68 applications of exponential and log

Applications of Log and Exponential Formulas

Recall the following compound interest formulas.
Pina (periodic compounding)

and Perta (continuous compounding)

P = principal, i = periodic rate,
N = total number of periods, A = accumulated value


then P(1 + i )N = A

then P(1 + i )N = A
P = principal, r = yearly or annual rate,
t = total number of years, A = accumulated value

then P(1 + i )N = A
then Pert = A

then P(1 + i )N = A
then Pert = A
The Pina formula has the base (1+ i) which varies
with i.

then P(1 + i )N = A
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.


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.


P = $500, A= $1,000, and r = 4% = 0.04. So
500e0.04t = 1000

To find the r or t in the formula Pert = A, divide both
sides by P.
P = $500, A= $1,000, and r = 4% = 0.04. So
500e0.04t = 1000

sides by P.
P = $500, A= $1,000, and r = 4% = 0.04. So
500e0.04t = 1000
e0.04t = 2 (=1000/500)

sides by P. We get ert = A so that r*t = ln( A ).
P P
P = $500, A= $1,000, and r = 4% = 0.04. So
500e0.04t = 1000
e0.04t = 2 (=1000/500)

P P
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)

P P
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

P P
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.

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

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)”.

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.

In example A. the doubling time D at r = 0.04 is
D = In(2) ≈ 17.5 yrs.
0.04

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)

D = In(2) ≈ 17.5 yrs.
0.04
In(2)
e0.05t
= ln(2) so the doubling time is
0.05 ≈ 14.0 yrs.

D = In(2) ≈ 17.5 yrs.
0.04
In(2)
e0.05t
0.05 ≈ 14.0 yrs.
Leaving the rate r as a variable, the same algebra
gives us the following.
The Doubling Time Formula

D = In(2) ≈ 17.5 yrs.
0.04
In(2)
e0.05t
0.05 ≈ 14.0 yrs.
For the exponential growth with rate r, i.e. Pe rt = A,
the doubling time D = r In(2) ( ≈ 0.69897 )
r .
Proof.

D = In(2) ≈ 17.5 yrs.
0.04
In(2)
e0.05t
0.05 ≈ 14.0 yrs.
r .
Proof. To double means to have an initial amount of
P = 1 unit to grow into A = 2 units.

D = In(2) ≈ 17.5 yrs.
0.04
In(2)
e0.05t
0.05 ≈ 14.0 yrs.
r .
P = 1 unit to grow into A = 2 units. Hence 1* er*t = 2.

D = In(2) ≈ 17.5 yrs.
0.04
In(2)
e0.05t
0.05 ≈ 14.0 yrs.
r .
So r * t = ln(2)

D = In(2) ≈ 17.5 yrs.
0.04
In(2)
e0.05t
0.05 ≈ 14.0 yrs.
r .
So r * t = ln(2) or t = ln(2)/r = D is the doubling time.

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

r
We use 0.72 because 72 can be divided by many
different values, hence convenient to use.

r
So at r = 4%, the doubling time ≈ 0.72 = 72 = 18
0.04 4
yrs,

r
0.04 4
if r = 8%, the doubling time ≈ 72 = 9 yrs,
8 yrs,

r
0.04 4
8 yrs,
72
if r = 12%, the doubling time ≈ 12 = 6 yrs.

r
0.04 4
8 yrs,
72
The equation D = In(2) sayrs that “the doubling time
r
D is inversely proportional to the growth rate r”.

r
0.04 4
8 yrs,
72
r
So if we double the growth rate r then D is shorten to
1/2 of the time before,

r
0.04 4
8 yrs,
72
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.

Likewise if D is known then the growth rate r is In(2) .
D

D
The Rule of 72 for the Growth Rate
Given the doubling time D, the approximate interest
rate r ≈ 0.72
D

D
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?

D
rate r ≈ 0.72
D
Using the Rule of 72, if D = 24 yrs, then r ≈0.72 = 3%
24

D
rate r ≈ 0.72
D
24
If D = 12 yrs, then r ≈ 0.72 = 6%
12

D
rate r ≈ 0.72
D
24
If D = 12 yrs, then r ≈ 0.72 = 6%
12
If D = 8 yrs, then r ≈0.72 = 9%
8

D
rate r ≈ 0.72
D
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,

D
rate r ≈ 0.72
D
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.

In a similar manner we may define tripling time or
quadrupling time,

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

r r
The quadrupling time = In(4) ≈ 1.44
r r

r r
r r
By the same algebra for deriving the doubling time
we obtain the following from the Perta formula.

r r
r r
It takes In(K) years to grow to K times of the original
r
size with continuous growth rate r.

r r
r r
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.

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.

68 applications of exponential and log

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