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

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. 

The table below shows the average weekly amount of electricity used in five Michigan cities in 2011.

y = 169.973(1.010x) where x = population and y = amount of electricity used.

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. 

1.0%

Approximately 416.2 thousand kw are used.

The table below shows the average weekly amount of electricity used in five Michigan cities in 2011.

y = 169.973(1.010x) where x = population and y = amount of electricity used.

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. 

The population is about 136,337.

y = 169.973(1.01x)

660 = 169.973(1.01x) 169.973 169.973

3.8830 = 1.01x

log(3.8830) = log(1.01x)

log(3.8830) = x log(1.01)

Therefore, the population of the Michigan city that generated 660 tons of recycled paper was about 136,338 people.

x = log(3.8830)

log(1.01)= 136.338

The population is about 136,337.

Answers to Common Logarithm Homework Handout

Find the value of each of the following. Round your answer to four decimal place accuracy.

1. log10450 2.6532 2. log10(.75) –.1249 3. log 29.8 1.4742

Write each of the following as an exponential equation and find the value of x to the nearest hundredth.

4. log10x = 3 5. log1030 = x 6. log x = -2

103 = x

x = 1000

10x = 30

x = 1.48

10-2 = x

x = .01

Solve each equation for x. Answer to the nearest thousandth. Show work.

7. 8. 9. 5910x )4.2(3.57500 x

x = log 59

x = 1.771

log log

x (log 8.3) = log 2030

x = = 3.599

2030(8.3 )x

log 2030log 8.3

119.0476 = 3.57x

log 119.0476 = log (3.57x)

log 119.0476 = x (log 3.57)

x = = 3.756log 119.0476 log 3.57

4.2 4.2

2030(8.3 )x

The average temperature of the gas emitted by stars decreases as the gas gets further from the surface of the star. Below is a table showing the distance D from the surface of a certain star (in thousands of miles) and the average temperature T of its gas emissions (in thousands of degrees Fahrenheit).

D = Distance in thousands of miles

0 5 10 15 20

T = Temperature in thousands of degrees

12.5 11.7 10.95 10.3 9.7

Use exponential regression to find an exponential model for the data.

10. What is the rate of decrease per thousand miles in the average temperature of the gas emitted by the star?

11. What is the average temperature of the gas emitted by the star when the gas is 12,500 miles from the star?

12. When the average temperature of the gas is 7,000 degrees, how many miles from the star has it traveled? Give a graphic solution.

13. Verify your answer to question 12 by giving an algebraic solution.

The average temperature of the gas emitted by stars decreases as the gas gets further from the surface of the star. Below is a table showing the distance D from the surface of a certain star (in thousands of miles) and the average temperature T of its gas emissions (in thousands of degrees Fahrenheit).

D = Distance in thousands of miles

0 5 10 15 20

T = Temperature in thousands of degrees

12.5 11.7 10.95 10.3 9.7

Use exponential regression to find an exponential model for the data.

The exponential regression model is T = 12.472(.987D)

The average temperature of the gas emitted by stars decreases as the gas gets further from the surface of the star. Below is a table showing the distance D from the surface of a certain star (in thousands of miles) and the average temperature T of its gas emissions (in thousands of degrees Fahrenheit).

D = Distance in thousands of miles

0 5 10 15 20

T = Temperature in thousands of degrees

12.5 11.7 10.95 10.3 9.7

Use exponential regression to find an exponential model for the data.

10. What is the rate of decrease per thousand miles in the average temperature of the gas emitted by the star?

T = 12.472(.987D)

1 – .987 = .013 The rate of decrease is 1.3%

The average temperature of the gas emitted by stars decreases as the gas gets further from the surface of the star. Below is a table showing the distance D from the surface of a certain star (in thousands of miles) and the average temperature T of its gas emissions (in thousands of degrees Fahrenheit).

D = Distance in thousands of miles

0 5 10 15 20

T = Temperature in thousands of degrees

12.5 11.7 10.95 10.3 9.7

11. What is the average temperature of the gas emitted by the star when the gas is 12,500 miles from the star?

The average temperature of gas 12,500 miles from the star is 10,590 degrees.

The average temperature of the gas emitted by stars decreases as the gas gets further from the surface of the star. Below is a table showing the distance D from the surface of a certain star (in thousands of miles) and the average temperature T of its gas emissions (in thousands of degrees Fahrenheit).

