Lesson 18
Expressed in Different Ways
18.1: Math Talk: Equal Expressions (5 minutes)
Warmup
This warmup prompts students to use properties of exponents to identify equal expressions. This skill will be critical as students investigate different ways to write expressions for compounded interest in the lesson.
Launch
Display one expression at a time. Give students quiet think time for each expression and ask them to give a signal when they have an answer and a strategy. Keep all expressions displayed throughout the talk. Follow with a wholeclass discussion.
Supports accessibility for: Memory; Organization
Student Facing
Decide if each expression is equal to \((1.21)^{100}\).
\(\left((1.21)^{10}\right)^{10}\)
\(\left((1.21)^{50}\right)^{50}\)
\(\left((1.1)^2\right)^{100}\)
\((1.1)^{200}\)
Student Response
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Activity Synthesis
Pay particular attention to the last expression, \((1.1)^{200}\). In order to identify this as equal to \((1.21)^{100}\) students need to work backward and write this as \(\left((1.1)^2\right)^{100}\). Another approach would be to work forward and rewrite \((1.1)^{200}\) as \(1.1^{2 \boldcdot 100}\) The third expression is intended to facilitate this thinking. All of these problems rely on an important property of exponents, \(\left(x^a\right)^b = x^{ab}\).
Design Principle(s): Optimize output (for explanation)
18.2: Population Projections (15 minutes)
Activity
This activity allows students to practice interpreting and using exponential expressions to solve problems, as well as writing equivalent expressions that allow new insights into the situation. It examines the percent change of the United States population from its beginning through the Civil War. During this time frame, the population happened to be modeled remarkably well by an exponential function.
Students are given a population model where time is measured in years. They then write different models where time is measured in decades or centuries. While the century model is primarily of theoretical interest, the decade model is very appropriate for showing the overall growth rate of the United States during this period, reinforcing the compounding effects of exponential growth.
In the activity synthesis, the term growth rate is defined.
Making graphing technology available gives students an opportunity to choose appropriate tools strategically (MP5).
Launch
Arrange students in groups of 2 and provide access to scientific calculators.
If time is limited, consider asking students to complete the first question together with their partner, divide up the work on the last two questions, and then discuss their responses.
Student Facing
 From 1790 to 1860, the United States population, in thousands, is modeled by the equation \(P = 4,\!000 \boldcdot (1.031)^t\) where \(t\) is the number of years since 1790.
 About how many people were living in the U.S. in 1790? What about in 1860? Show your reasoning.
 What is the approximate annual percent increase predicted by the model?
 What does the model predict for the population in 2017? Is it accurate? Explain.

 What percent increase does the model predict each decade? Explain.
 Suppose \(d\) represents the number of decades since 1790. Write an equation for \(P\) in terms of \(d\) modeling the population in the US (in thousands).

 What percent increase does the model predict each century? Explain.
 Suppose \(c\) represents the number of centuries since 1790. Write an equation for \(P\) in terms of \(c\) modeling the population in the United States (in thousands).
Student Response
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Anticipated Misconceptions
Students may miss that the problem specifies the \(P\) value from the equation represents thousands of people. Suggest a closer reading and thinking about whether a population for the country in 1790 of 4,000 makes sense.
Activity Synthesis
Focus the discussion on how students found the percent increase per decade and per century. Ask students:
 "By what percentage did the U.S. population increase per decade?" (\((1.031)^{10} \approx 1.36\), so the population grew by around 36% each decade.)
 "By what percentage did it increase per century?" (The growth factor, \((1.031)^{100}\), is about 21.18. Subtracting 1 for the original 100% of the population gives 20.18. This means it grew by 2,018%.)
Explain that in situations characterized by exponential change by a percentage, people sometimes refer to the percent change as a growth rate. (In previous lessons, this idea was called interest rate in the specific context of interest on savings or debt, but growth rate is a more general term.)
