Data + Intervention Packet — Teacher Reference

Core Data — High-school past-30-day e-cig use (NYTS, %)

2021 omitted — methodology shifted to web-based collection during COVID, breaking comparability.

Sources

Decay Fit (2019 → 2024, used in lesson)

From 27.5% (2019) to 7.8% (2024) over 5 years, a continuous-rate fit gives:

P(t) = 27.5 · e−0.252 t   (t = years since 2019; r ≈ −25.2%/yr)

Equivalent annual factor ≈ 0.777 (so the rate is "about 22% of vapers stopping each year, net").

Interventions with measured effect sizes

FDA cartridge flavor enforcement (Feb 2020) Restricted flavored cartridge-based e-cigs (e.g. JUUL pods); allowed menthol + tobacco. Coincided with HS decline 27.5% → 19.6% (2019→2020). BUT disposable use jumped from 2.4% → 26.5% of users — partial loophole.
Federal Tobacco 21 (Dec 20, 2019) Raised minimum sales age 18 → 21. IOM model projects 25% reduction in smoking initiation among 15–17-year-olds. Perception "easy to buy" dropped 67.2% → 58.9% (2019→2020).
Truth Initiative — "This is Quitting" texts RCT in JAMA (2024). 7-month abstinence: treatment 37.8% vs control 28% — absolute +9.8 pp. ~750k teens enrolled.
State flavor ban — California (Dec 21, 2022) Total e-cig nicotine sold ↓ 37% vs control states; menthol cig sales ↓ 95.4%. But youth past-30-day e-cig prevalence ≈ unchanged year-1 (90% used flavored anyway via online/black market). Sales lever > survey lever in year 1.
State flavor ban + 75% excise tax — Massachusetts (Jun 2020) MA HS vaping 32.2% (2019) → 18.3% (2023) — a 43% drop. CT (no ban) saw similar decline (11.5% by 2023). Effect not cleanly attributable to MA tax alone.
Disposable-vape enforcement (2023→ FDA warning letters) Elf Bar use among HS vapers: 56.7% (2023) → 36.1% (2024) — concurrent with HS prevalence 10.0% → 7.8%. Mechanism: shrinking distribution channel.

Trauma awareness

Some students vape; some have peers/family who do; some have had vape-related conflicts at home. The framing: "The math describes a population. We're not asking who in this room vapes — we're asking what would change the trajectory." If a student opts out of an activity, accept silently. Don't single-out by name. Cool-down can be completed individually.

Reading the Curve
on Teen Vaping

Graphs & Logs · Decay since 2019

A pile of colorful disposable e-cigarettes (vapes) — the product class whose use among US high schoolers peaked in 2019 and has fallen sharply since.
Disposable electronic cigarettes — the device class that re-shaped the youth vaping market after the 2020 cartridge flavor ban. Photo by Wikimedia user Pannet, 2022 — CC BY-SA 4.0.

In 2019, more than 1 in 4 high schoolers vaped. In 2024, fewer than 1 in 12.

The curve is dropping fast.

What rate is it actually falling at — and what would it take to push it the rest of the way to zero?

Standards: HSF-LE.A.4 (use logs to solve exponential equations), HSA-REI.D.11 (intersection of two graphs as solution). This redesigned lesson keeps the IM math — reading a graph to estimate a solution, and using \(\ln\) to solve algebraically — but trades the unrealistic millionaire-account context for real CDC data students recognize. Today: fit the post-peak decay, find the half-life of the trend, and ask what trajectory would finish it.
Prep Checklist
  • ☐ Open the orange Data + Intervention Packet (top-right) before class — it has the citations and intervention cards.
  • ☐ Scientific calculators (with ln button) for every student.
  • ☐ Graphing technology (Desmos / Geogebra) projected for Activity 2.
  • ☐ Trauma: Vaping touches some students directly. Use third-person framing (“the population,” not “people in this room”). Allow silent opt-out without comment.
Warm-Up 5 min

Do Now — Reading a curve

f(t) = e^t 200 150 100 50 0 0 1 2 3 4 5 6 t f(t) y = 100 t = ?
On your own — then confirm with a partner.
  1. Use the graph to estimate \(\ln(100)\). What value of t makes \(e^{t} = 100\)?
  2. Why does the graph let you read \(\ln(100)\) at all? What does \(\ln(100)\) mean?
  3. How would you check your estimate on a calculator?
5 min total. Students work alone (~3 min), then pair-share. The conceptual move: \(\ln(100)\) is the answer to "to what power do we raise e to get 100?" — that is, the input t for which \(e^{t} = 100\). Visually, that is the t-coordinate where the curve crosses the dashed line \(y = 100\).