D = Distance in thousands of miles

0 5 10 15 20

T = Temperature in thousands of degrees

12.5 11.7 10.95 10.3 9.7

12. When the average temperature of the gas is 7,000 degrees, how many miles from the star has it traveled? Give a graphic solution.

The temperature is 7,000 degrees at a distance of 44,140 miles from the star.

13. Verify your answer to question 12 by giving an algebraic solution.

T = 12.472(.987D)

7 = 12.472(.987D)

.5613 = (.987D)

log(.5613) = log(.987D)

log(.5613) = Dlog(.987)

D = = 44.13 log(.5613)

log(.987)

The temperature is 7,000 degrees at a distance of 44,130 miles from the star.

131.77 mg remain after 2 hours

1-.989=.011 or 1.1%

y = 496.888(.989x) where x = time in minutes and y = # of mg remaining in blood

Time(in minutes)

10 20 30 40 50 60

Amount of active ingredient left in bloodstream (in mgs)

445 402 360 320 290 260

14. Write an equation for an exponential model for the amount of active ingredient left in the bloodstream. Round all coefficients to three decimal places (nearest thousandth).

15. What is the rate of decrease of the active ingredient in the bloodstream?

16. How much of the 500 mg of the active ingredient is left in the bloodstream after 2 hrs?

17. After how many minutes will exactly 100 mg of the active ingredient be left in the bloodstream? Round your answer to the nearest hundredth of a minute.

18. Solve question 17 algebraically.

144.94 min

Squibb Pharmaceuticals is testing a new pain reliever that comes in pill form, each pill containing 500 mg of the active ingredient. In order to determine how long it takes for the active ingredient to leave the blood stream and be absorbed into the body, the company conducts clinical trials. In the trials, blood samples are taken from subjects who have taken one of the pills. The table below shows the average amount of the active ingredient that remains in the bloodstream after each 10 minute interval.

100

100 = 496.888(.989x)

.2013 = .989x

log(.2013) = log(.989x)

Time (in minutes) 10 20 30 40 50 60

Amount of active ingredient left in bloodstream (in mgs)

445 402 360 320 290 260x = 144.94

496.888496.888

)989log(.

)2013log(.x = = 144.92

log(.2013) = x (log.989)

Natural Logarithms use a different base, the number e.

Logarithms using base e are called natural logarithms. The abbreviation “ln” is generally used for natural logarithms. Thus,

Last time, we examined common logarithms (logarithms using base 10).

Use your calculator to find the value of e3.69 to the nearest whole number.

e3.69 40 Express this equation using logarithms

ln 40 3.69

ln x means loge x.

yx 10log xy 10 log

Use the ln button on the calculator to verify this last answer.

100 = 496.888(.989x)

.2013 = .989x

log(.2013) = log(.989x)

Time (in minutes) 10 20 30 40 50 60

Amount of active ingredient left in bloodstream (in mgs)

445 402 360 320 290 260x = 144.94

496.888496.888

)989log(.

)2013log(.x = = 144.92

log(.2013) = x (log.989)

100 = 496.888(.989x)

.2013 = .989x

log(.2013) = log(.989x)

Time (in minutes) 10 20 30 40 50 60

Amount of active ingredient left in bloodstream (in mgs)

445 402 360 320 290 260x = 144.94

496.888496.888

)989log(.

)2013log(.x = = 144.92

log(.2013) = x (log.989)ln

ln

ln

ln ln

ln

Year Census Population

1850 1,6101860 4,3851870 5,7281880 11,1831890 50,3951900 102,4791910 319,1981920 576,6731930 1,238,048

Source: http://en.wikipedia.org/wiki/Los_Angeles

Shown is a table of the population of the City of Los Angeles as reported in the census from 1850 to 1930.

1.Make a scatter plot of the population using years since 1850.

2.Would an exponential model be appropriate for the data?

3. Find an exponential model for the data.

r is called the correlation coefficient.

r2 is called the coefficient of determination.

Both r and r2 are measures of how good a fit a model is to a set of data.

r will always be between -1 and 1, and r2 will always be between 0 and 1.

The closer they are to -1 or 1, the better the fit.

Year 2002 2003 2004 2005 2006

Surface Area in sq.

miles

26,579 29,104 31,868 34,896 38,211

Homework:

Download, print, and complete the Practice Test for Chapter 4.