Module 7 Lecture Notes
Contents
7.1
7.2
7.3
Solving Exponential and Logarithmic Equations . . . . . . . . . . . . .
2
7.1.1
Solving Exponential Equations . . . . . . . . . . . . . . . . . . . . . . . .
3
7.1.2
Solving Logarithmic Equations . . . . . . . . . . . . . . . . . . . . . . . .
8
7.1.3
Solving Equations Using a Graphing Calculator . . . . . . . . . . . . . . . 14
Financial Applications: Compound Interest . . . . . . . . . . . . . . .
15
7.2.1
Simple Interest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
7.2.2
The Formula for Compound Interest . . . . . . . . . . . . . . . . . . . . . 16
7.2.3
The Continuous Interest Rate Formula . . . . . . . . . . . . . . . . . . . . 20
7.2.4
Comparing Effective Interest Rates . . . . . . . . . . . . . . . . . . . . . . 22
Biology Applications: Exponential Growth and Decay . . . . . . . . .
26
7.3.1
Doubling Time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
7.3.2
Half-Life . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
Properties of Logarithms:
For any real number x and any positive real numbers M , N , and a (a 6= 1), it holds that
• loga (1) = 0
• loga (a) = 1
• loga (M N ) = loga (M ) + loga (N )
• loga M
= loga (M ) − loga (N )
N
• loga (ax ) = x
• loga (M r ) = r loga (M )
• aloga (x) = x, x > 0
• loga (M ) =
1
log(M )
log(a)
or loga (M ) =
ln(M )
ln(a)
Math 111 Module 7 Lecture Notes
7.1
Solving Exponential and Logarithmic Equations
Recall the following homework problem:
Suppose the formula D = 5e−0.4h can be used to find the number of milligrams D of a certain
drug in a patient’s bloodstream h hours after the drug was administered. When the number of
milligrams reaches 2, the drug is to be administered again. What is the time between injections?
To solve 2 = 5e−0.4h , we isolate e−0.4h and use the natural log function to solve for h:
2 = 5e−0.4h
2
= e−0.4h
5
2
ln
= −0.4h
5
ln 25
=h
−0.4
h ≈ 2.291
Thus the drug needs to be readministered after about 2.291 hours.
What if the base was not 10 or e? What if we had the following instead: 3 = 5 · 2−0.4h ?
To solve this equation, we could isolate the expontial expression, use the log2 function, and then
the change of base function to approximate h:
2 = 5 · 2−0.4h
2
= 2−0.4h
5
2
log2
= −0.4h
5
log2 52
=h
−0.4
log( 25 )
log(2)
h=
−0.4
h ≈ 1.842
However, we could also just apply the the common log function or the natural log function to each
side, as shown below:
Instructor: A.E.Cary
Page 2 of 31
Math 111 Module 7 Lecture Notes
Common Log
Natural Log
2 = 5 · 2−0.4h
2 = 5 · 2−0.4h
2
= 2−0.4h
5
2
log
= log 2−0.4h
5
2
log
= −0.4h log(2)
5
log 25
=h
−0.4 log(2)
2
= 2−0.4h
5
2
ln
= ln 2−0.4h
5
2
ln
= −0.4h ln(2)
5
ln 25
=h
−0.4 ln(2)
h ≈ 1.842
h ≈ 1.842
The advantage to this approach is that it will work in all cases. The more complicated the
equation becomes, the more appropriate this method becomes.
7.1.1
Solving Exponential Equations
In the equations that follow, this alternate method will be demonstrated. Note that in each case,
either the natural log function or the common log function could be used. The only time one is
preferable to the other is when an exponential expression with either base e or base 10 occurs; in
this case, it makes sense to use the logarithm with the same base.
Example 1: Solve the following exponential equations for x. Express irrational solutions in exact
form and clearly state the solution set.