Listen for students who say "about 4.6" or "a little less than 5." Promote those. Calculator check: ln(100) ≈ 4.605, or e^4.6 ≈ 99.5.

No need to introduce vaping yet — the warm-up is the math primer that Activity 1 will then apply.
Warm-Up · Synthesis included

What does \(\ln(100)\) mean?

Two true statements about \(\ln(100)\). One in algebra, one in geometry. Both answer the same question.

“What does \(\ln(100)\) mean in algebra?” \(\ln(100)\) is the exponent we have to put on e to get 100. So if we write \(e^{t} = 100\), then by definition \(t = \ln(100)\). That is exactly what the natural log does — it pulls the exponent out.
“What does \(\ln(100)\) mean on the graph?” \(\ln(100)\) is the t-coordinate where the curve \(f(t) = e^{t}\) crosses the horizontal line \(y = 100\). The graph trades the algebra question for a geometry question: "where does the curve hit 100?"
“Best estimate?” From the graph: a little above 4.5. From the calculator: \(\ln(100) \approx 4.605\). Both say the same thing — \(e^{4.605} \approx 100\). The graph rounds; the log is exact.
click to advance discussion ▶
Try Saying \(\ln(100)\) means ___. On the graph, that shows up as ___.
Make sure the two interpretations land side-by-side — algebraic ("the exponent that makes \(e^{t} = 100\)") and geometric ("the input where the curve hits 100"). The whole lesson rests on that bridge. If students are stuck, prompt: "What does \(\ln\) do to a number? Where does it 'live' in an equation?"

Don't skip the calculator step. Students need to feel the equivalence: graph estimate ≈ 4.6, calculator says 4.605. Same answer, two tools.
Activity 1 · Launch 3 min

Teen Vaping — the post-2019 fall

Each year, the CDC runs the National Youth Tobacco Survey — a nationally representative, school-based survey. One question they ask: "Have you used an e-cigarette in the past 30 days?" Here is the high-school answer, year by year. Vaping rose to a peak in 2019 and has been falling since.
Year % of HS students vaping
201711.7%
201820.8%
2019 (peak)27.5%
202019.6%
202214.1%
202310.0%
20247.8%
HS past-30-day e-cig use, 2017–2024 30% 25% 20% 15% 10% 5% 0 2017 2018 2019 2020 2021 2022 2023 2024 peak 27.5% 7.8%
What stands out? What looks like a pattern, and what looks like a break in the pattern?
Launch (~3 min). Display the data. Let students notice and wonder. Likely noticings: rose fast 2017→2019; fell fast since; 2021 missing; the falling part looks like decay. 2021 was COVID-era methodology change — CDC says it isn't comparable; we drop it.

Don't fit anything yet. The work-time will fit a continuous-decay model to the post-peak years (2019→2024). Source: NYTS 2024 MMWR; full citations in the orange packet (top-right corner).
Reference · How To Activity 1 · Work Time 15 min
Reference · How To — Solve \(a \cdot e^{kt} = y\) with \(\ln\)
Step What to do Example: \(27.5 \cdot e^{-0.252\,t} = 5\)
1 Isolate the exponential. Divide both sides by the leading coefficient. \(e^{-0.252\,t} = \dfrac{5}{27.5} \approx 0.1818\)
2 Take the natural log of both sides. \(\ln\bigl(e^{-0.252\,t}\bigr) = \ln(0.1818)\)
3 Use \(\ln(e^{x}) = x\) to pull the exponent out. \(-0.252\,t = \ln(0.1818) \approx -1.704\)
4 Divide by the coefficient on \(t\). \(t = \dfrac{\ln(0.1818)}{-0.252} \approx 6.76\) years
★ Interpret in context. t is years since 2019. \(2019 + 6.76 \approx 2025.8\) — if the trend holds, the curve hits 5% in late 2025.
Activity 1 · Work Time
Fitting only the post-peak years (2019→2024), a good continuous-decay model is \(P(t) = 27.5 \cdot e^{-0.252\,t}\), where t is years since 2019 and \(P\) is the % of HS students vaping.
Problems
  1. Predict \(P(5)\). What % does the model give for 2024? How close is it to the real 7.8%?
  2. What is the model's continuous decay rate? What does the negative sign mean?
  3. Set up an equation to find when the curve drops to 5%. Solve it with \(\ln\).
  4. Use the same method to find when the model predicts 1% — effectively gone.
15 min · Activity 1 Work Time. Leave on screen. Students may use calculators throughout.