Make sure students understand how growth rate is related to growth factor. For example, in the equation \(P = 4,\!000 \boldcdot (1.031)^t\), the growth factor is 1.031, and the growth rate is 0.031, which can also be expressed as 3.1%.
To connect the key ideas in this lesson and the past few ones, discuss questions such as:
 "Is the growth rate per decade ten times the growth rate per year? Why or why not?" (No. The growth is compounded over time. Between 1790 and 1791, the population grows by 3.1% of 4,000 so, in thousands, that's 12.4, giving 4,012.4. Between 1791 and 1792, the population in thousands will grow by 3.1% of 4,012.4. So for the second year, the growth is 3.1% of 4,000 plus 3.1% of 12.4 from the first year. And so on. So the growth rate per decade is more than \(3.1 \boldcdot 10\).)
 "Which growth rate is easiest or most helpful for understanding the population change in that period? Why might that be?" (Perhaps the easiest to grasp is the growth rate per decade, about a third per decade. The yearly growth is relatively small so it is hard to appreciate. The growth per century is very large.)
 "If we use the model by year to find population in 2017, what would it be?" (About 4 billion people. In thousands, it is \(4,\!000 \boldcdot (1.031)^{227} \approx 4,\!090,\!505.78\). It is off by more than a factor of 10.)
 "What might be a reason that the population calculated using the model is so far off from the actual number?" (A variety of reasons, but one factor is that limits in land and resources made it unlikely for population to continue to grow at the same rate over time.)
Design Principle(s): Support sensemaking
Supports accessibility for: Conceptual processing; Language
18.3: Interest Calculations (15 minutes)
Activity
In the previous activity, students saw that there are different ways to express the same exponential growth pattern. Here they identify and interpret expressions that represent different compoundinterest situations.
The expressions are difficult to calculate and so students will need to rely on structure (MP7). From this point of view, the key part of the expression is what is on the inside of the innermost parentheses. The decimal part of this (0.07) represents the annual interest rate while the denominator (2 and 12) represents how frequently the interest is compounded.
Look for students who focus on the structure of the expressions and what each part of the expression means in context. Invite these students to share during the discussion.
Launch
Arrange students in groups of 2. Ask students to take turns: the first partner identifies a match and explains why they think it is a match, while the other listens and works to understand. Then they switch roles.
Design Principle(s): Support sensemaking; Maximize metaawareness
Student Facing
Here are three expressions and three descriptions. In each case, $1,000 has been put in an interestbearing bank account. No withdrawals or other deposits (aside from the earned interest) are made for 6 years.
 \(1,\!000 \boldcdot \left(1+\dfrac{0.07}{12}\right)^{72}\)
 \(1,\!000 \boldcdot \left(1+\dfrac{0.07}{2}\right)^{12}\)
 \(1,\!000 \boldcdot \left(\left(1+\dfrac{0.07}{12}\right)^{12}\right)^6\)
 7% annual interest compounded semiannually
 7% annual interest compounded monthly
 7% annual interest compounded every two months
Sort the expressions that represent the same amounts of interest into groups. One group contains more than two expressions. One of the descriptions does not have a match. Write an expression that matches it.
Student Response
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Student Facing
Are you ready for more?
Investing $1,000 at a 5% annual interest rate for 6 years, compounded every two months, yields $1,348.18. Without doing any calculations, rank these four possible changes in order of the increase in the interest they would yield from the greatest increase to the least increase:
 Increase the starting amount by $100.
 Increase the interest rate by 1%.
 Let the account increase for one more year.
 Compound the interest every month instead of every two months.
Once you have made your predictions, calculate the value of each option to see if your ranking was correct.
Student Response
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Anticipated Misconceptions
Some students may need a reminder that semiannually means twice per year, or every 6 months.
Students may not understand the denominator of the fraction included in the expressions. Help them to see that the 7% interest for the year is split into smaller percentages calculated more than once per year.