(a) 3x = 15
3x = 15
log (3x ) = log(15)
x log(3) = log(15)
x=
Solution Set:
Instructor: A.E.Cary
log(15)
log(3)
log(15)
log(3)
Page 3 of 31
Math 111 Module 7 Lecture Notes
(b) 3 · 2x = 15
3 · 2x = 15
3 · 2x
15
=
3
3
x
2 =5
log (2x ) = log(5)
x log(2) = log(5)
x=
Solution Set:
n
log(5)
log(2)
log(5)
log(2)
o
(c) 23x = 15
23x = 15
log 23x = log(15)
3x log(2) = log(15)
x=
Solution Set:
n
log(15)
log(3)
log(15)
3 log(2)
o
(d) 2−x = 15
2−x = 15
ln 2−x = ln(15)
−x ln(2) = ln(15)
x=−
ln(15)
ln(2)
n
o
ln(15)
Solution Set: − ln(2)
Instructor: A.E.Cary
Page 4 of 31
Math 111 Module 7 Lecture Notes
In solving the exponential equations below, the natural log or common log function will be used.
Once this function is applied to each side of the equation, the resulting equation will need to be
solved. The key is to identify what type of equation will be left. In many cases, the equation will
just be linear.
It may not look linear–but remember that coefficients and constants such as 2 log(5)
5
and ln 4 are just numbers! In other cases, the resulting equaiton may be quadratic. In these
instances, use any of the following that are appropriate: factoring, the square root property, the
quadratic formula, or completing the square.
Example 2: Solve the following exponential equations for x. Express irrational solutions in exact
form and rounded to three decimal places.
(a) 4 · 23x = 5
4 23x = 5
23x =
5
4
5
ln 2
= ln
4
5
3x ln(2) = ln
4
ln 45
x=
3 ln(2)
3x
x ≈ 0.107
(
Solution Set:
Instructor: A.E.Cary
)
ln 54
3 ln(2)
Page 5 of 31
Math 111 Module 7 Lecture Notes
2
(b) 10x = 8
2
10x = 8
2
log 10x = log(8)
x2 = log(8)
p
x = ± log(8)
x ≈ 0.950
Solution Set:
o
n p
p
− log(8), log(8)
(c) 73x−4 = 39
73x−4 = 39
ln 73x−4 = ln(39)
(3x − 4) ln(7) = ln(39)
3x ln(7) − 4 ln(7) = ln(39)
3x ln(7) = ln(39) + 4 ln(7)
x=
ln(39) + 4 ln(7)
3 ln(7)
x=
ln(39) + ln(2401)
ln(343)
x=
ln(93639)
ln(343)
(Sufficient simpification)
x ≈ 1.961
ln(93639)
Solution Set:
ln(343)
Instructor: A.E.Cary
Page 6 of 31
Math 111 Module 7 Lecture Notes
(d) 2x+3 = 52x+1
2x+3 = 52x+1
log 2x+3 = log 52x+1
(x + 3) log(2) = (2x + 1) log(5)
x log(2) + 3 log(2) = 2x log(5) + log(5)
This is linear!
x log(2) + 3 log(2) − x log(2) = 2x log(5) + log(5) − x log(2)
3 log(2) = 2x log(5) + log(5) − x log(2)
3 log(2) − log(5) = 2x log(5) + log(5) − x log(2) − log(5)
3 log(2) − log(5) = 2x log(5) − x log(2)
3 log(2) − log(5) = x(2 log(5) − log(2))
3 log(2) − log(5)
x(2 log(5) − log(2))
=
(2 log(5) − log(2))
(2 log(5) − log(2))
3 log(2) − log(5)
=x
(2 log(5) − log(2))
(Sufficient simplification)
log (23 ) − log(5)
(log (52 ) − log(2))
log 85
x=
log 25
2
x=
x ≈ 0.186
(
Solution Set:
Instructor: A.E.Cary
log
log
)
8
5 25
2
Page 7 of 31
Math 111 Module 7 Lecture Notes
7.1.2
Solving Logarithmic Equations
If loga (x) = loga (y), then x = y.
Example 3: Solve the following logarithmic equations for x. Express irrational solutions in exact
form and clearly state the solution set.
(a) log4 (x + 1) = log4 (25)
log4 (x + 1) = log4 (25)
x + 1 = 25
x = 24
Solution Set: {24}
(b) 2 log4 (x + 1) = log4 (25)
2 log4 (x + 1) = log4 (25)
log4 (x + 1) = log4 (25)
(x + 1)2 = 25
√
x + 1 = ± 25
x + 1 = ±5
x = −6 or x = 4
As log4 (−6 + 1) is undefined, the proposed solution -6 is extraneous. Therefore the only
solution is 4 and the solution set is {4}.