Problem 1: \(P(5) = 27.5 \cdot e^{-1.26} \approx 27.5 \cdot 0.284 \approx 7.81\%\). Within 0.01 pp of the real 7.8% — the fit is excellent for 2024.
Problem 2: \(-0.252\) per year, so a continuous decay rate of about \(25.2\%\) per year. Negative = decreasing. Equivalent annual factor: \(e^{-0.252} \approx 0.777\), so each year retains about 77.7% of the previous year's vapers.
Problem 3: \(t = \ln(5/27.5)/(-0.252) \approx 6.76\) yr ⇒ about late 2025.
Problem 4: \(t = \ln(1/27.5)/(-0.252) \approx 13.15\) yr ⇒ about early 2032.

Watch for: students who try to take \(\ln\) before isolating; students who divide by \(-0.252\) before taking \(\ln\). The How To controls for both.
Amplify · 4 inputs

Order per problem: Text → Math Response (named answer1, answer2, answer3, answer4) → Note (CL in </>).

Format examples: Example: 7.81 · Example: -0.252 · Example: 6.76 · Example: 13.15

Problem 1: P(5) = ? (% in 2024 from the model)
content: when answer1.submitted and answer1.numericValue > 7.78 and answer1.numericValue < 7.84 "Correct! P(5) = 27.5 · e^(-1.26) ≈ 7.81%. The model basically nails 2024 (real value 7.8%)." when answer1.submitted and answer1.numericValue > 7.79 and answer1.numericValue < 7.81 "Correct! P(5) ≈ 7.8% — the model matches the actual 2024 NYTS reading." when answer1.submitted and answer1.numericValue > 0.077 and answer1.numericValue < 0.079 "You're close — 0.078 is the decimal form. The data are reported as percents (7.8%, not 0.078). Try multiplying by 100." when answer1.submitted and answer1.numericValue > 27.49 and answer1.numericValue < 27.51 "That is P(0), the 2019 value. Substitute t = 5 (years since 2019) into 27.5 · e^(-0.252 · t)." when answer1.submitted and not isBlank(answer1.latex) "Almost — substitute t = 5 into 27.5 · e^(-0.252 · t). What does your calculator give?" otherwise ""
Problem 2: Continuous decay rate (per year, with sign).
content: when answer2.submitted and answer2.numericValue > -0.253 and answer2.numericValue < -0.251 "Correct! The continuous rate is -0.252 per year — the negative sign tells us the curve is decreasing." when answer2.submitted and answer2.numericValue > 0.251 and answer2.numericValue < 0.253 "Almost — the magnitude is right (0.252), but the curve is decaying. The exponent is negative: -0.252." when answer2.submitted and answer2.numericValue > -25.3 and answer2.numericValue < -25.1 "You're close — -25.2 is the rate as a percent per year. The coefficient in the exponent is -0.252 (the decimal form)." when answer2.submitted and not isBlank(answer2.latex) "Not quite. Look at the exponent in P(t) = 27.5 · e^(k·t). What is k?" otherwise ""
Problem 3: Year when P(t) = 5% (t = years since 2019).
content: when answer3.submitted and answer3.numericValue > 6.74 and answer3.numericValue < 6.78 "Correct! t = ln(5/27.5) / (-0.252) ≈ 6.76 years. So the model predicts 5% around late 2025 (2019 + 6.76)." when answer3.submitted and answer3.numericValue > 2025.7 and answer3.numericValue < 2025.9 "Correct! 2025.8 is the year-form. In years-since-2019 form, t ≈ 6.76." when answer3.submitted and answer3.numericValue > 1.694 and answer3.numericValue < 1.714 "Almost — 1.704 is ln(5/27.5) (the value of -0.252 · t). One more step: divide by -0.252 to get t." when answer3.submitted and answer3.numericValue > -1.714 and answer3.numericValue < -1.694 "You're close — -1.704 is ln(5/27.5). Don't forget the negative on -0.252 — the t value should be positive." when answer3.submitted and answer3.numericValue > 0.1817 and answer3.numericValue < 0.1819 "I see what happened — 0.1818 is 5/27.5. That's the value of e^(-0.252·t), not t itself. Take ln of that, then divide by -0.252." when answer3.submitted and not isBlank(answer3.latex) "Not quite — set 27.5 · e^(-0.252·t) = 5, divide by 27.5, take ln of both sides, then divide by -0.252." otherwise ""
Problem 4: Year when P(t) = 1%.
content: when answer4.submitted and answer4.numericValue > 13.13 and answer4.numericValue < 13.17 "Correct! t = ln(1/27.5) / (-0.252) ≈ 13.15 years. Model predicts 1% around early 2032 (2019 + 13.15)." when answer4.submitted and answer4.numericValue > 2032.0 and answer4.numericValue < 2032.3 "Correct! 2032.15 in calendar form — about February 2032 if the model holds." when answer4.submitted and answer4.numericValue > 3.30 and answer4.numericValue < 3.32 "Almost — 3.31 is ln(27.5) (or close to ln(1/27.5) without the sign). After dividing by -0.252 you should get about 13.15." when answer4.submitted and answer4.numericValue > -3.32 and answer4.numericValue < -3.30 "You're close — -3.314 is ln(1/27.5). Now divide by -0.252 to get t." when answer4.submitted and not isBlank(answer4.latex) "Not quite — same method as Problem 3, but with 1 in place of 5." otherwise ""
Activity 1 · Synthesis 5 min

