Lesson 4

Understanding Decay

4.1: Notice and Wonder: Two Tables (5 minutes)


So far, students have seen patterns and relationships involving mostly whole numbers. This warm-up encourages students to observe linear and exponential patterns that involve fractions, preparing them for upcoming work.

Student Facing

What do you notice? What do you wonder?

Table A

\(x\) \(y\)
0 2
1 \(3\frac12\)
2 5
3 \(6\frac{1}{2}\)
4 8

Table B

\(x\) \(y\)
0 2
1 3
2 \(\frac92\)
3 \(\frac{27}{4}\)
4 \(\frac{81}{8}\)

Student Response

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Activity Synthesis

Invite students to share their observations and questions. To encourage more students to participate, after each person shares, ask if others noticed or wondered about the same thing.

If not already mentioned by students, highlight how the values in the two tables change from row to row (Table A by equal differences and Table B by equal factors). If no students wonder about whether the values in the two tables will ever agree again, ask, “Will a linear growth and an exponential growth that start with the same value have the same value again at some later point?” Consider posting the question for students to think about throughout the unit.

4.2: What's Left? (15 minutes)


Students have explored quantities that increase by the same factor (e.g., doubling or tripling). Up to this point, that factor has always been greater than 1. In this lesson, they will look at situations where quantities change exponentially, but by a positive factor that is less than 1. The factor is still called the growth factor (the growth factor can be less than 1) though students should learn that sometimes people refer to it as a decay factor.

Multiplying a quantity by a factor less than 1 means that the quantity is decreasing. Students may naturally think of a decrease as subtraction. For example, a quantity \(x\) that loses \(\frac13\) of its value may be represented as \(x - \frac{1}{3}x\). Because exponential relationships are characterized by multiplication by a factor, however, it will be helpful to view this instead as \(x \boldcdot \left(1- \frac{1}{3}\right)\), or as multiplying \(x\) by \(\frac{2}{3}\).


Ask students to close their books or devices. Pose this question:

Diego has $100 and spends \(\frac14\) of it. How much does he have left?

Give students a minute to think about it, and then ask a few students to share their answer and their reasoning.

Conversing: MLR8 Discussion Supports. Arrange students in groups of 2. Students should take turns explaining the steps of the first problem. Invite Partner A to begin with the first step, while Partner B listens. Alternate roles for the remaining steps. Encourage students to refer to explicit parts of the expression, such as the operation, specific numbers, the use of parentheses, or the order of the numbers. Encourage students to challenge each other when they disagree. This will help students connect mathematical work to situations in which quantities change by a factor that is less than 1.
Design Principle(s): Support sense-making; Maximize meta-awareness
Representation: Internalize Comprehension. Use color and annotations to illustrate student thinking. As students share their reasoning about Diego’s strategy, scribe their thinking on a visible display.
Supports accessibility for: Visual-spatial processing; Conceptual processing

Student Facing

  1. Here is one way to think about how much Diego has left after spending \(\frac14\) of $100. Explain each step.
    • Step 1: \(100 - \frac14 \boldcdot 100\)
    • Step 2: \(100\left(1 - \frac14\right)\)
    • Step 3: \(100 \boldcdot \frac34\)
    • Step 4: \(\frac34 \boldcdot 100\)
  2. A person makes $1,800 per month, but \(\frac13\) of that amount goes to her rent. What two numbers can you multiply to find out how much she has after paying her rent?
  3. Write an expression that only uses multiplication and that is equivalent to \(x\) reduced by \(\frac18\) of \(x\).

Student Response

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Anticipated Misconceptions

Students may have trouble making the connection between the fraction that is being subtracted and the fraction that remains when the subtraction is rewritten as multiplication. Invite these students to think through the steps in Diego's calculation. What does the number 100 represent? What about the \(\frac{1}{3}\)? Once students identify the analogous quantities in the second and third questions, encourage them to revisit the subsequent steps of Diego's calculation.