Instructor: A.E.Cary
Page 8 of 31
Math 111 Module 7 Lecture Notes
(c) 20 log4 (x) = 10
20 log4 (x) = 10
1
2
=x
log4 (x) =
41/2
2=x
Solution Set: {2}
Alternately:
20 log4 (x) = 10
log4 x20 = 10
x20 = 410
x20
1/20
= 410
1/20
x = 41/2
x=2
Instructor: A.E.Cary
Page 9 of 31
Math 111 Module 7 Lecture Notes
(d) log6 (x + 4) + log6 (x + 3) = 1
log6 (x + 4) + log6 (x + 3) = 1
log6 (x + 4)(x + 3) = 1
(x + 4)(x + 3) = 61
It’s quadratic!
x2 + 7x + 12 = 6
x2 + 7x + 6 = 0
(x + 6)(x + 1) = 0
x + 6 = 0 or x + 1 = 0
x = −6
or
x = −1
As log6 (−6 + 4) is undefined, -6 is an extraneous solution. The solution set is {−1}.
(e) ln(x + 1) − ln(x) = 2
ln(x + 1) − ln(x) = 2
x+1
ln
=2
x
x+1
= e2
x
x + 1 = xe2
It’s a rational equation!
It’s a linear equation!
1 = xe2 − x
1 = x(e2 − 1)
e2
Solution Set:
Instructor: A.E.Cary
1
1
=x
−1
e2 −1
Page 10 of 31
Math 111 Module 7 Lecture Notes
(f) log2 (x + 1) + log2 (x + 7) = 3
log2 (x + 1) + log2 (x + 7) = 3
log2 (x + 1)(x + 7) = 3
(x + 1)(x + 7) = 23
x2 + 8x + 7 = 8
x2 + 8x − 1 = 0
This equation is quadratic but cannot be factored. It can be solved by completing the square
or by using the quadratic formula. Below, completing the square is used:
x2 + 8x − 1 = 0
x2 + 8x + 16 − 16 − 1 = 0
(x2 + 8x + 16) − 17 = 0
(x + 4)2 − 17 = 0
(x + 4)2 = 17
√
x + 4 = ± 17
√
x = −4 ± 17
√
√
x = −4 + 17 or x = −4 − 17
x ≈ 0.123 or x ≈ −8.123
√ √
The solution −4 − 17 is extraneous, so the solution set is −4 + 17 .
Note that the equation x2 + 8x − 1 = 0 could also be solved using the quadratic formula, as
shown below:
x2 + 8x − 1 = 0
p
82 − 4(1)(−1)
x=
2(1)
√
−8 ± 68
x=
2
√
−8 ± 4 · 17
x=
2
√
−8 ± 2 17
x=
2
√
x = −4 ± 17
−8 ±
Instructor: A.E.Cary
Page 11 of 31
Math 111 Module 7 Lecture Notes
Example 4: The loudness L(x) (in decibels)
of a sound of intensity x (measured in watts per
square meter) is defined by L(x) = 10 log Ix0 , where I0 = 10−12 watts per square meter and
represents the least intense sound that a human ear can detect.
(a) Normal conversation has an loudness level of 60 dB. What is the intensity of this sound?
To determine the loudness level of 60 dB, we will need to solve L(x) = 60:
x
60 = 10 log
I0
x
6 = log
I0
x
106 =
I0
106 · I0 = x
106 · 10−12 = x
x = 10−6
The intensity of sound for a normal conversation is 10−6 watts per square meter.
(b) A jet takeoff has an loudness level of 140 dB. What is the intensity of this sound?
To determine the loudness level of 140 dB, we will need to solve L(x) = 140:
x
140 = 10 log
I0
x
14 = log
I0
x
1014 =
I0
1014 · I0 = x
1014 · 10−12 = x
x = 102
The intensity of sound when a jet takes off is 100 watts per square meter.
Instructor: A.E.Cary
Page 12 of 31
Math 111 Module 7 Lecture Notes
(c) The threshold of human hearing is 0 dB. What is the intensity of this sound?
To determine the loudness level of 0 dB, we will need to solve L(x) = 0:
x
0 = 10 log
I0
x
0 = log
I0
x
100 =
I0
x
1=
I0
I0 = x
x = 10−12
The intensity of sound when a jet takes off is 10−12 watts per square meter.