The model's predictions — and what \(\ln\) buys us

Question Equation Solved with \(\ln\) In years
When does \(P\) hit 5%?\(27.5\,e^{-0.252\,t} = 5\)\(t = \dfrac{\ln(5/27.5)}{-0.252}\)≈ 6.76 (late 2025)
When does \(P\) hit 1%?\(27.5\,e^{-0.252\,t} = 1\)\(t = \dfrac{\ln(1/27.5)}{-0.252}\)≈ 13.15 (early 2032)
When does \(P\) halve?\(27.5\,e^{-0.252\,t} = 13.75\)\(t = \dfrac{\ln(0.5)}{-0.252}\)≈ 2.75 (the half-life)

Notice the pattern: every "when does it reach y?" question collapses into the same shape, \(t = \dfrac{\ln(y / 27.5)}{-0.252}\). The natural log is the operation that pulls the exponent out. That's all it does.

“Why does \(\ln\) work here? Why not \(\log_{10}\)?” Because the model is built with base e. The natural log is exactly the inverse of \(e^{x}\): \(\ln(e^{x}) = x\). If our model used \(10^{x}\), we would use \(\log_{10}\). The point: the log we pick has to match the base of the exponential.
“What is a half-life here, and why is it the same for any starting value?” The half-life is how long it takes for the % vaping to fall by half — 2.75 years in this model. It does not depend on whether you start at 27.5% or 7.8% — that is the equal-factors-over-equal-intervals property from Lesson 5. Decay halves are a property of the rate, not the starting amount.
“Should we trust this model out to 2032?” Honest answer: probably not blindly. We fit five data points. If interventions weaken or new products appear, the rate changes. The model is a baseline — what happens if nothing changes. Comparing reality to that baseline is how you tell whether something did change.
click to advance discussion ▶
Try Saying The natural log lets us solve \(e^{kt} = y\) for t because ___.
Try Saying Extrapolating to 2032 is/isn't fair because ___.
Sequence the discussion: (1) why \(\ln\) is the right tool here, (2) what the half-life means, (3) honest model limits. The third question matters — this is the lesson where students start treating models as baselines for comparison rather than crystal balls. Lesson 16 (Part 2) will push further on this.
Math As A Second Language