Activity Synthesis

Ask previously selected students to share their explanations of how a subtraction expression can be written as multiplication of two numbers. Make sure that students understand that a decreasing quantity can be expressed both ways, using subtraction or multiplication.

If time permits, consider asking “There were \(x\) dogs at the park and \(\frac{1}{5}\) of the dogs were on a leash. How many dogs were not on a leash?” The number of dogs not on a leash could be expressed using subtraction as \(x - \frac{1}{5}x\) or using multiplication as \(\frac{4}{5}x\).

Emphasize that it is important to be able to express this situation using multiplication because exponential situations involve quantities changing by a common factor (or being multiplied by a number).

4.3: Value of a Vehicle (15 minutes)


In this activity, students examine a situation where a quantity decreases by repeated multiplication by the same factor. They represent the pattern of decrease with an equation and interpret the different parts of the equation in terms of the context (the depreciation of a car). The definition of growth factor is expanded to accommodate situations involving decrease.


Explain the scenario to the class, or just display the first sentence of the task: Every year after a new car is purchased, it loses \(\frac13\) of its value. Let’s say that a new car costs $18,000.

Ask the class to estimate: Without calculating, what do you think the value of the car will be 4 years after it was purchased as a new car? Poll the class for their estimates.

Clarify to students that they may complete the table with numerical expressions that show how to find the values rather than with the results of evaluating the expressions.

Reading, Conversing, Writing: MLR5 Co-Craft Questions. To help students make sense of the language of exponential decay, start by displaying only the context for the question “Every year after a new car is purchased, it loses \(\frac{1}{3}\) of its value. Let’s say that a new car costs $18,000.” Give students 1–2 minutes to write their own mathematical questions about the situation, then invite students to share their questions with the class. Listen for and amplify any questions about the future value of the car, especially those using language such as “losing value” and “decrease.” This will build student understanding of the language of exponential decay and help ensure students interpret the task correctly.
Design Principle(s): Maximize meta-awareness; Cultivate conversation
Representation: Internalize Comprehension. Activate or supply background knowledge. Allow students to use calculators to ensure inclusive participation in the activity.
Supports accessibility for: Memory; Conceptual processing

Student Facing

Every year after a new car is purchased, it loses \(\frac13\) of its value. Let’s say that a new car costs $18,000.

  1. A buyer worries that the car will be worth nothing in three years. Do you agree? Explain your reasoning.
  2. Write an expression to show how to find the value of the car for each year listed in the table.
    year value of car (dollars)
    0 18,000
  3. Write an equation relating the value of the car in dollars, \(v\), to the number of years, \(t\).
  4. Use your equation to find \(v\) when \(t\) is 0. What does this value of \(v\) mean in this situation?
  5. A different car loses value at a different rate. The value of this different car in dollars, \(d\), after \(t\) years can be represented by the equation \(d = 10,\!000 \boldcdot \left( \frac45 \right)^t\). Explain what the numbers 10,000 and \(\frac45\) mean in this situation.

Student Response

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Student Facing

Are you ready for more?

Start with an equilateral triangle with area 1 square unit, divide it into 4 congruent pieces as in the figure, and remove the middle one. Then, repeat this process with each of the remaining pieces. Repeat this process over and over for the remaining pieces. The figure shows the first two steps of this construction.

3 figures showing construction of a pattern

What fraction of the area is removed each time? How much area is removed after the \(n\)-th step? Use a calculator to find out how much area remains in the triangle after 50 such steps have been taken.

Student Response

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Activity Synthesis

Make sure that students understand the expression for the value of cars after \(t\) years came from multiplying by \(\frac23\) repeatedly (\(t\) times). Tell students that we still say this quantity changes exponentially, though sometimes people use the more specific exponential growth and exponential decay to indicate whether the amount is increasing or decreasing. The multiplier is still called the growth factor, but when it is a positive number less than 1, the result decreases with every iteration. Sometimes people use decay factor to indicate that a quantity that decreases exponentially involves a positive factor less than 1.