Instructor: A.E.Cary
Page 13 of 31
Math 111 Module 7 Lecture Notes
7.1.3
Solving Equations Using a Graphing Calculator
Example 5: The equation e2x = x + 2 cannot easily be solved algebraically through the means
we’ve studied in this course. The approximate solutions will need to be found graphically.
To solve this equation graphically, we will determine where the graphs of y = e2x and y = x + 2
intersect:
The points of intersection are approximately (−1.9810, 0.0190) and (0.4475, 2.4475). The solution
set is {x | x ≈ −1.9810, 0.4475}.
Example 6: Use a graphing calculator solve the equation below, rounding all solutions accurately
to four decimal places.
log2 (x − 1) − log6 (x + 2) = 2
To solve this equation, we will find the point of intersection of y1 (x) = log2 (x − 1) − log6 (x + 2)
and y2 (x) = 2. To input the first function into a graphing calculator, the change-of-base formula
is needed. This will appear as:
log(x-1)/log(2)-log(x+2)/log(6)
The point of intersection is about (12.1485, 2). The solution set is {x|x ≈ 12.1485}.
Instructor: A.E.Cary
Page 14 of 31
Math 111 Module 7 Lecture Notes
7.2
Financial Applications: Compound Interest
When money is invested into an account, it often earns interest at a compound interest rate. The
idea behind compound interest is this: If you invest a sum of money and don’t remove it, then the
next time you earn interest you are earning interest on your interest. This idea of “earning interest
from your interest” is the basis for compound interest. Simple interest is interest that does not
compound. We will briefly explore this before moving on to compound interest. These concepts
apply to both investments and loans.
7.2.1
Simple Interest
Simple Interest: The amount of interest for an investment or loan, P (for principal), with a
simple interest rate account r for a period of t years is given by:
I = P rt
Example 7: Suppose you invest $20,000 into an account earning 5% simple interest. What will
be the value of the investment after 6 years? How much interest will you have earned after 6 years?
This interest will not compound, and we will use the formula for simple interest:
I = (20, 000)(0.05)(6)
= 6, 000
The amount of interest earned after 6 years is $6,000. The value of the investment will be $26,000.
Instructor: A.E.Cary
Page 15 of 31
Math 111 Module 7 Lecture Notes
7.2.2
The Formula for Compound Interest
Example 8: You again invest $20,000, but this time into an account with an interest rate of 5%
compounded annually. How much will you owe after 6 years? What is the effective annual interest
rate?
• After 1 year:
A = 20, 000 + (0.05)20, 000
= 20, 000(1.08)
= 21, 000
• After 2 years:
A = 21, 000 + (0.05)21, 000
= 21, 000(1.05)
= 22, 050
Note that in the above, 21, 000(1.05) = 20, 000(1.05) (1.05) = 20, 000(1.05)2 . We can use
this to determine the value after 6 years.
• After 6 years:
A = 20, 000(1.05)6
≈ 26801.91
The value of the account after 6 years will be $26801.91. The amount of interest earned after 6
years will be $6801.91. Note that this amount is signifcantly more than the account that earned
simple interest at a rate of 6%.
The effective rate of interest will be 5%. This can be determined by the formula or by the amount
of interest earned in one year compared to the investment. Using the formula, we see that the base
of this exponential function is 1.05, and therefore the effective interest rate is 0.05, or 5%. Using
1000
the amount of interest, we could compute the following: 20000
= 0.05
Instructor: A.E.Cary
Page 16 of 31
Math 111 Module 7 Lecture Notes
Compound Interest Formula:
The amount A after t years due to a principal P invested at an annual interest rate r compounded n times per year is
r nt
A=P · 1+
n
In the formula above, the stated interest rate r is broken up into n pieces and compounds n times
each year. Suppose you invest $20,000 into an account that earns an interest at a rate of 5%. If
this compounds annually, then the formula would just be:
1·t
0.05
P = 20, 000 1 +
1
If the 5% compounds quarterly, the formula would be:
4·t
0.05
P = 20, 000 1 +
4
In other words, you will multiply the initial $20,000 by 1 + 0.05
four times each year. In other
4
words, you multiply by 1.0125 four times each year. As we will see shortly, this is NOT the same
as earning 5% compounded annually.
The effective rate of interest is the equivalent annual simple interest that would yield the
same amount as compounding n times per year, or continuously, after 1 year.