Solving \(a \cdot e^{kt} = y\) with the natural log

Math · We Say · Meaning
★ Math (given)
\(t = \dfrac{\ln(y/a)}{k}\)
▲ We Say
“t equals the natural log of y-over-a, divided by k”or“t is the log of the ratio, scaled by one over k”
● Meaning
To solve \(a \cdot e^{kt} = y\) for the time t: divide by a, take \(\ln\) of both sides, then divide by k. The natural log pulls the exponent out; k pops out as a coefficient.
For our vaping model — \(a = 27.5\), \(k = -0.252\) \[t = \dfrac{\ln(y/27.5)}{-0.252}\quad\Rightarrow\quad y = 5 \;\Rightarrow\; t \approx 6.76\;\text{yr}\]
Standalone MASL slide. The general form \(a \cdot e^{kt} = y\) was modeled in Activity 1; this slide names the solved-for-t formula. Connect: "Whenever you see \(a \cdot e^{kt}\) anywhere — bacteria, money, cooling, decay — the time-to-reach-y question always lands at this formula."
Activity 2 · Launch 3 min

Same answer, different tool

Activity 1 used \(\ln\) to solve \(27.5\,e^{-0.252\,t} = 5\) and got \(t \approx 6.76\) years.

Now we'll graph it. We'll plot the curve \(y = 27.5\,e^{-0.252\,t}\) and a horizontal line \(y = 5\), and find where they cross. The t-coordinate of the crossing should match what \(\ln\) gave us.
A system in disguise

\(\begin{cases} y = 27.5\,e^{-0.252\,t} \\ y = 5 \end{cases}\)   — intersection point gives the time when vaping rate hits 5%.

Why does the intersection of two graphs give the answer to a single equation?
Launch (~3 min). The big move from Activity 1: solving \(f(t) = c\) is the same as finding where the graphs of \(y = f(t)\) and \(y = c\) intersect. Algebraically: any \((t, y)\) on both curves satisfies both equations — so the system's solution is the equation's solution.

Activity 2 demos this in Desmos / Geogebra. Project a graphing tool now if possible. Two graphs, one intersection, one answer.
Activity 2 · Work Time 10 min
P(t) = 27.5 · e^(-0.252 t) vs. y = 5 30 25 20 15 10 5 0 0 2 4 6 8 10 12 t (years since 2019) % HS vaping y = 5 (6.76, 5) ≈ late 2025 2019: 27.5% 2022: 14.1%
Problems
  1. From the graph, estimate when the curve crosses \(y = 5\) (5% of HS students vaping). Match this to your Activity 1 \(\ln\) answer.
  2. Sketch (or imagine) the line \(y = 1\) on the same axes. About when does the curve cross it?
  3. What's the half-life of the curve — when does \(P\) drop from 27.5% to 13.75%? Read it off the graph; check with \(\ln\).
  4. Which tool — the graph or \(\ln\) — is more useful, and when?
Try Saying Reading the graph gives me ___; using \(\ln\) gives me ___.
10 min. Students compare graph readings to log answers.

Problem 1: Graph: about t = 6.7 to 6.8. Log (Act 1): 6.76. Match.
Problem 2: Graph extends past 12 years. Log: 13.15. Note: the curve is so flat past t = 10 that the graph estimate is worse than the log there — that's the lesson.
Problem 3: Half-life ≈ 2.75. Graph reading at 13.75% is approximate; log gives \(\ln(0.5)/(-0.252) \approx 2.751\).
Problem 4: Discussion target. Graph: faster, gives intuition, easy when target is in clear part of curve. Log: exact, works in flat regions, doesn't need a window choice. Both correct — different tools for different jobs.
Amplify · 3 inputs

Order per problem: Text → Math Response (named answer1, answer2, answer3) → Note. Problem 4 → Text Response only (named answer4).

