Lesson 15

This warm-up is students’ first opportunity to make an explicit connection between the graph of an exponential function with base \(e\) and the natural logarithm. Students are prompted to explain why the graph representing an exponential function can be used to estimate a natural logarithm (MP3) and then to make an estimate. They also practice using a calculator to evaluate a natural logarithm.

As students discuss, monitor for those who can articulate the connections between the given graph and logarithm. Invite them to share during the whole-class discussion.

Arrange students in groups of 2. Give them a moment of quiet time to think about the questions. Then, ask them to discuss their thinking with a partner. Follow with a whole-class discussion. 

Focus the discussion on the first question. Invite previously identified students to share their explanations. An important takeaway for students is seeing that \(\ln 100\) is the exponent to which we raise a base \(e\) to produce 100. The graph shows the value of the function \(e^x\), so we can trace the graph and estimate the value of \(x\) when it reaches 100.

In this activity, students use the natural logarithm to solve problems in a financial context. They first evaluate an exponential function with base \(e\) for different values of the input. After making sense of what a logarithm means in the given situation, they use the natural logarithm to solve an exponential equation by rewriting it as a logarithm and answer questions about the situation.

The financial context presented here is unrealistic and the numbers chosen are simple. The intent is to keep the function very straightforward at this stage so that students can better understand what a natural logarithm means in the given situation (MP2). Later, students will encounter more realistic contexts and numbers.

Students engage in aspects of mathematical modeling as they make some assumptions about the account and the age of retirement in the last question (MP4). If students choose to use graphing technology to solve \(1 \boldcdot e^{(0.06t)}=1000\), they are using appropriate tools strategically (MP5).

Arrange students in groups of 2. Give students a few minutes of quiet think time for the first two questions and then ask them to discuss their responses with their partner.

Pause the class for a discussion before students proceed to the last question. Make sure students understand that the expression \(\ln 5\) does not represent the time when the account will have 5 thousand dollars, because the exponent that we’re trying to find is \(0.06t\) and not just \(t\). In other words, we’re trying to solve \(e^{(0.06t)} = 5\), rather than \(e^t=5\). Once we know the value of \(0.06t\), which is \(\ln 5\), it will need to be divided by 0.06 to find \(t\).

The final question about the account balance reaching a million dollars before retirement requires some assumptions about the age of retirement. For students who struggle to get started on this question, two good approaches to encourage include:

• making an estimate of retirement age and checking the balance of the account at that time
• using technology to find out about when the account balance reaches one million dollars

Invite students to share their responses and strategies for the last question. Make sure students see these approaches:

• writing \(1 \boldcdot e^{(0.06t)}=1,\!000\) and solving using a natural log
• using technology to graph \(y=1 \boldcdot e^{(0.06t)}\) and estimating from the graph the \(t\)-coordinate for which the \(y\)-coordinate is 1,000

If no students chose to graph to solve the question, demonstrate how to do so. Highlight the connection between the point of intersection of the graphs and the solution to \(1 \boldcdot e^{(0.06t)}=1,\!000\) calculated using a natural log.

Conclude the discussion by noting that if it were possible to retire a millionaire by depositing 1,000 dollars and leaving it untouched until retirement, it is likely that there would be many more retired millionaires. If time allows, ask students to determine what interest rate is necessary to turn $1,000 into $1,000,000 over 50 years. (Almost 14%.)

This activity gives students a chance to solve exponential equations in context by using a logarithm and by graphing.

Students have previously used graphs to estimate solutions to exponential equations. To find the input of a function that produces a certain output, they have primarily relied on visual inspection of the point when the graph reaches that value. Here they see that the estimation can be made more explicit and precise by graphing a horizontal line with a particular value, locating the intersection of the exponential function and that line, and then finding the coordinates of that intersection (by estimating or by using technology). This work previews further exploration on using the intersection of graphs to solve exponential problems in upcoming lessons.

Some students may need help understanding the equation for \(f\). Ask questions like, “What does the 250 of \(f(w) = 250 \boldcdot e^{(0.5w)}\) tell us about the initial population?” (It was first measured to be 250.) and “How is the population changing? By what factor?” (It is growing by a factor of \(e^{0.5}\) or about 1.65 every week.) to help them get started.

While students are not asked to find the intersection of the two graphs in the final question, it is important to see this intersection point in order to appreciate the value of the graph for determining when the population reaches a certain level. Encourage students to choose a different domain or range if they cannot see where the two graphs meet, such as by adjusting the graphing window to the following boundaries: \(0\lt w \lt 16\) and \(0\lt y\lt 150,\!000\).

Focus the discussion on the last question. Make sure students see why the intersection of the two graphs tells us the \(w\)-value when \(y\) is 100,000. We can use graphing technology to identify the \(w\)-coordinate of that intersection and obtain a more precise estimate than if we had just visually inspected the \(w\)-value when \(f(w) = 100,\!000\).