Example 9: Once again, you are going to invest $20,000. You have the following choices. For
each, find the value of the investment after 6 years and the effective rate of interest.
• 5% compounded quarterly
4t
0.05
Formula: A = 20, 000 1 +
4
After 6 years:
4·6
0.05
A = 20000 1 +
4
≈ 26947.02
The value of the account after 6 years will be $26947.02.
The effective rate can be determined by the formula or by the amount of interest earned
in one year compared to the investment. Using the formula, we see that the base of this
Instructor: A.E.Cary
Page 17 of 31
Math 111 Module 7 Lecture Notes
exponential function is 1 +
effective interest rate is:
0.05 4
.
4
This is what we multiply by each year. Therefore the
4
0.05
− 1 ≈ 0.050945
1+
4
Thus the effective interest rate is 5.0945%.
We could also use the amount of interest earned over one year and compare this to the
original $20,000: Using the amount of interest, we could compute the following:
0.05 4
4
20000 1 +
− 20000
=
20000
20000
=
1+
0.05 4
4
−1
20000
4
0.05
1+
4
≈ 0.05945
• 5% compounded monthly
12t
0.05
Formula: A = 20, 000 1 +
12
After 6 years:
12·6
0.05
A = 20000 1 +
12
≈ 26980.35
The value of the account after 6 years will be $26980.35.
The effective rate can be determined by the formula or by the amount of interest earned
in one year compared to the investment.
Using the formula, we see that the base of this
0.05 12
. This is what we multiply by each year. Therefore the
exponential function is 1 + 12
effective interest rate is:
12
0.05
1+
− 1 ≈ 0.051162
12
Thus the effective interest rate is 5.1162%.
Instructor: A.E.Cary
Page 18 of 31
Math 111 Module 7 Lecture Notes
• 5% compounded daily
0.05
Formula: A = 20, 000 1 +
365
After 6 years:
365t
365·6
0.05
A = 20000 1 +
365
≈ 26996.62
The value of the account after 6 years will be $26996.62.
The effective rate can be determined by the formula or by the amount of interest earned
in one year compared to the investment.
Using the formula, we see that the base of this
0.05 365
. This is what we multiply by each year. Therefore the
exponential function is 1 + 365
effective interest rate is:
365
0.05
1+
− 1 ≈ 0.051267
365
Thus the effective interest rate is 5.1267%.
Instructor: A.E.Cary
Page 19 of 31
Math 111 Module 7 Lecture Notes
7.2.3
The Continuous Interest Rate Formula
Continuous Interest Formula:
The amount A after t years due to a principal P invested at an annual interest rate r compounded continuously is
A = P ert
Example 10: If $20,000 was invested into an account earning 5% that compounds continuously,
what would the value of the investment be after 6 years? What is the effective annual interest
rate?
Formula: A = 20, 000e0.05t
After 6 years:
A = 20000e0.05∗6
≈ 26997.18
The value of the account after 6 years will be $26997.18.
The effective rate can be determined by the formula or by the amount of interest earned in one
year compared to the investment. Using the formula, we see that the base of this exponential
function is e0.05 . This is what we multiply by each year. Therefore the effective interest rate is:
e−.05 − 1 ≈ 0.051271
Thus the effective interest rate is 5.1271%.
Instructor: A.E.Cary
Page 20 of 31
Math 111 Module 7 Lecture Notes
Example 11: Table 7.1 summarizes the previous examples.
Table 7.1
Compounding Frequency
Annual
Quarterly
Monthly
Daily
Annual Growth
1
1 + 0.05
1
4
1 + 0.05
4
12
1 + 0.05
12
365
1 + 0.05
365
e0.05
Continuously
Factor
Effective Annual Rate
= 1.05
5%
≈ 1.050945
5.0945%
≈ 1.051162
5.1162%
≈ 1.051267
5.1267%
≈ 1.051271
5.1271%
Example 12: Now assume that you have $1 and it earns 100% annual interest. Complete Table 7.2 below for the various compounding frequencies. (This is utterly silly in reality–but will
show you exactly where e comes from!!)
