Lab 12 Solutions
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Course
1018
Subject
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Date
Jan 9, 2024
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7
Uploaded by EarlHeatBeaver31
Fall 2023
MATH 1018: Pre-Calculus in Practice
Lab 12
More Exponential Functions and Logarithmic Functions
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Instructions
•
Form groups of 1–4 and write the group-members names in the space above.
Only one
submission per group is allowed. All group members should agree on an answer before it is
written down. Groups can change from week to week. All group members will receive the
same mark.
•
Make sure to write down the steps of your work unless otherwise instructed.
Final answers
without the steps shown are not sufficient to earn full completion credit.
•
You may choose to use a calculator for any lab problems. However you are encouraged not
to use a calculator for most lab problems since you will not be allowed to use a calculator
during quizzes or the final exam.
Intended Learning Outcomes
In this lab, you will be practicing the following skills in a modeling setting.
•
Solve simple equations involving exponential and logarithmic expressions
•
Graph logarithmic functions
•
Work with common models build on exponential functions
MATH 1018: Pre-Calculus in Practice
Page 1
Examples
1. Tony invested $40000 at the beginning of 2010. Unfortunately his investment has been losing
value at the rate of 2.5% per year.
(a) Find a formula for the value of his investment as a function of time.
Solution:
Let
V
(
t
)
be the value of Tony’s investment
t
years after the beginning of
2010. Then
V
(
t
) =
ab
t
for some
a
̸
= 0
, b >
0
, b
̸
= 1
.
The initial investment was $40000, so
a
= 40000
. Since the investment loses 2.5% of
its value each year, its value would get multiplied by a factor of 0.975 each year, so
b
= 0
.
975
. Therefore
V
(
t
) = 40000(0
.
975)
t
.
(b) Change the base of the exponential function found in part (a) to
e
. Base on the result,
if we consider the decrease of value as a continuous process, what would the continuous
decay rate be?
Solution:
We are trying change the base of the exponential function
V
(
t
)
, so we want
to rewrite it as
V
(
t
) = 40000
e
rt
for some value
r
instead.
Then we need to convert
0
.
975
to a base-
e
number first. Note that
0
.
975 =
e
ln(0
.
975)
.
Therefore,
V
(
t
) = 40000
e
ln(0
.
975)
t
.
The continuous decay rate is
−
ln(0
.
975)
, or approximately
2
.
53%
.
(c) Now suppose Steve invested $40000 at the beginning of 2015. This investment has also
been decreasing in value at the rate of 3.1% per year. When will the value Tony’s invest-
ment and Steve’s investment equal in value?
Solution:
Recall that back in part (a),
t
was defined as the number of years after
2010. Here we use the same definition of
t
, then by the same logic, the value of Steve’s
investment,
S
(
t
)
, can be modelled as
S
(
t
) = 40000(0
.
969)
t
−
5
, for
t
≥
5
.
Then we are trying to solve for
t
when
40000(0
.
975)
t
= 40000(0
.
969)
t
−
5
.
MATH 1018: Pre-Calculus in Practice
Page 2
40000(0
.
975)
t
= 40000(0
.
969)
t
−
5
(0
.
975)
t
= (0
.
969)
t
−
5
(0
.
969
log
0
.
969
(0
.
975)
)
t
= (0
.
969)
t
−
5
0
.
969
log
0
.
969
(0
.
975)
t
= (0
.
969)
t
−
5
log
0
.
969
(0
.
975)
t
=
t
−
5
(1
−
log
0
.
969
(0
.
975))
t
= 5
t
=
5
1
−
log
0
.
969
(0
.
975)
Their investments would have the same value
5
1
−
log
0
.
969
(0
.
975)
years after the be-
ginning of 2010.
2. Noise level in decibels (dB) is measured by comparing its sound intensity, I, to a benchmark
sound
I
0
with intensity
10
−
16
watts/cm
2
.
In particular, the following formula is used to
calculate the decibel rating of a sound.
Noise Level of sound in dB
= 10 log
I
I
0
(a) If a sound doubles in sound intensity, how much would its its decibel rating increase?
Simplify you answer as much as possible. The final answer should be one single term.
Solution:
Suppose the noise has intensity
I
1
originally, then after doubling, its in-
tensity would become
2
I
1
.
Its original decibel rating is
10 log
I
1
I
0
, and its new decibel rating is
10 log
2
I
1
I
0
.
Then the noise level increased by
MATH 1018: Pre-Calculus in Practice
Page 3
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10 log
2
I
1
I
0
−
10 log
I
1
I
0
=10
log
2
I
1
I
0
−
log
I
1
I
0
=10 log
2
I
1
I
0
·
I
0
I
1
=10 log(2)
Therefore, the noise level increased by
10 log(2)
dB.
(b) Loud music can measure 110 dB whereas normal conversation measures 50 dB. How many
times more intense is loud music than normal conversation?
Solution:
Let
I
L
be the sound intensity of loud music, and let
I
N
be the sound
intensity of normal conversation.
Then
10 log
I
L
I
0
= 110
and
10 log
I
N
I
0
= 50
.
Subtracting these two requations gives
10 log
I
L
I
0
−
10 log
I
N
I
0
= 60
10
log
I
L
I
0
−
log
I
N
I
0
= 60
10 log
I
L
I
0
·
I
0
I
N
= 60
10 log
I
L
I
N
= 60
log
I
L
I
N
= 6
I
L
I
N
= 10
6
Hence loud music is
10
6
times as intense as normal conversation.
