Mid-Module Assessment Review

Today, we will review for the Module 1 Mid- Module Assessment

1. The number of Direct TV subscribers has increased significantly since 2005. In fact, the number of subscribers has doubled each year. It was reported that in 2009 there were 2 million subscribers to Direct TV. a. Assuming the number of subscribers continues to double each year, for the next three years, determine the number of subscribers in 2010, 2011, 2012. 2010=2 2 =4 million subscribers. 2011=2 4 =8 million subscribers. 2012=2 8 =16 million subscribers.

# of Subscribers in millions b. Assume the trend in numbers of subscribers doubling each year was true for all years from 2005 to 2013. Complete the table below using 2009 as year 1 with 2 million as the number of subscribers that year. Year -3 -2 -1 1 2 3 4 5 # of Subscribers in millions 1 8 1 4 1 2 8 16 32

c. Given only the number of subscribers in 2009 and the assumption that the number of subscribers doubles each year, how did you determine the number of subscribers for years 2, 3, 4, and 5? Since 2009 is considered year 1, year 2 would be 2010, year 3 would be 2011 and so on. We’re told the number of subscribers doubles each year, this means in year 2 there are two times as many subscribers or 2(2) =4. To find the number of subscribers in the next year you just need to multiply the number of subscribers in the previous year by 2. d. Given only the number of subscribers in 2009 and the assumption that the number of subscribers doubles each year, how did you determine the number of users for years 0, -1, -2, and -3? Again, since 2009 is considered year 1, then the year before that, 2008 would be considered year 0, the year before that, 2007 would be year −1 and so on. To find out the number of subscribers in year 0 we would divide the number of subscribers in year 1 by 2, so 2÷2=1. To find the number of subscribers for each previous year, just divide the number of subscribers from the year after by 2.

e. Write an equation to represent the number of subscribers in millions, 𝑁, for year, 𝑡, 𝑡≥−3. f. Using the context of the problem, explain whether or not the equation you came up with in part (e) would work for finding the number of subscribers in millions in year t, for all 𝑡≤0. 𝑁= 2 𝑡 , 𝑡≥−3 In the context of this problem, the value of 𝑡≤0 represents the years before 2009. Since eventually you would get a value of t that would correspond to a year before Directv was around, this equation wouldn’t make sense for all 𝑡≤0.

*g. Assume that the total number of subscribers continues to double each year after 2013. Determine the number of subscribers in 2015. Given that the U.S. population at the end of 2014 was approximately 319 million, is this assumption reasonable? Explain your reasoning. Assuming the total number of subscribers continues to double after the year 2013, Directv would have 𝑁= 2 7 =128 million subscribers. Given the population of the U.S. at the end of 2014 was 319 million, this scenario is entirely possible, but not likely.

2. Let m be a whole number. a. Use the properties of exponents to write an equivalent expression that is a product of unique primes, each raised to an integer power. 15 24 ∙ 21 8 35 8 = (3∙5) 24 ∙ 3∙7 8 (5∙7) 8 = 3 24 ∙ 5 24 ∙ 3 8 ∙ 7 8 5 8 ∙ 7 8 = 3 32 ∙ 5 24 ∙ 7 8 5 8 ∙ 7 8 = 3 32 1 ∙ 5 24 5 8 ∙ 7 8 7 8 = 3 32 ∙ 5 24−8 ∙ 7 8−8 = 3 32 ∙ 5 16 ∙ 7 0 = 3 32 ∙ 5 16

b. Use the properties of exponents to prove the following identity: 15 3𝑚 ∙ 21 𝑚 35 𝑚 = 3 4𝑚 ∙ 5 2𝑚 LHS = (3∙5) 3𝑚 3∙7 𝑚 5∙7 𝑚 = 3 3𝑚 ∙ 5 3𝑚 ∙ 3 𝑚 ∙ 7 𝑚 5 𝑚 ∙ 7 𝑚 = 3 4𝑚 ∙ 5 3𝑚 ∙ 7 𝑚 5 𝑚 ∙ 7 𝑚 = 3 4𝑚 ∙ 5 3𝑚−𝑚 ∙ 7 𝑚−𝑚 = 3 4𝑚 ∙ 5 2𝑚 *c. What value of m could be substituted into the identity in part (b) to find the answer to part (a)? Since 4𝑚=32 and 2𝑚=16 that means 𝑚=8

3. a. James writes 3 4 ∙ 9 2 = 27 6 and the teacher marked it wrong 3. a. James writes 3 4 ∙ 9 2 = 27 6 and the teacher marked it wrong. Explain James’ error. It appears that James multiplied the base of 3 and 9 together which is incorrect. Also, you can not add exponents of bases that are not the same. James should have recognized that 9 2 = (3 2 ) 2 = 3 4 . Therefore he should have written: 3 4 ∙ 3 4 = 3 8 .

b. Find n so that the equation below is true: 3 3 ∙ 9 3 = 3 3 ∙ 3 𝑛 = 3 9 Since 9 3 = (3 2 ) 3 = 3 6 , we can substitute 3 6 in for 9 3 . Therefore 𝑛=6 would make the equation true.

c. Use the definition of exponential notation to demonstrate why 3 3 ∙ 9 3 = 3 9 . LHS =3 3 ∙ (3 2 ) 3 = 3 3 ∙ 3 6 =3∙3∙3∙3∙3∙3∙3∙3∙3 = 3 9

d. You write 5 3 ∙ 5 −8 = 5 −5 . Brent doesn’t believe you and screams “Prove it!” Show directly why your answer is correct without referencing the Laws of Exponents for integers, i.e. 𝑥 𝑎 ∙ 𝑥 𝑏 = 𝑥 𝑎+𝑏 for positive numbers x and integers a and b. LHS RHS = 5 3 ∙ 1 5 8 = 1 5 5 =5∙5∙5∙ 1 5 ∙ 1 5 ∙ 1 5 ∙ 1 5 ∙ 1 5 ∙ 1 5 ∙ 1 5 ∙ 1 5 = 1 5∙5∙5∙5∙5 = 5∙5∙5 5∙5∙5∙5∙5∙5∙5∙5 = 1 5∙5∙5∙5∙5