Table 7.2
Compounding Frequency
Annual
Semi-annual
Quarterly
Monthly
Daily
Hourly
Each minute
Each second
Continuously
Instructor: A.E.Cary
Annual Growth
1
1 + 11
2
1 + 12
4
1 + 14
1 12
1 + 12
1 365
1 + 365
8760
1
1 + 8760
525600
1
1 + 525600
31536000
1
1 + 31536000
e1
Factor
=2
≈ 2.25
≈ 2.441406
≈ 2.613035
≈ 2.714567
≈ 2.718127
≈ 2.718279
≈ 2.718282
≈ 2.718282
Page 21 of 31
Math 111 Module 7 Lecture Notes
7.2.4
Comparing Effective Interest Rates
Example 13: Determine which of the following interest rates for an investment is a better deal:
• 6% compounded quarterly
4t
0.06
Formula: A = P 1 +
4
4
− 1 ≈ 0.06136
Effective Rate: 1 + 0./06
4
The effective interest rate is 6.136%.
• 5.95% compounded continuously
Formula: A = P e0.0595t
Effective Rate: e0.0595 − 1 ≈ 0.0631
The effective interest rate is 6.131%.
The investment with a 6% interest rate compounded quarterly is a better deal than the account
with an interest rate of 5.95% compounded continuously.
Example 14: Determine which of the following interest rates for an investment is a better deal:
• 9% compounded quarterly
4t
0.09
Formula: A = P 1 +
4
4
Effective Rate: 1 + 0.09
− 1 ≈ 0.09308
4
The effective interest rate is 9.308%.
• 8.95% compounded continuously
Formula: A = P e0.0895t
Effective Rate: e0.0895 − 1 ≈ 0.09363
The effective interest rate is 9.363%.
The investment with a 8.95% interest rate compounded continuously is a better deal than the
account with an interest rate of 9% compounded continuously.
Instructor: A.E.Cary
Page 22 of 31
Math 111 Module 7 Lecture Notes
Example 15: You invest $9,000 into an account with interest rate of 7% compounded monthly.
How long will it take for the account value to reach $15,000?
12t
0.07
The formula for this account will be A = 9, 000 1 +
. To determine when this account’s
12
value will reach $15,000, we will solve A = 15, 000:
12t
0.07
15000 = 9000 1 +
12
12t
15000
0.07
= 1+
9000
12
12t
0.07
5
= 1+
3
12
12t !
5
0.07
ln
= ln
1+
3
12
5
0.07
ln
= 12t ln 1 +
3
12
ln 53
=t
12 ln 1 + 0.07
12
t ≈ 7.3
It will take approximately 7.3 years for this account’s value to reach $15,000.
Example 16: You invest $5,000 into an account that earns 2.25% interest compounded continuously. How long will it take for the account value to triple?
The formula for this account will be A = 5, 000e0.0225t . To determine when this account’s value
will double, we will solve A = 3(5000):
3(5000) = 5000e0.0225t
3 = e0.0225t
ln(3) = 0.0225t
ln(3)
=t
0.0225
t ≈ 48.8
It will take about 48.8 years for this account’s value to triple.
Instructor: A.E.Cary
Page 23 of 31
Math 111 Module 7 Lecture Notes
Example 17: What interest rate (compounded continuously) is required for the value of an investment to double in 15 years?
An account whose interest rate compounds continuously is modeled with the formula A = P ert .
Here, we know that A = 2P when t = 15. We will use this to determine the necessary rate:
2P = P er(15)
2 = er(15)
ln(2) = r(15)
ln(2)
=r
15
r ≈ 0.0462
An interest rate of approximately 4.62% compounded continuously will double in 15 years.
Example 18: What interest rate (compounded annually) is required for the value of an investment
to triple in 15 years?
nt
An account whose interest rate compounds continuously is modeled with the formula A = P 1 + nr .
Here, we know that A = 3P when t = 15 and n = 1. We will use this to determine the necessary
rate:
r 1(15)
3P = P 1 +
1
15
3 = (1 + r)
√
15
3=1+r
√
15
3−1=r
r ≈ 0.0756
An interest rate of approximately 7.56% compounded annually will triple in 15 years.
Instructor: A.E.Cary
Page 24 of 31
Math 111 Module 7 Lecture Notes
Example 19: Julie and Mia each made investments in 1999. Let J(t) be Julie’s investment t
years after 1999 and let M (t) be Mia’s investment t years after 1999. Their respective investments
can be modeled by the functions
J(t) = 50000(1.092)t
and
M (t) = 47000e0.09t
(a) When will Julie’s investment have doubled in value?
