Lesson 3
Understanding Rational Inputs
3.1: Keeping Equations True (5 minutes)
Warmup
In an earlier unit, students learned the meaning of rational exponents and how to evaluate them. This warmup reviews that work, preparing students to explore exponential functions with rational input values by focusing on two basic fractional exponents, \(\frac12\) and \(\frac13\), and some of their properties.
Student Facing
 Select all solutions to \(x \boldcdot x = 5\). Be prepared to explain your reasoning.
 \(\frac{1}{25}\)
 \(\sqrt{5}\)
 \(\frac{5}{2}\)
 \(5^{\frac{1}{2}}\)
 \(\frac{\sqrt{5}}{2}\)
 \(\sqrt{25}\)
 Select all solutions to \(p \boldcdot p \boldcdot p = 10\). Be prepared to explain your reasoning.
 \(10^{\frac{1}{3}}\)
 \(\sqrt{10}\)
 \(\frac{10}{3}\)
 \(\frac{\sqrt{10}}{3}\)
 \(\sqrt[3]{10}\)
 \(\frac{1}{3} \sqrt{10}\)
Student Response
Teachers with a valid work email address can click here to register or sign in for free access to Student Response.
Activity Synthesis
Briefly survey the class on whether each expression is a solution to the given equation. If there is a disagreement about whether an expression is a solution, ask students with contrasting views to share their reasoning so agreement can be reached. For an expression that everyone agrees is a solution, invite 2–3 students to share their reasoning.
Make sure students recall that a number raised to an exponent of \(\frac{1}{2}\), as in \(x^{\frac{1}{2}}\), is \(\sqrt{x}\). Highlight that squaring each number, \(x^{\frac{1}{2}}\) and \(\sqrt{x}\), results in \(x\). \(\displaystyle (x^{\frac{1}{2}})^2 = x^1 = x\) \(\displaystyle (\sqrt{x})^2 = \sqrt{x} \boldcdot \sqrt{x} = x\)
Likewise, a number raised to an exponent of \(\frac13\), as in \(p^{\frac13}\), is defined to be \(\sqrt[3]{p}\). Cubing each number, \(p^{\frac13}\) and \(\sqrt[3]{p}\), results in \(p\). \(\displaystyle (p^{\frac{1}{3}})^3 = p^1 = p\) \(\displaystyle (\sqrt[3]p)^3 = \sqrt[3]{p} \boldcdot \sqrt[3]{p} \boldcdot \sqrt[3]{p} = p\)
3.2: Florida in the 1800’s (20 minutes)
Activity
In this activity, students construct an exponential function to model the population growth in Florida. Then, they interpret the function when evaluated at certain rational numbers. They do so quantitatively, in terms of the population, and abstractly, in terms of an expression with a fractional exponent (MP2).
As students work, identify those who express the population of a given year as an exponential expression and those who use a calculator to estimate a numerical value. Both approaches are important: the first, because it gives an opportunity to interpret a fractional power, and the second, because it gives a concrete sense of the population growth.
Launch
Provide access to devices that can run Desmos or other graphing technology.
Design Principle: Support sensemaking
Supports accessibility for: Conceptual processing; Memory
Student Facing
In 1840, the population of Florida was about 54,500. Between 1840 and 1860, the population grew exponentially, increasing by about 60% each decade.
 Find the population of Florida in 1850 and in 1860 according to this model.
 The population is a function \(f\) of the number of decades \(d\) after 1840. Write an equation for \(f\).

 Explain what \(f(0.5)\) means in this situation.
 Graph your function using graphing technology and estimate the value of \(f(0.5)\).
 Explain why we can find the value of \(f(0.5)\) by multiplying 54,500 by \(\sqrt{1.6}\). Find that value.
 Based on the model, what was the population of Florida in 1858? Show your reasoning.
Student Response
Teachers with a valid work email address can click here to register or sign in for free access to Student Response.
Student Facing
Are you ready for more?
Andre said, “The population of Florida increased by the same percentage between 1842 and 1852 and between 1847 and 1857.” Do you agree with his statement? Explain or show your reasoning.
Student Response
Teachers with a valid work email address can click here to register or sign in for free access to Extension Student Response.
Anticipated Misconceptions
If students struggle with graphing the function, ask them to think about what horizontal and vertical dimensions to set for the graphing window so the time in decades and the population in thousands can be shown meaningfully. For example, graphing an input range from 0 to 20 would mean assuming the population of Florida grew the same percent for 200 years, which is unlikely.
Activity Synthesis
Focus the discussion on the meaning of \(f(0.5)\) and \(f(1.8)\) in this context and on how to reason about these values. Display a table for all to see, like the one shown here, to fill in during the discussion:
time (decades since 1840) 
population (thousands)  approximate population (thousands) 

0  54,500  54,500 
0.5  
1  \(54,\!500 \boldcdot (1.6)\)  87,200 
1.8  
2  \(54,\!500 \boldcdot (1.6) \boldcdot (1.6)=54,\!500 \boldcdot (1.6)^2\)  139,520 
3  \(54,\!500 \boldcdot (1.6) \boldcdot (1.6) \boldcdot (1.6)=54,\!500 \boldcdot (1.6)^3\)  223,232 
Select previously identified students to share their responses for the population of Florida with the inputs 0.5 and 1.8, written as expressions (\(54,\!500 \boldcdot (1.6)^{0.5}\) and \(54,\!500 \boldcdot (1.6)^{1.8}\)) and as numerical values (about 68,938 and 127,003). Help students see that:
 We can think of the population in 1860, 2 decades after 1840, as \(54,\!500 \boldcdot (1.6)\boldcdot (1.6)\) because the 60% increase is applied twice. Although we can’t write the same kind of expanded expression for the population 0.5 decade after 1840, we can use properties of exponents to reason about its value.