Format examples: Example: 6.76 · Example: 13.15 · Example: 2.75

Problem 1: When does P(t) hit 5? (graph estimate, in years since 2019)
content: when answer1.submitted and answer1.numericValue > 6.5 and answer1.numericValue < 7.0 "Correct! About 6.7-6.8 years from the graph — matches the ln answer of 6.76 from Activity 1." when answer1.submitted and answer1.numericValue > 6.74 and answer1.numericValue < 6.78 "Correct! 6.76 — exactly the ln value. Graph reading at this point is precise." when answer1.submitted and answer1.numericValue > 4.99 and answer1.numericValue < 5.01 "Hmm — 5 is the y-value (the 5% target), not the t-value. The intersection's t-coordinate is what we want." when answer1.submitted and answer1.numericValue > 2025.7 and answer1.numericValue < 2025.9 "Correct in calendar form! 2025.8 ≈ late 2025, which equals 2019 + 6.76." when answer1.submitted and not isBlank(answer1.latex) "Almost — find where the blue curve and the red dashed line cross. Read the t-coordinate (the input)." otherwise ""
Problem 2: When does P(t) hit 1?
content: when answer2.submitted and answer2.numericValue > 13.0 and answer2.numericValue < 13.3 "Correct! About 13.1-13.2 years — matches the ln answer of 13.15. (The curve is very flat by then — graph reading gets fuzzier.)" when answer2.submitted and answer2.numericValue > 11.0 and answer2.numericValue < 12.5 "Close — but the curve hits 1 a little later. Try ln(1/27.5) / (-0.252) and see if your graph estimate matches." when answer2.submitted and answer2.numericValue > 9.99 and answer2.numericValue < 10.01 "Hmm — at t = 10, the curve is around 2.2%, not 1%. Look further right; the curve gets very flat." when answer2.submitted and not isBlank(answer2.latex) "Almost — extend the graph past t = 12 if needed, or solve 27.5 e^(-0.252 t) = 1 with ln." otherwise ""
Problem 3: Half-life of the curve (years).
content: when answer3.submitted and answer3.numericValue > 2.74 and answer3.numericValue < 2.76 "Correct! Half-life = ln(0.5) / (-0.252) ≈ 2.75 years. From 27.5% to 13.75% takes about 2.75 yr — and the same time again to halve again." when answer3.submitted and answer3.numericValue > 2.7 and answer3.numericValue < 2.8 "Correct! Roughly 2.75 years — matches the algebraic value." when answer3.submitted and answer3.numericValue > 13.7 and answer3.numericValue < 13.8 "Hmm — 13.75% is the y-value (half of the starting %), not the half-life. The half-life is the t-value when y reaches 13.75." when answer3.submitted and answer3.numericValue > 5.4 and answer3.numericValue < 5.6 "Almost — 5.5 might be when the curve halves twice (27.5 → 13.75 → ~6.9), which is two half-lives. One half-life is about 2.75." when answer3.submitted and not isBlank(answer3.latex) "Not quite — the half-life is t such that P(t) = 13.75. Solve 27.5 e^(-0.252 t) = 13.75." otherwise ""
Problem 4: Which tool — graph or ln — is more useful, and when? [Text Response only]
content: when answer4.submitted and not isBlank(answer4.content) "Thank you — share your reasoning during the synthesis discussion." when answer4.submitted and isBlank(answer4.content) "Please write a short comparison before submitting." otherwise ""
Activity 2 · Synthesis 5 min

Two tools, same answer — when to reach for each

Graph (intersection) Natural log
What you doPlot \(y = f(t)\) and \(y = c\); read the intersection's t.Isolate the exponential, take \(\ln\), divide.
StrengthVisual; shape-of-curve intuition; easy in the steep region.Exact; works when the curve is flat; no graphing tool needed.
WeaknessImprecise where the curve flattens; depends on the window.Hides the shape; one bad sign error and the answer is silently wrong.
When to useQuick estimate; checking reasonableness; teaching / explaining.Final answer; very flat regions; communicating exact values.

Both tools answer the same question. The smart move is to use one to check the other.

Try Saying For this question I would reach for ___ first because ___, then check with ___.
Don't let students treat this as one is right, the other is wrong. They are different lenses on the same equation. The graph makes the geometry visible; the log makes the algebra exact. Practitioners use both. Tomorrow (Lesson 16) builds on this when the equations get harder to isolate.
Agency Move · Pick a rate 5 min

What rate kills it by 2030?

The setup The current model says \(P(t) = 27.5\,e^{-0.252\,t}\) with t = years since 2019. At −25.2% per year (continuous), 1% arrives in early 2032.

Suppose the goal is 1% by January 2030 — \(t = 11\) years from 2019. What continuous rate k would we need?
Solve for k \[27.5 \cdot e^{k \cdot 11} = 1 \;\;\Longrightarrow\;\; k = \dfrac{\ln(1/27.5)}{11} \approx -0.301\]
A rate of −0.301 instead of −0.252 — that is the extra push we are asking for. It is bigger than today's rate by about 0.049 per year — the model needs to be pulled down faster.