To determine when Julie’s investment will have doubled, we will solve J(t) = 2(50, 000):
J(t) = 2(50000)
50000(1.092)t = 2(50000)
(1.092)t = 2
log (1.092)t = log(2)
t log(1.092) = log(2)
t=
log(2)
log(1.092)
t ≈ 7.8757
Therefore Julie’s investment will have doubled in approximately 7.8757 years, or in 2006.
(b) When will Julie and Mia’s investments be equal?
M (t) = J(t)
47000e0.09t = 50000(1.092)t
ln 47000e0.09t = ln 50000(1.092)t
ln(47000) + ln e0.09t = ln(50000) + ln (1.092)t
ln(47000) + 0.09t = ln(50000) + t ln (1.092)
ln(47000) − ln(50000) = t ln (1.092) − 0.09t
47000
= t (ln (1.092) − 0.09)
ln
50000
ln(0.94) = t (ln (1.092) − 0.09)
ln(0.94)
=t
ln (1.092) − 0.09
t ≈ 31.1
In about 31.1 years, Julie and Mia’s investments will have the same value.
Instructor: A.E.Cary
Page 25 of 31
Math 111 Module 7 Lecture Notes
7.3
Biology Applications: Exponential Growth and Decay
7.3.1
Doubling Time
Populations that obey uninhibited growth grow exponentially according to the formula
A(t) = A0 ekt
where k is the continuous growth rate and A0 is the initial amount.
The doubling time for a population is the amount of time it takes a population growing
exponentially to double in size.
Example 20: The colony of Lactobacillus acidophilus bacteria obeys the law of uninhibited
growth. One particular colony is modeled by the function N (t) = 1000e0.009242t , where N (t) is
the number of bacteria after t minutes.1
(a) State the initial amount of bacteria and the continuous rate of growth of the bacteria.
The initial number of bacteria is 1000. The continuous rate of growth is 0.009242, or
0.9242% per minute.
(b) What is the size of the population after 10 minutes?
N (10) = 1000e0.009242(10)
≈ 1097
After 10 minutes, there will be approximately 1097 bacteria.
(c) How long will it take the population to double?
2000 = 1000e0.009242t
2 = e0.009242t
ln(2) = 0.009242t
ln(2)
=t
0.009242
t ≈ 74.9997
The population will double in about 75 minutes.
1
http://textbookofbacteriology.net/growth_3.html
Instructor: A.E.Cary
Page 26 of 31
Math 111 Module 7 Lecture Notes
Example 21: The fruit fly drosphilia have a doubling time of 10 days.2 There are initially 8 fruit
flies.
(a) The population of fruit flies is modeled by N (t) = N0 ekt . Use the doubling time to find the
value of k.
Since it takes 10 days for the population to double, we know that N (t) = 2N0 when t = 10.
We can use these values to solve for t, which is the doubling time:
2N0 = N0 ek(10)
2 = ek(10)
ln(2) = 10k
ln(2)
=k
10
k ≈ 0.069315
(b) What is the continuous growth rate?
The continuous growth rate is about 6.9315% per day.
(c) Write the full formula for N (t).
Exact: N (t) = 8e
ln(2)
·t
10
Approximate: N (t) = 8e0.069315t
(d) How many fruit flies will there be after 30 days?
N (30) = 8e0.069315(30)
≈ 64
After 30 days, there will be approximately 64 fruit flies.
(e) When will there be 1000 fruit flies?
1000 = 8e0.069315t
1000
8e0.069315t
=
8
8
0.069315t
125 = e
ln(125) = 0.069315t
ln(125)
=t
0.069315
t ≈ 69.66
After about 70 days, there will be 100 fruit fruit flies.
2
https://www.lscore.ucla.edu/hhmi/performance/VickiHahmFinal.pdf
Instructor: A.E.Cary
Page 27 of 31
Math 111 Module 7 Lecture Notes
7.3.2
Half-Life
Substances that undergo uninhibited radioactive decay do so exponentially according to
the formula
N (t) = N0 ekt
where k is the continuous decay rate and N0 is the initial amount.
The half-life for a radioactive substance is the amount of time it takes for the quantity of the
substance to be one half its original amount.
