Year Budget in Billions of Dollars
1997 3551999 3902001 4422002 4852004 554
m 0 2 4 6T 12.02 8.08 5.44 3.66
For questions 1 – 3, do NOT use exponential regression.
1.The table below shows the temperature T of a certain liquid after it has been cooled in a freezer for m minutes. Show that the data can be modeled with an exponential function. 2. What is the decay factor per minute? Briefly show work. 3.Write an exponential model for the data.
4.The defense budget for the United States in various years is show in the table at the right. Use exponential regression to find a model for this data using years since 1997. Round all values to the nearest thousandth.
T = 12.02(.82m)
B = 348.454(1.067t) where B = budget in billions of dollars and t = number of years since 1997
.671/2 .82 The decay factor per minute is .82
02457
Remember to download from D2L and print a copy of the Final Group Project.
Date SectionOctober 30 4.4November 4 4.4 Continued
November 6 Review for test 3November 11 Test 3November 13 5.2November 18 5.4November 20 5.5November 25 No Class (Fall Break)November 27 No Class (Fall Break)December 2 5.6December 4 Last Day of Class
Final Group Project dueDec 9 – 15
Final exam week – specific dates and times will be announced in future
Uranium 239 (U239) is an unstable isotope of uranium that decays rapidly. In order to determine the rate of decay, 10 grams were placed in a container, and the amount remaining was measured at 5-minute intervals and recorded in the table below.
Timein minutes
Grams
remaining
0 10
5 8.63
10 7.45
15 6.43
20 5.55
1. Show that an exponential model would be appropriate for this data.
2. What is the decay factor per minute (nearest thousandth)?
3. Find an exponential model for the data.
4. What is the decay rate each minute? What does this number mean in practical terms?
5. Use functional notation to express the amount remaining after 13 minutes and then calculate that value.
6. The half-life of a radioactive element is the time it takes for the mass to decay by half. Use the graphing calculator to find the half-life of U239 to the nearest hundredth of a minute.
U
Periodic Table of Elements
G(t) = 10 )971(. t
Uranium 239 (U239) is an unstable isotope of uranium that decays rapidly. In order to determine the rate of decay, 10 grams were placed in a container, and the amount remaining was measured at 5-minute intervals and recorded in the table below.
Timein minutes
Grams
remaining
0 10
5 8.63
10 7.45
15 6.43
20 5.55
6. The half-life of a radioactive element is the time it takes for the mass to decay by half. Use the graphing calculator to find the half-life of U239 to the nearest hundredth of a minute.
U
Periodic Table of Elements
23.55 minutes
G(t) = 10 )971(. t
Our goal for the rest of this class period is to find an algebraic way of solving the equation
(and others like it)
x = 23.55
5 = 10(.971x)
Answers to even-numbered HW problemsSection 4.3
S-2 Exponential
S-10 y = 5.592(.814x)
Ex 6 t N T N
0 21.3 4 204.4
1 37.5 5 359.7
2 66.0 6 633.1
3 95.3
76.13.21
5.37
76.15.37
0.66
44.10.66
3.95
76.14.204
7.359
76.17.359
3.633
Suspect data point
Our goal for the rest of this class period is to find an algebraic way of solving the equation
(and others like it)
x = 23.55
.5 = .971x
1010
Since the variable is part of the exponent, the only way to isolate the variable is to use logarithms.
5 = 10(.971x)
10x = 10
101 = 10
101 = 10
10x = 100
101 = 10
102 = 100
101 = 10
102 = 100
10x = 40 What is the value of x?
101 = 10
102 = 100
10x = 40
The question can be rephrased:
If 10 is raised to a power and the result is 40, what is the exponent?
What is the value of x?
101 = 10
102 = 100
10x = 40
The question can be rephrased:
If 10 is raised to a power and the result is 40, what is the exponent?
The word logarithm is a synonym for exponent.
101 = 10
102 = 100
10x = 40
The question can be rephrased:
If 10 is raised to a power and the result is 40, what is the exponent?
This question can be rephrased:
If 10 is the base, what is the logarithm that will give a value of 40?
101 = 10
102 = 100
10x = 40
The question can be rephrased:
If 10 is raised to a power and the result is 40, what is the exponent?
This question can be rephrased:
What is log1040 ?
x = log1040
exponent base 10 result
1.6
The common logarithm of a, written log10a, is defined as the power of 10 that gives a.
The word logarithm is a synonym for exponent.
The common logarithm of 40 is 1.6 because using base 10, the exponent needed to get a value
of 40 is 1.6.
4010 6.1 log1040 = 1.6
Can we get a more accurate value for log1040 ?
log1040 1.602059991
Is it true that 25 = 32 ?
log232 = 5
Write an equivalent statement using logarithms.
Identify the exponent, the base, and the result.
5 2 32
Is it true that 102.68 = 478.63 (approximately)?
log10 478.63 = 2.68
Write an equivalent statement using logarithms.
Identify the exponent, the base, and the result.
log 478.63 = 2.68
When no base is indicated, it is understood to be 10.
2.68 10 478.63
When the base of a logarithm is 10, we call it a common logarithm, and we do not indicate the base.
log1040 is the same as log 40.
Find each of the following common logarithms on a calculator. Round to four decimal places.
a) log 723,456
b) log 0.0000245
c) log (4)
When the base of a logarithm is 10, we call it a common logarithm, and we do not indicate the base.
log1040 is the same as log 40.
Write an equivalent exponential equation for each.
ERR: nonreal ans
104.6108 = .0000245= 4.6108105.8594 = 723,456= 5.8594
An important property of logarithms:
log (ab) = b (log a)
Illustration:
Does log (27) = 7(log 2)?
An important property of logarithms:
log (ab) = b (log a)
Illustration:
Does log (27) = 7(log 2)?
Take the logarithm on both sides
.5 = .971x
log(.5) = log(.971x)
log(.5) = x log (.971)
-.3010
x = 23.55
Divide both sides by 105 = 10(.971x)
Now let’s solve the equation
Apply the property
x = = 23.52-.3010
-.0128
Compute the values
Solve for x
Take the logarithm on both sides
.5 = .971x
log(.5) = log(.971x)
log(.5) = x log (.971)
x = 23.55
Divide both sides by 105 = 10(.971x)
Now let’s solve the equation
log(.971)
log(.5)= x
Apply the property
Solve for x
23.55 = x
Solve algebraically to the nearest thousandth
)7.3(484.734 x
Solve algebraically to the nearest thousandth
)7.3(484.734 x 48 48
15.3 = )7.3( x
log(15.3) = log )7.3( x
log(15.3) = x log (3.7)
log(15.3)
log (3.7)= x
2.085 = x
Section 4.4 (Do not read)
Handout – Common Logs
(correction to question 18. It should read
“Solve question 17 algebraically.”)
The table below shows the average weekly amount of electricity used in five Michigan cities in 2011.
Population of city in thousands
Amount of electricity
in 1000’s of kilowatts
73.5 365
85.4 395
97.2 442
108.0 503
120.9 584
Additional Practice
1. Make a scatter plot of the data. Based on the graph, would an exponential model be appropriate?
2.Write an equation for an exponential model for the weekly amount of electricity used versus the population. Round coefficients to three decimal places (nearest thousandth).
3. Graph the exponential model.
4. What is the growth rate in weekly electricity use per 1,000 people?
5. Use your model to estimate the weekly amount of electricity used in 2011 in a Michigan city with a population of 90,000 people. 6. The weekly amount of electricity used by the people of another Michigan city in 2011 was 660 thousand kilowatts. Use your graph to estimate the population of the city that year. 7. Solve question 6 algebraically.