Where does that extra push come from? The intervention cards on the next slide show what's been measured to actually move the rate. Pick a bundle of two or three. Estimate how much extra k they buy.
Setup slide for the agency move (2 min). The math is one line: \(k = \ln(1/27.5)/11 \approx -0.301\). What matters is what the next slide asks: which interventions, in combination, plausibly produce the −0.049 push?
Agency Move · Defend a bundle included

Pick two or three. Defend the rate bump.

Each card lists a real, measured intervention and the actual effect documented in the literature. Pick two or three to combine. Argue what extra k you think the bundle produces, and why. There is no "right answer" — only honest ones.

FDA cartridge flavor ban (2020) Past use: HS dropped 27.5% → 19.6% in one year, but disposables jumped 2.4% → 26.5% of users. Net: large but leaky.
Federal Tobacco 21 (Dec 2019) Modeled −25% on initiation among 15–17-yr-olds. "Easy to buy" perception 67% → 59% in year 1. Effect: real but slow.
"This is Quitting" texts (Truth) RCT: 7-mo abstinence 37.8% vs 28% control. +9.8 pp absolute. Per-user effect: strong.
State flavor ban (CA, 2022) Total e-cig nicotine sold ↓37% vs control states. Year-1 youth prevalence ≈ flat. Sales lever > survey lever.
Disposable enforcement (FDA letters, 2023→) Elf Bar share among HS vapers 56.7% → 36.1% (one year). Distribution-channel effect.
Vape excise tax (MA-style, 75%) MA HS prevalence 32.2% → 18.3% (2019→23). Neighboring CT (no ban) saw similar drop. Effect: confounded.
Try Saying I would push for ___ + ___. I think it shifts k from −0.252 to about ___ because ___.
Agency move (3 min, popcorn discussion). Two or three voices. Don't grade defensibility on this slide — the unit assessment will. The point now: students see they have a rate-knob, not just a number-knob. Scaffold with the sentence frame. Acknowledge skeptical answers ("None of these would shift it that much, because the leveling-off is structural") — that is a defensible position.

Closing line: "The math doesn't decide what to push for. But it tells you whether what you picked could plausibly do the job."
Lesson Synthesis 3 min

Two tools, one trajectory

Today we asked the same question in two different ways — "when does \(27.5\,e^{-0.252\,t} = 5\)?" — and got the same answer (\(t \approx 6.76\)) two different times, with two different tools.

“How do we solve an exponential equation by graphing?” Graph the curve \(y = f(t)\) and the horizontal line \(y = c\). Find the t-coordinate where they intersect. Any \((t, y)\) on both curves makes both equations true — that is the system's solution, and it's our equation's solution too.
“How do we solve it with \(\ln\)?” Isolate the exponential, take \(\ln\) of both sides, divide by the coefficient on t. The natural log pulls the exponent out of the equation. Result: a single closed-form expression for t.
“Each tool has its job — what's the job split?” Graph: shape, intuition, quick estimates, easy in the steep part of the curve. \(\ln\): exact answer, works in flat regions, written form for communication. Real practitioners use both — one to check the other.
“What was the agency move today?” A trajectory has a rate. Real interventions move the rate. We can compute what rate would kill teen vaping by 2030 and ask which interventions, in combination, are big enough to do it. The math doesn't pick the policy — but it tells you whether your pick has a chance.
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I Can… (SWBAT)
Use a graph to identify solutions to exponential equations.
Use the natural logarithm to calculate solutions to exponential equations.
Compare the two methods and pick the right tool for the job.
Lesson Synthesis (3 min). Cold-call through the four points. Make sure students name both methods explicitly. SWBAT closes the lesson — do not put it on the title slide.
Cool-Down 5 min

A 12th-grade vape model

Among 12th-grade boys specifically, vaping has fallen even faster than the overall HS rate. Suppose for this group the model is \(B(t) = 30 \cdot e^{-0.30\,t}\), where t is years since 2019 and \(B\) is % vaping.
B(t) = 30 · e^(-0.30 t) 30% 20% 10% 0 0 2 4 6 8 10 t (years since 2019) y = 5
Problems
  1. Use the graph to estimate when \(B(t) = 5\). Round to the nearest whole year.
  2. Write and solve an equation with \(\ln\) to find t precisely. Show your reasoning.
Cool-Down · 5 min. Mirrors the IM cool-down ("A Moldy Surface") with the lesson's data context. Both methods, both answers.