 Here are three different ways to write the growth factor for the population 5 years after 1840: \((1.6)^{0.5}\) or \({1.6}^{\frac12}\) is \(\sqrt{1.6}\).
Conclude by asking students to write three different ways to express the growth factor for the population 18 years after 1840 (\((1.6)^{1.8}\) or \({1.6}^{\frac{18}{10}}\) is \(\sqrt[18]{(1.6)^{10}}\)).
3.3: Disappearing Medicine (10 minutes)
Activity
In the previous activity, students interpreted a description of an exponential function, wrote an equation, and produced a graph. In this activity, they work the other way around. Given a graph of an exponential relationship, they produce an equation and reason about the context. Students then express the change in the function in terms of its decay rate, recalling the connection between decay factor and decay rate.
Launch
Supports accessibility for: Memory; Organization
Student Facing
The amount of a medicine in the bloodstream of a patient decreases roughly exponentially. Here is a graph representing \(f\), an exponential function that models the medicine in the body of a patient, \(t\) hours after an injection is given.
 Use the graph to estimate \(f \left(\frac13\right) \) and explain what it tells us in this situation.
 After one hour, 0.75 mg of medicine remains in the bloodstream. Find an equation that defines \(f\).
Student Response
Teachers with a valid work email address can click here to register or sign in for free access to Student Response.
Student Facing
Are you ready for more?
By what percentage does the amount of medicine in the body decay every 10 minutes? Explain or show your reasoning.
Student Response
Teachers with a valid work email address can click here to register or sign in for free access to Extension Student Response.
Activity Synthesis
Begin the discussion by inviting students to share how they derived an expression representing the function \(f\) and the meaning of 12 and \(\frac{1}{16}\) (or 0.0625) in the expression \(12 \boldcdot \left(\frac{1}{16}\right)^t\) (or \(12 \boldcdot (0.0625)^t\)).
Next, focus the discussion on the interpretation of \(f(\frac13)\) in this context and how its value could be estimated graphically and reasoned algebraically. Highlight:
 \(\frac13\) an hour is the same as 20 minutes.
 Using the equation, \(f(\frac13)=12 \boldcdot (0.0625)^{\frac13}\), which is approximately 4.8 milligrams of medicine.
Conclude the discussion by asking students to identify the decay factor for 20 minutes and what it means. (\((0.0625)^{\frac13}\), or about 39.7% of the drug remains in the body after 20 minutes.)
Lesson Synthesis
Lesson Synthesis
The goal of this discussion is to summarize how fractional exponents and exponential functions come together in this lesson. Display the graph for all to see and tell students it is the function \(B=2^h\), which models the population of a certain bacteria \(B\), in thousands, \(h\) hours after being counted.
 “What are the coordinates of \(P\) and what do they mean in this context?” (\(\left(\frac{1}{4},2^{\frac{1}{4}}\right)\) or about \((0.25, 1.19)\). 15 minutes after the hour, the population is about 1,189.)
 “What are the coordinates of \(Q\) and what do they mean in this context?” (\(\left(\frac{5}{4},2^{\frac{5}{4}}\right)\) or about \((1.25,2.38)\). 1 hour and 15 minutes after being counted, the population is about 2,378.)
If students notice that the value of the function at \(Q\) is twice the value at \(P\), invite students to reason about why that is true. (The interval between \(\frac14\) and \(\frac54\) is 1 hour, which means the outputs are related by a factor of 2.) Reasoning more deeply about why this is true is a focus of a future lesson, so it is okay to skip discussing this feature of exponential functions now.
Throughout this unit, we will encounter exact expressions like \(2^{\frac{1}{3}}\) and their approximations like 1.26. Exact expressions are important for revealing mathematical structure while approximations are important for estimating values, especially in concrete situations. When a value is approximate, rather than exact, it is important to note this: \(2^{\frac{1}{3}} \approx 1.26\).
3.4: Cooldown  Flea Treatment (5 minutes)
CoolDown
Teachers with a valid work email address can click here to register or sign in for free access to CoolDowns.
Student Lesson Summary
Student Facing
Some exponential functions can have inputs that are any numbers on the number line, not just integers.
Suppose the area of a pond covered by algae \(A\), in square meters, is modeled by \(A = 200 \boldcdot \left(\frac{1}{2}\right)^w\) where \(w\) is the number of weeks since a treatment was applied to the pond. How could we use this equation to determine the area covered after 1 day?
Well, since \(w=1\) is one week and each week has 7 days, \(w=\frac17\) is 1 day. So after 1 day, the algae covers \(200 \left(\frac{1}{2}\right)^{\frac{1}{7}}\) square meters, or about 181 square meters. Using a calculator, we know that the expression \(\left(\frac{1}{2}\right)^{\frac{1}{7}}\), which is equivalent to \(\sqrt[7]{\frac{1}{2}}\), is about about 0.906. This means that after 1 day, only 91% of the algae from the previous day remains.
This information can also be seen on a graph representing the area. The point at \((1, 100)\) marks the area covered by the algae after 1 week. Point \(P\) marks the covered area after \(\frac{1}{7}\) of a week or one day.
The graph can be used to estimate the vertical coordinate of \(P\) and shows that it is close to 180.