Questions for discussion:

  • “Why does it make sense to multiply an entry in one row by \(\frac{2}{3}\) to get the entry for the next row?” (Each year, only \(\frac23\) of the car value remains.)
  • “Why not write the expression for each year as subtraction, for instance, \(18,\!000 - \frac13 \boldcdot 18,\!000\)?” (We could, but writing subtraction can get unwieldy quickly. For year 2, the expression would be \((18,\!000 - \frac13 \boldcdot 18,\!000) - \frac13 (18,\!000 - \frac13 \boldcdot 18,\!000)\). If we calculate the value of each term it would be simpler (\(12,\!000 - \frac13 \boldcdot 12,\!000\)), but it would still not allow us to see a pattern and write a general expression as easily.)
  • “Does the \(\frac45\) in the last equation mean losing \(\frac45\) of the value each year? Why or why not?” (No. Multiplying by \(\frac 45\) repeatedly means each time only \(\frac45\) of the value remains, so the car is losing \(\frac15\) of its value each year.)
  • “Will the value of the second car (starting at $10,000 and losing \(\frac{1}{5}\) of it each year) ever be worth more than the first car (starting at $18,000 and losing\(\frac{1}{3}\) of it each year)? Explain.” (Yes. After 4 years, the value of the second car will be $4,096 while the value of the first car will be $3,555.56)

Make sure that students understand the exponential expression for the value of cars after \(t\) years came from multiplying by \(\frac23\) repeatedly (\(t\) times).

Lesson Synthesis

Lesson Synthesis

We studied quantities that decrease or decay by the same factor repeatedly. Use examples from activities in the lesson, or a new example like this one.

Suppose a digital camera worth \$400 loses \(\frac{1}{3}\) of its value each year.

  • How can we express the value of the camera, in dollars, after one year? After two years?
  • Why might it make sense to use only multiplication (instead of subtraction and multiplication) to show the value of the camera over time?
  • When we use only multiplication, why doesn’t the \(\frac13\) show up in the expression?
  • What is the value of the camera, in dollars, after \(t\) years?

4.4: Cool-down - The Depreciating Phone (5 minutes)


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Student Lesson Summary

Student Facing

Sometimes a quantity grows by the same factor at regular intervals. For example, a population doubles every year. Sometimes a quantity decreases by the same factor at regular intervals. For example, a car might lose one third of its value every year. 

Let's look at a situation where the quantity decreases by the same factor at regular intervals. Suppose a bacteria population starts at 100,000 and \(\frac{1}{4}\) of the population dies each day. The population one day later is \(100,\!000 - \frac14 \boldcdot 100,\!000\), which can be written as \(100,\!000 \left(1 - \frac14\right)\). The population after one day is \(\frac{3}{4}\) of 100,000 or 75,000. The population after two days is \(\frac{3}{4} \boldcdot 75,\!000\). Here are some further values for the bacteria population

of days
bacteria population
0 100,000
1 75,000 (or \(100,\!000 \boldcdot \frac34\))
2 56,250 (or \(100,\!000 \boldcdot \frac34 \boldcdot \frac34\), or \(100,\!000 \boldcdot \left(\frac34\right)^2\))
3 about 42,188 (or \(100,\!000 \boldcdot \frac34 \boldcdot \frac34 \boldcdot \frac34\), or \(100,\!000 \boldcdot \left(\frac34\right)^3\))

In general, \(d\) days after the bacteria population was 100,000, the population \(p\) is given by the equation: \(\displaystyle p = 100,\!000 \boldcdot \left(\frac{3}{4}\right)^d,\) with one factor of \(\frac{3}{4}\) for each day.

Situations with quantities that decrease exponentially are described with exponential decay. The multiplier (\(\frac34\) in this case) is still called the growth factor, though sometimes people call it the decay factor instead.