Problem 1: Graph estimate ≈ t = 6 (just before 6, actually).
Problem 2: \(30\,e^{-0.3\,t} = 5\) ⇒ \(e^{-0.3 t} = 1/6\) ⇒ \(t = \ln(1/6)/(-0.3) = \ln(6)/0.3 \approx 5.97\) years — about late 2024 for the 12th-grade-boys subgroup.

Common errors: dividing by 0.3 before taking ln; forgetting the negative sign; treating 5 (the y-target) as the answer.
Amplify · Cool-Down Setup

Problem 1: Text component (full problem text + Example: 6) → Math Response named answer1 → Note with CL.

Problem 2: Text component (full problem text + Example: 5.97) → Math Response named answer2 → Note with CL.

Use numericValue ranges for real-number checks.

Problem 1: Graph estimate of t when B(t) = 5 (whole year).
content: when answer1.submitted and answer1.numericValue > 5.5 and answer1.numericValue < 6.5 "Correct! About t = 6 — the curve crosses y = 5 just before t = 6, so 6 is the right whole-year estimate." when answer1.submitted and answer1.numericValue > 4.99 and answer1.numericValue < 5.01 "Hmm — 5 is the y-value (the 5% target), not the t-value. Read the t-coordinate of the intersection point." when answer1.submitted and answer1.numericValue > 4.4 and answer1.numericValue < 5.4 "Almost — at t = 5, B(5) ≈ 6.7, which is above 5. The intersection is closer to t = 6." when answer1.submitted and answer1.numericValue > 6.5 and answer1.numericValue < 8.0 "You're close — but the curve crosses y = 5 before t = 7. Try estimating again or check with ln." when answer1.submitted and not isBlank(answer1.latex) "Read where the purple curve crosses the red dashed line y = 5. Round the t-coordinate to the nearest whole year." otherwise ""
Problem 2: Solve 30 · e^(-0.3 t) = 5 with ln.
content: when answer2.submitted and answer2.numericValue > 5.96 and answer2.numericValue < 5.98 "Correct! t = ln(6) / 0.3 ≈ 5.97 years — so about late 2024 for the 12th-grade-boys subgroup." when answer2.submitted and answer2.numericValue > 5.9 and answer2.numericValue < 6.0 "Correct! Approximately 5.97 years — well done." when answer2.submitted and answer2.numericValue > 1.78 and answer2.numericValue < 1.80 "Almost — 1.79 is ln(6). One more step: divide by 0.3 to get t ≈ 5.97." when answer2.submitted and answer2.numericValue > -1.80 and answer2.numericValue < -1.78 "You're close — -1.79 is ln(1/6). Don't forget the negative on -0.3 when you divide; the answer should be positive." when answer2.submitted and answer2.numericValue > 0.166 and answer2.numericValue < 0.168 "I see what happened — 1/6 ≈ 0.167 is the value of e^(-0.3 t), not t itself. Take ln, then divide by -0.3." when answer2.submitted and answer2.numericValue > -5.98 and answer2.numericValue < -5.96 "You're close — got the magnitude right (5.97) but with the wrong sign. Re-check: dividing ln(1/6) by -0.3 gives a positive number." when answer2.submitted and isBlank(answer2.latex) "Enter your answer in years, rounded to the hundredths place. Example: 5.97" when answer2.submitted "Not quite. Set 30·e^(-0.3 t) = 5, divide by 30, take ln of both sides, then divide by -0.3." otherwise ""
Lesson Summary

Two paths to the same time

To find the input t that makes \(a \cdot e^{kt} = c\), you have two reliable methods. They give the same number; they answer the same question. Use whichever is sharper for the situation in front of you — or use both and check.
By graphing
Plot \(y = a \cdot e^{kt}\) and \(y = c\) on the same axes. The t-coordinate of their intersection is the solution.
By natural log
Isolate the exponential, take \(\ln\) of both sides, then divide: \[t = \dfrac{\ln(c/a)}{k}\]
From today's data Real CDC data: HS vaping fell from 27.5% (2019) to 7.8% (2024). Continuous rate ≈ −25.2% per year; half-life ≈ 2.75 years. Real interventions — flavor bans, T21, "This is Quitting," disposable enforcement — have already moved this rate. The next move is decided.
Lesson Summary slide. Read aloud. Tomorrow's lesson (Part 2) extends this to comparison contexts — two exponentials at once, two intersection points, and what their crossing year tells you about which trajectory wins.