In(2) sayrs that “the doubling time
r
D is inversely proportional to the growth rate r”.
So if we double the growth rate r then D is shorten to
1/2 of the time before, and if we triple (3x) the rate r
then D is shorten to 1/3 as before as shown above.
43. Applications of Log and Exponential Formulas
Likewise if D is known then the growth rate r is In(2) .
D
44. Applications of Log and Exponential Formulas
Likewise if D is known then the growth rate r is In(2) .
D
The Rule of 72 for the Growth Rate
Given the doubling time D, the approximate interest
rate r ≈ 0.72
D
45. Applications of Log and Exponential Formulas
Likewise if D is known then the growth rate r is In(2) .
D
The Rule of 72 for the Growth Rate
Given the doubling time D, the approximate interest
rate r ≈ 0.72
D
Example B. What is the approximate interest rate r if
the doubling time D = 24 yrs? D =12 yrs? D= 6 yrs?
46. Applications of Log and Exponential Formulas
Likewise if D is known then the growth rate r is In(2) .
D
The Rule of 72 for the Growth Rate
Given the doubling time D, the approximate interest
rate r ≈ 0.72
D
Example B. What is the approximate interest rate r if
the doubling time D = 24 yrs? D =12 yrs? D= 6 yrs?
Using the Rule of 72, if D = 24 yrs, then r ≈0.72 = 3%
24
47. Applications of Log and Exponential Formulas
Likewise if D is known then the growth rate r is In(2) .
D
The Rule of 72 for the Growth Rate
Given the doubling time D, the approximate interest
rate r ≈ 0.72
D
Example B. What is the approximate interest rate r if
the doubling time D = 24 yrs? D =12 yrs? D= 6 yrs?
Using the Rule of 72, if D = 24 yrs, then r ≈0.72 = 3%
24
If D = 12 yrs, then r ≈ 0.72 = 6%
12
48. Applications of Log and Exponential Formulas
Likewise if D is known then the growth rate r is In(2) .
D
The Rule of 72 for the Growth Rate
Given the doubling time D, the approximate interest
rate r ≈ 0.72
D
Example B. What is the approximate interest rate r if
the doubling time D = 24 yrs? D =12 yrs? D= 6 yrs?
Using the Rule of 72, if D = 24 yrs, then r ≈0.72 = 3%
24
If D = 12 yrs, then r ≈ 0.72 = 6%
12
If D = 8 yrs, then r ≈0.72 = 9%
8
49. Applications of Log and Exponential Formulas
Likewise if D is known then the growth rate r is In(2) .
D
The Rule of 72 for the Growth Rate
Given the doubling time D, the approximate interest
rate r ≈ 0.72
D
Example B. What is the approximate interest rate r if
the doubling time D = 24 yrs? D =12 yrs? D= 6 yrs?
Using the Rule of 72, if D = 24 yrs, then r ≈0.72 = 3%
24
If D = 12 yrs, then r ≈ 0.72 = 6%
12
If D = 8 yrs, then r ≈0.72 = 9%
8
So if D is shorten to 1/2 of the time before, then the
growth rate r must be doubled,
50. Applications of Log and Exponential Formulas
Likewise if D is known then the growth rate r is In(2) .
D
The Rule of 72 for the Growth Rate
Given the doubling time D, the approximate interest
rate r ≈ 0.72
D
Example B. What is the approximate interest rate r if
the doubling time D = 24 yrs? D =12 yrs? D= 6 yrs?
Using the Rule of 72, if D = 24 yrs, then r ≈0.72 = 3%
24
If D = 12 yrs, then r ≈ 0.72 = 6%
12
If D = 8 yrs, then r ≈0.72 = 9%
8
So if D is shorten to 1/2 of the time before, then the
growth rate r must be doubled, and if D is to be shorten
to 1/3 as before, then the growth rate r must be tripled.
51. Applications of Log and Exponential Formulas
In a similar manner we may define tripling time or
quadrupling time,
52. Applications of Log and Exponential Formulas
In a similar manner we may define tripling time or
quadrupling time, i.e. the time it takes for the growth to
become three times or four times of its original size.
53. Applications of Log and Exponential Formulas
In a similar manner we may define tripling time or
quadrupling time, i.e. the time it takes for the growth to
become three times or four times of its original size.
The tripling time = In(3) ≈ 1.08
r r
54. Applications of Log and Exponential Formulas
In a similar manner we may define tripling time or
quadrupling time, i.e. the time it takes for the growth to
become three times or four times of its original size.
The tripling time = In(3) ≈ 1.08
r r
The quadrupling time = In(4) ≈ 1.44
r r
55. Applications of Log and Exponential Formulas
In a similar manner we may define tripling time or
quadrupling time, i.e. the time it takes for the growth to
become three times or four times of its original size.
The tripling time = In(3) ≈ 1.08
r r
The quadrupling time = In(4) ≈ 1.44
r r
By the same algebra for deriving the doubling time
we obtain the following from the Perta formula.
56. Applications of Log and Exponential Formulas
In a similar manner we may define tripling time or
quadrupling time, i.e. the time it takes for the growth to
become three times or four times of its original size.
The tripling time = In(3) ≈ 1.08
r r
The quadrupling time = In(4) ≈ 1.44
r r
By the same algebra for deriving the doubling time
we obtain the following from the Perta formula.
It takes In(K) years to grow to K times of the original
r
size with continuous growth rate r.
57. Applications of Log and Exponential Formulas
In a similar manner we may define tripling time or
quadrupling time, i.e. the time it takes for the growth to
become three times or four times of its original size.
The tripling time = In(3) ≈ 1.08
r r
The quadrupling time = In(4) ≈ 1.44
r r
By the same algebra for deriving the doubling time
we obtain the following from the Perta formula.
It takes In(K) years to grow to K times of the original
r
size with continuous growth rate r.
In(10
So it would takes 0.08 ≈ 27.8 years at the continuous
)
rate 8% to grow to to 10 times of the original amount.
58. Applications of Log and Exponential Formulas
The variable K is called the factor.
• For the exponential growth (r > 0), the factor K is
independent of the initial size.
• It would take In(K)/r yrs to reach the size of factor K.
If r < 0 then we have a process of exponential decay,
contraction, depreciation, etc.. and as time goes on,
the factor K is less than 1. For example, if r = –0.05
then after 10 year, the initial unit would shrink to the
factor K = e–0.05(10) = e–0.5 ≈ 3/5 of the original.
• For the exponential decay (r < 0), the factor K is
independent of the initial size.
• It would take In(K)/r yrs to reach the size of factor K.