Lesson 4

This Math Talk encourages students to think about rules of exponents in order to solve exponential equations. These exponential rules elicited here will be helpful later in the lesson and in future lessons when students study the values of exponential functions at rational inputs and use these functions to model different phenomena.

Display one problem at a time. Give students quiet think time for each problem and ask them to give a signal when they have an answer and a strategy. Keep all problems displayed throughout the talk. Follow with a whole-class discussion.

Ask students to share their strategies for each problem. Record and display their responses for all to see. To involve more students in the conversation, consider asking:

• “Who can restate \(\underline{\hspace{.5in}}\)’s reasoning in a different way?”
• “Did anyone have the same strategy but would explain it differently?”
• “Did anyone solve the problem in a different way?”
• “Does anyone want to add on to \(\underline{\hspace{.5in}}\)’s strategy?”
• “Do you agree or disagree? Why?”

If not mentioned in students’ explanations, emphasize that these equations can all be solved by writing each expression as a power of the same base, and then examining the exponents of those expressions. For example, \(\frac{1}{125}\) is the same as \(\frac{1}{5^3}\).

In this activity, students practice constructing an exponential function and evaluating it at a fractional input value as they translate between decades and years. They also think about how to express a percent increase over a fractional interval of input—specifically, how to express the annual growth rate of a population given the growth rate for one decade. They analyze and articulate why the growth rate for one year is not one tenth of that of one decade (MP3).

Tell students that, as of 2017, Nigeria is the seventh-most populous country in the world. It has grown at a fairly steady (and relatively high) rate since 1980.

If students struggle to identify the input to their function where \(d\) is not an integer, remind them that there are 10 years in a decade. Then ask them how many decades is 6 years? What about one year? 

Focus the discussion on why the annual growth rate is not \(\frac1{10}\) of 28.4%. Consider asking students:

• “Why might someone think the annual population growth rate is 2.84%?” (Because 28.4 divided by 10 is 2.84, and they assume that population was growing linearly, or by the same amount each year, rather than by the same factor each year.)
• “How can we determine the growth factor for one year?” (We know that when the annual growth factor, say \(b\), is raised to an exponent of 10, it must equal 1.284, or \(b^{10} = 1.284\), so \(b\) must be \(1.284^\frac{1}{10}\) or \(\sqrt[10]{1.284}\) because raising that number to an exponent of 10 gives 1.284.)
• “How do we express the factor as a growth rate? ( \(1.284^\frac{1}{10} \approx 1.0258\), so the growth rate is 2.58%.)

This task prompts students to carefully interpret a description and some expressions representing exponential decay. It further reinforces the distinction between how exponential and linear functions change over intervals of time, which was explored in the previous activity.

The decay factor is not given for a familiar unit of input (that is, the caffeine decays by half every 6 hours, instead of every hour). To correctly represent the decay with functions, students need to notice that the \(\frac12\) does not correspond to the change in 1 hour, and that the decay factor in 1 hour is not \(\frac16\) of \(\frac12\).

Ask students to close their books or devices. Tell the class they are going to do some calculations around how the amount of caffeine in the body changes over time after consuming a certain amount. Before students read the task statement or do any calculations, ask “How much caffeine do you think is left in a healthy adult 24 hours after they consume 100 mg (if no additional caffeine is consumed)?” Poll the class for their estimates and display the result for all to see. Students will consider this question again during the whole-class discussion.

Tell students to open their books or devices and read the opening statement of the activity. Explain that the term half-life refers to the time it takes for a certain quantity (such as the concentration of a substance in a body) that is decaying exponentially to decrease by half. In this context, a half-life of 6 hours means that the amount of caffeine in the body is cut in half 6 hours after it is consumed.

Pause students after they complete the first set of questions. Discuss what the 100, the \(\frac12\), and the exponent in each expression represent, and review students’ responses to the questions before they continue with the task.

Students may struggle viewing groups of 6 hours as a unit for the half-life of caffeine. Identifying that \(100 \boldcdot \left(\frac{1}{2}\right)^{\frac{1}{6}}\) is the number of mg of caffeine left after 1 hour is the critical step. If students do not see what this expression means in the situation, ask them how any times the scale factor \(\left(\frac{1}{2}\right)^{\frac{1}{6}}\) has to be applied for the caffeine to be cut in half. Then ask them how many hours it takes for the caffeine to be cut in half.

Begin the discussion by repeating the earlier question, “How much caffeine do you think is left in a healthy adult 24 hours after they consume 100 mg (if no additional caffeine is consumed)?” Invite students to share their reasoning, making sure they see that “24 hours later” is 4 periods of 6 hours, so the amount of caffeine will have been halved 4 times: \(100 \boldcdot \left(\frac12\right)^4 = 100 \boldcdot \frac{1}{16} = 6.25\).

Next, focus the discussion on the distinction between expressing the amount of caffeine as a function of time in groups of 6 hours (function \(f\)) and time in groups of 1 hour (function \(g\)). Discuss with students:

• “What decay factor did you use to write the expression for function \(g\)? What is the approximate decay rate per hour?” (Raise \(\frac12\) to an exponent of \(\frac 16\), which is approximately 0.89, so the decay rate is about 11%.)
• “Why not divide \(\frac12\) by 6 (or multiply it by \(\frac16\)) to get the decay factor for an hour?” (Because when that factor is raised to an exponent of 6, it needs to equal \(\frac12\). \(\left(\frac{1}{12}\right)^6\) is not \(\frac12\).)

• “How can we show that the expression for \(g\), which is \(100 \boldcdot \left(\left(\frac12\right)^{\frac16}\right)^h\), is equivalent to the expression for \(f\), which is \(100 \boldcdot \left(\frac12\right)^t\)?” (We know that \(h=6t\). If we substitute \(6t\) in the expression for \(g\), we have \(100 \boldcdot \left(\frac12\right)^{(\frac16)^{6t}}\), which equals \(100 \boldcdot \left(\frac12\right)^t\).)

Time permitting, consider showing the two graphs representing the two functions.

Function \(g\)

Function \(f\)

Here are some possible questions for discussion about the graphs:

• “How are these two graphs alike and different?” (They seem to be the same curve. They have the same vertical intercept. The scales for the horizontal axis are different. One curve goes through \((1,50)\) and the other goes through \((6,50)\).)

• “On the graph representing \(f\), what does the point at \((1,50)\) mean in the context? What is the corresponding point on the graph of \(g\)?” (The amount of caffeine in the body after 1 period of 6 hours is 50 milligrams. This is the same as the point \((6,50)\) on the graph of \(g\).)

• “How could these two graphs represent the same decay, if the coordinates they go through are not the same?” (They use different units for the input. 1 unit of the input in \(f\) is 6 units of the input in \(g\).)

• “To find the amount of caffeine 3 hours after consumption, what input value should we use for each function?” (\(\frac12\) for \(f\) and 3 for \(g\)) “What about for 30 minutes after consumption?” (\(\frac {1}{12}\) for \(f\) and \(\frac12\) for \(g\))