Example 22: The half-life of carbon-14 is 5600 years. Write the percentage of carbon-14, A(t),
remaining after t years of decay. Round the value you find for k accurate to six decimal places.
The formula for A(t) will be A(t) = A0 ekt where k is the continuous decay rate and A0 is the
initial value. We know that A(t) = 21 A0 when t = 5600 as the half-life is 5760. We will use this to
determine k:
1
A0 = A0 ek(5600)
2
1
= ek(5600)
2
1
ln
= 5600k
2
ln 12
=k
5600
k =≈ −0.000124
The percentage of carbon-14 remaining after t years can be modeled by A(t) = 100e
approximated, this is written as A(t) = 100e−0.000124t .
Instructor: A.E.Cary
ln(1/2)
t
5600
. When
Page 28 of 31
Math 111 Module 7 Lecture Notes
Example 23: In 1991, two hikers discovered a historic iceman in the Ötztal Alps in Italy.3 Assuming 46% of his carbon-14 was found remaining in the sample, how many years before the sample
was obtained did the iceman die? Use the formula you found in the previous example.
We need to solve A(t) = 46. Using the approximated formula, we would obtain:
46 = 100e−0.000124t
46
= e−0.000124t
100
0.46 = e−0.000124t
ln(0.46) = −0.000124t
ln(0.46)
=t
−0.000124
t ≈ 6262.33
The iceman died about 6262.33 years before the sample was obtained.
Using the exact formula, we would obtain:
46 = 100e
ln(1/2)
t
5600
ln(1/2)
46
= e 5600 t
100
0.46 = e
ln(0.46) =
ln(0.46)
ln(1/2)
5600
ln(1/2)
t
5600
ln(1/2)
t
5600
=t
t ≈ 6273.65
3
http://www.nupecc.org/iai2001/report/B44.pdf
Instructor: A.E.Cary
Page 29 of 31
Math 111 Module 7 Lecture Notes
Example 24: The radioisotope Sodium-24 decays at a continuous rate of about 4.5% per hour.
What is the half-life of this radioactive substance?4
We will use the formula for continuous exponential decay: A(t) = A0 ekt . The value of k is given
indirectly; as Sodium-24 decays at a rate of 4.5% per hour, we know that k = −0.045. Since we
need to determine the half-life, we need to find the value of t when A(t) = 12 A0 :
1
A0 = A0 e−0.045t
2
1
= e−0.045t
2
1
ln
= −0.045t
2
ln 21
=t
−0.045
t ≈ 15.4
The half-life of Sodium-24 is about 15.4 hours.
Example 25: The radioisotope Barium-139 has a half-life of 86 minutes. Find the continuous
rate of decay.
We will use the formula for continuous exponential decay: A(t) = A0 ekt . The value of k is what
we need to determine. As the half-life of Barium-139 is 86 minutes, we know that A(t) = 21 A0
when t = 86. We can use this to solve for k:
1
A0 = A0 ek(86)
2
1
= ek(86)
2
1
ln
= 86k
2
ln 12
=k
86
k ≈ −0.0081
Barium-139 decays at a rate of 0.81% per minute.
4
http://www.ndt-ed.org/EducationResources/HighSchool/Radiography/halflife2.htm
Instructor: A.E.Cary
Page 30 of 31
Math 111 Module 7 Lecture Notes
Example 26: The half-life of Cobalt-60 is 5.27 years.5 . If 15 grams are present now, how many
grams will be present in 100 years?
The first thing we need is the formula for how this radioactive substance decays. The general
formula will be A(t) = A0 ekt . Once we have the formula, we can evaluate that formula for t = 100.
To determine the formula, we first need to find k. We will use the half-life to do this:
1
A0 = A0 ek(5.27)
2
1
= ek(5.27)
2
1
ln
= 5.27k
2
ln 21
=k
5.27
k ≈ −0.1315
As this substance initially has 15 grams, we know A0 = 15. Thus the formula is A(t) = 15e−0.1315t .
Now that the formula has been determined, we can use it to find how many grams of Cobalt-60
would remain after 100 years:
A(100) = 15e−0.1315(100)
≈ 0.000029
After 100 years, only about 0.000029 grams of Cobalt-60 would remain.
5
http://www.bt.cdc.gov/radiation/isotopes/cobalt.asp
Instructor: A.E.Cary
Page 31 of 31