Section 6.9 solutions
1) The data shows the cooling temperatures of a freshly brewed cup of coffee after it is poured from the
brewing pot into a serving cup. The brewing pot temperature is approximately 180º F.
Time
0
(minutes)
Temperature 179.5
(Fahrenheit)
5
8
11
15
18
22
25
30
34
164
154
149.2
141.7
134.6
125.4
123.5
120
118.5
1a) plot the data using a graphing calculator
1b) Decide the type of regression model that is appropriate
This looks like exponential decay. The data is not really linear as it starts to flatten out at the end. I will
use Exponential regression. (ExpReg on my TI-84)
Answer: Exponential Regression
1c) Find the appropriate regression equation, and write it in the form P(t) = P0ekt
expReg with my TI-84 gives the equation:
y = 172.544(.988x) , the x stands for time, so I will change the x to t to make the next steps easier to
follow
y = 172.454(.988t)
I need to rewrite this equation in the form P(t) = P0ekt
to do this I need to find values for P0 and k.
These are the formal steps to rewrite the equation
in the desired form.
Step 1: find a value for P0
Set the calculator equation equal to the desired
equation
172.544(.988)t = P0ekt
If we let t = 0 we can solve for P0
172.544(.988)0 = P0ek*0
172.544(1) = P0e0
172.544(1) = P0(1)
172.544 = P0
We can now change the P0 in equation marked *
P(t) = 172.544ekt
Step 2 find a value for k
Set the calculator equation equal to the desired
form with the correct P0
172.544(.988)t = 172.544ekt (divide by 172.544 to
get
.988t = ekt (this must be true for all values of t, so
now let t = 1)
.9881 = ek*1
.988 = ek (take ln of each side)
Ln(.988) = ln ek
-.012 = k
Desired equation P(t) = 172.544e-.012t
Answer: P(t) = 172.544e-.012t
There are shortcuts that can be used to find the P0
and k.
Shortcut method:
Step 1: find the value of P0
P0 equals the value of a from the exponential
regression equation obtained on my calculator.
P0 = 172.454
Step 2:
k equals ln(b), where b is the value from the
exponential regression obtained on my calculator
k = ln(.988)
k = -.012
1d) Predict when the coffee temperature will be 115 degrees Fahrenheit.
Replace the P(t) with 115 and solve for t.
115 = 172.544e-.012t (divide by 172.544 and round to 3 decimals)
.666 = e-.012t (take ln of each side)
Ln(.666) = ln e-.012t (use power to product rule)
Ln(.666) = -.012t(ln e) (ln e = 1_
Ln(.666) = -.012t
33.87
Answer: 33.87 minutes
Section 6.9: Building Exponential and Logarithmic models
3) The following is a table comparing the number of people who got sick after exposure to a certain
virus.
Days
1
after
exposure
Number 21
of
people
sick
2
3
4
5
6
45
96
208
450
971
3a) Plot the data using a graphing calculator
3b) Decide the type of regression model that is appropriate
This looks like an exponential growth function
Answer: exponential regression
3c) Find the appropriate regression equation
Calculator answer: y = 9.696(2.154t) (a = 9.696 and b = 2.154)
I want to write this in the form P(t) = P0ekt
I will use the shortcut approach
P0 = k (so P0= 9.696)
k = ln(b) (so k = ln(2.154) = .767)
Answer: P(t) = 9.696e.767t
3d) Predict when 2000 people will be sick
Replace P(t) with 2000 and solve for t.
2000 = 9.696e.767t (divide by 9.696)
206.271 = e.767t (take ln of each side)
Ln(206.271) = ln e.767t (use power to product rule)
Ln(206.271) = .767t(ln e )
Ln(206.271) = .767t
ln⁡(206.271)
.767
=𝑡
Answer: 6.95 days
5) A chemist has a 50-gram sample of a radioactive material. He records the amount of radioactive
material present every week for 6 weeks and obtains the following data
week
Weight
(grams)
0
50
1
44
2
38
3
33
4
29
5
26
6
23
5a) Plot the data using a graphing calculator
5b) Decide the type of regression model that is appropriate
The data almost looks linear. However, I know that the exponential regression is appropriate for
radioactive decay problems. The linear regression might by okay.
Answer: exponential regression
5c) Find the appropriate regression equation
y = 49.616(.878t)
now use shortcut to find P0 and k
P0 = a = 49.616
k = ln(b) = ln(.878) = -.130
Answer: P(t) = 49.616e-.130t
5d) Predict when there will be 10 grams of the material left.
10 = 49.616e-.130t (divide by 49.616)
.202 = e-.130t
Ln(.202) = ln(e-.130t)
Ln(.202) = -.130t(ln e)
Ln (.202) = -.130t
ln⁡(.202)
−.130
=𝑡
Answer: 12.30 weeks
7) The data below show the average growth rates of 12 Weeping Higan cherry trees planted in
Washington, D.C. At the time of planting, the trees were one year old and were all 6 feet in height.
Age of
tree in
years
height in
feet
1
2
3
4
5
6
7
8
9
10
6
9.5
13
15
16.7
17.5
18.5
19
19.5
19.7
7a) Plot the data using a graphing calculator
7b) Decide the type of regression model that is appropriate
This is not exponential in nature, logarithmic regression is appropriate
Answer: logarithmic regression
7c) Find the appropriate regression equation
Answer: y = 5.946 + 6.286ln(t) or P(t) = 5.946 + 6.286ln(t)
7d) Predict when the trees will be 20 feet in height.
20 = 5.946 + 6.286ln(t) (subtract 5.946 from each side)
14.054 = 6.286ln(t) (divide by 6.286)
2.236 = ln (t)
2.236 = loge(t)
e2.236 = t
Answer: 9.36 years