MATH 1018: Pre-Calculus in Practice
Page 4
Problems
General Guidelines for Solving a Modeling Problem
(i) Identify what the problem is asking for. The quantity in the final question usually needs to
be set up as a variable.
(ii) Define variables in full sentence(s), with units, if not given by the problem.
(iii) Interpret the conditions given in the problem as a formula/equation.
(iv) Solve the problem mathematically.
(v) Write the final conclusion in a full sentence in response to the question asked by the problem.
1. A cheesecake is taken out of the oven with an ideal internal temperature of 165 degrees Fahren-
heit, and is placed into a 35 degree refrigerator. After 10 minutes, the cheesecake has cooled
to 150 degrees. If you must wait until the cheesecake has cooled to 70 degrees before you eat
it, how long will you have to wait?
Use Newton’s law of cooling,
T
(
t
) =
ae
kt
+
T
s
to model this problem, where
T
is the temperature
of an object,
T
s
is the temperature of the surrounding air, and
a
and
k
are constants.
Solution:
By Newton’s law of cooling, we get
T
(
t
) =
ae
kt
+ 35
.
Then
T
(0) =
ae
0
+ 35 = 165
, or
a
= 165
−
35 = 130
. So we have
T
(
t
) = 130
e
kt
+ 35
.
Also
T
(10) = 130
e
10
k
+ 35 = 150
, which gives
e
10
k
=
115
130
=
23
26
.
This means
10
k
=
ln
23
26
, or
k
=
1
10
ln
23
26
.
This gives
T
(
t
) = 130
e
t
ln(23
/
26)
/
10
+ 35
.
Now we set
T
(
t
) = 70
and solve for
t
.
130
e
t
ln(23
/
26)
/
10
+ 35 = 70
130
e
t
ln(23
/
26)
/
10
= 35
e
t
ln(23
/
26)
/
10
=
7
26
1
10
t
ln
23
26
= ln
7
26
t
=
10 ln(7
/
26)
ln(23
/
26)
MATH 1018: Pre-Calculus in Practice
Page 5
You must wait for
10 ln(7
/
26)
ln(23
/
26)
minutes.
2. The population of a city is increasing at the rate of 3.2% per year, since the year 2000. Its
population in 2015 was 235000 people.
(a) Find a formula for the population of this city as a function of time.
Solution:
Let
f
(
t
)
be the population of this city
t
years after 2000. Then
f
(
t
) =
ab
t
for some
a
̸
= 0
, b >
0
, b
̸
= 1
.
The population increases at the rate of 3.2% per year, so
b
= 1 + 3
.
2% = 1
.
032
, and
f
(
t
) =
a
(1
.
032)
t
.
Moreover
f
(15) =
a
(1
.
032)
15
= 235000
, so
a
=
235000
(1
.
032
15
≈
146512
. (The approxima-
tion is not required when you have no access to a calculator.)
Therefore
f
(
t
) =
235000
1
.
032
15
(1
.
032)
t
.
(b) Change the base of the exponential function found in part (a) to
e
. Base on the result,
if we consider population increase as a continuous process, what would the continuous
growth rate be?
Solution:
Note that
1
.
032 =
e
ln(1
.
032)
, then
f
(
t
) =
235000
1
.
032
15
(
e
ln(1
.
032)
)
t
=
235000
1
.
032
15
e
ln(1
.
032)
t
.
Hence the continuous growth rate would be
ln(1
.
032)
.
3. Kelly and Rorie are participating in a 5-kilometre run. The distance between Kelly and the
finish line can be modelled using
f
(
t
) = 2 log
4
(
−
t
+ 32)
and the distance between Rorie and
the finish line can be modelled using
g
(
t
) =
−
2 log
4
(2
t
+ 2) + 6
, where
t
is the number of
minutes since the run started.
Assume Rorie is always ahead of Kelly during the run. At what time(s) are they exactly 2
kilometres apart?
Solution:
If Rorie is ahead, then Rorie’s distance to the finish line is less than Kelly’s
distance to the finish line.
We would be looking for the time
t
(in minutes) when
f
(
t
)
−
g
(
t
) = 2
.
MATH 1018: Pre-Calculus in Practice
Page 6
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2 log
4
(
−
t
+ 32)
−
(
−
2 log
4
(2
t
+ 2) + 6) = 2
2 log
4
(
−
t
+ 32) + 2 log
4
(2
t
+ 2)
−
6 = 2
2 log
4
(
−
t
+ 32) + 2 log
4
(2
t
+ 2) = 8
log
4
(
−
t
+ 32) + log
4
(2
t
+ 2) = 4
log
4
((
−
t
+ 32)(2
t
+ 2)) = 4
(
−
t
+ 32)(2
t
+ 2) = 256
−
2
t
2
+ 62
t
+ 64 = 256
−
2
t
2
+ 62
t
−
192 = 0
t
2
−
31
t
+ 96 = 0
By the quadratic formula,
t
=
31
±
√
961
−
384
2
=
31
±
√
577
2
.
Therefore, the distance between them is 2km when it was
31 +
√
577
2
minutes and
31
−
√
577
2
minutes after the run started.
MATH 1018: Pre-Calculus in Practice
Page 7