Activity Synthesis
Select previously identified students to share how they made their matches. For each expression, prompt them to discuss the meaning of each part. For example, for \(1,\!000 \boldcdot\left(\left(1 + \frac{0.07}{12}\right)^{12}\right)^6\), ask:
 What does \(1+ \frac{0.07}{12}\) mean? (It is the percent interest for each month, or the portion of the 7% yearly rate that is applied monthly.)
 What does \(\left(1 + \frac{0.07}{12}\right)^{12}\) mean? (It is the monthly rate compounded over 12 months, or the factor by which the account grows after one year. It is the effective annual interest rate.)
 What does the exponent 6 mean in \(\left(\left(1 + \frac{0.07}{12}\right)^{12}\right)^6\)? (It is the number of years that the account earns interest, which is 6 years.)
Lesson Synthesis
Lesson Synthesis
Today we looked at how exponential functions can be written in different ways in order to be more meaningful. Invite students to articulate how this is so, using an example such as the following.
After taking a dose of 0.2 mg, the amount of medicine \(m\) in a person's body, in mg, can be modeled by three different equations based on the units of time used to measure the change. \(h\), \(d\), and \(w\) represent the number of hours, days, and weeks, respectively, since taking the medicine. \(\displaystyle m = (0.2) \boldcdot (0.996)^h\), \(\displaystyle m = (0.2) \boldcdot \left((0.996)^{24}\right)^d\), \(\displaystyle m = (0.2) \boldcdot \left((0.996)^{168}\right)^w\)
 What does each equation tell us? What does the 0.996 represent? What do the 24 and 168 represent?
 Which of the equations, if any, gives us a clear picture of how the body breaks down the medicine? Why is that?
 \((0.996)^{24} \approx 0.91\) and\((0.996)^{168}\approx 0.51\). What do 0.91 and 0.51 tell us? Do they give us a clearer idea of how the medicine breaks down? How?
18.4: Cooldown  Printing Business (5 minutes)
CoolDown
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Student Lesson Summary
Student Facing
Expressions can be written in different ways to highlight different aspects of a situation or to help us better understand what is happening. A growth rate tells us the percent change. As always, in percent change situations, it is important to know if the change is an increase or decrease. For example:

A population is increasing by 20% each year. The growth rate is 20%, so after one year, 0.2 times the population at the beginning of that year is being added. If the initial population is \(p\), the new population is \(p + 0.2p\), which equals \((1 + 0.2)p\) or \(1.2p\).

A population is decreasing by 20% each year. The growth rate is 20%, so after one year, 0.2 times the population at the beginning of that year is being lost. If the initial population is \(p\), the new population is \(p – 0.2p\), which equals \((1 – 0.2)p\) or \(0.8p\).
Suppose the area \(a\) covered by a forest is currently 50 square miles and it is growing by 0.2% each year. If \(t\) represents time, from now, in years we can express the area of the forest as:
\(\displaystyle a = 50 \boldcdot (1+0.002)^t\)
\(\displaystyle a = 50 \boldcdot (1.002)^t\)
In this situation, the growth rate is 0.002, and the growth factor is 1.002. Because 0.002 is such a small number, however, it maybe difficult to tell from this function how quickly the forest is growing. We may find it more meaningful to measure the growth every decade or every century. There are 10 years in a decade, so to find the growth rate in decades, we can use the expression \((1.002)^{10}\), which is approximately 1.02. This means a growth rate of about 2% per decade. Using \(d\) for time, in decades, the area of the forest can be expressed as:
\(\displaystyle a=50 \boldcdot \left((1+0.002)^{10}\right)^d\)
\(\displaystyle a \approx50 \boldcdot (1.02)^d\)
If we measure time in centuries, the growth rate is about 22% per century because \(1.002^{100} \approx 1.22\). Using \(c\) to measure time, in centuries, our equation for area becomes:
\(\displaystyle a = 50 \boldcdot \left((1+0.002)^{100}\right)^c\)
\(\displaystyle a \approx 50 \boldcdot (1.22)^c\)