📋 Data + Intervention Packet — Teacher Reference
UAW 2023 Stand-Up Strike — verified data points
  1. Strike fund balance entering strike: ~$825 million (May 2023, before strike began Sept 15). Source: Morningstar, Oct 2023; Wikipedia · 2023 UAW strike (accessed May 2026).
  2. Total UAW members at Big 3: ~146,000. Source: NPR, Oct 2023.
  3. Strike pay: $500 / week / striking worker, starting day 8. (Increased from $400 in Aug 2023.) Source: UAW press release; ABC News.
  4. Stand-Up Strike strategy (rolling/targeted, not full plant): began Sept 15 with ~13,000 workers; peaked at ~46,000 (~one-third of membership). Source: Wikipedia; CNBC.
  5. Length: 44 days (Sept 15 – Oct 30, 2023). Final fund spend ≈ $58M (per Evercore ISI estimate via Morningstar) — well under the projected burn for a full-membership strike.
  6. Final contract: 25% raises over 4.5 years (11% in year 1), COLA restored, EV plant pathway, end of wage tiers. Long-time worker pay rises ~33%; some new/temp workers see pay more than double. Source: CBS News; NPR.
  7. Morningstar projection (had it been a full strike of all 146,000): fund estimated to last 6–11 weeks depending on healthcare. Source: Morningstar.
Intervention menu (with measured effect sizes) Math behind the lesson: All sources accessed May 2026 from articles published Sept–Nov 2023.

Solving Exponential Equations
— or — How long does $825M of leverage last?

President Joe Biden walks along the UAW picket line at the GM Willow Run Distribution Center on September 26, 2023, in Belleville, Michigan. First sitting US president to join a picket line.
UAW Stand-Up Strike, Belleville MI, Sept 26 2023 — first sitting US president to join a picket line. The strike lasted 44 days and won the largest auto contract since 2007. The fund that paid for it: about $825M, $500/week per striker.
Photo: Adam Schultz / Official White House Photo, public domain.

You can solve 2x = 18 in one step.

Why can't you solve 2x = 18 the same way —
and what do you do instead?

Same question as: "given $825M and a $73M-per-week burn rate, exactly how many weeks does our leverage last?"

Standards: HSF-LE.A.4. Lesson 14 closes the section on logarithms. Today: students solve \(a \cdot b^{t} = y\) for \(t\) using log and ln. The wrench lens: this is the structure of strike fund leverage math — how long can the union outlast the company?

Trauma-aware note: Some students may have family in unions or in plants that have struck (Philly has SEIU, AFSCME, AFT, UAW Local 1069 at Boeing-Helicopter, etc.). Briefly invite, do not require, students to share. Sample teacher line: "Some of you may know people who've been on strike or in a union. You can share, ask, or just listen — all three are fine today."

Prep checklist
Warm-Up 5 min

Do Now — A Valid Solution?

A student solves \(5 \cdot 2^{3x} = 18\) like this. Each step is on its own line. Be ready to explain what changed between consecutive lines.
\[5 \cdot 2^{3x} = 18\] \[2^{3x} = \tfrac{18}{5}\] \[3x = \log_{2}\!\left(\tfrac{18}{5}\right)\] \[x = \tfrac{1}{3}\,\log_{2}\!\left(\tfrac{18}{5}\right)\]
Complete on your own. Then confirm your answers with a partner.
  1. Explain what happened between line 1 and line 2.
  2. Explain what happened between line 2 and line 3.
  3. Is line 4 a finished solution — or do we still need to "evaluate" the log?
5 min for warm-up (slides 2–3). Independent ~3 min, then pair-share. Listen for: students who recognize that the product \(3x\) can be treated as a single object (it is "the exponent"), and who see lines 2 and 3 as the same equation written two different ways — exponential form and log form.

Watch-for: some students will say line 4 isn't done because there's still a log in it. That's the misconception to surface in synthesis: writing the exact answer using log notation IS finishing. Evaluating it with a calculator gives an approximation.
Warm-Up · Synthesis 5 min

A Valid Solution — Discussion

The four lines are the four moves you make on every exponential equation we'll see today. Once you can name what changed, you can run the move on harder equations — including the strike-fund equation in Activity 2.

“Line 1 → Line 2: what changed?” Both sides were divided by 5 to isolate the exponential part: \(\tfrac{5 \cdot 2^{3x}}{5} = \tfrac{18}{5}\), leaving \(2^{3x} = \tfrac{18}{5}\). The form is now (base)(exponent) = (constant) — exactly the format we need.
“Line 2 → Line 3: what changed?” The equation was rewritten in logarithmic form. \(2^{3x} = \tfrac{18}{5}\) means \(3x\) is the exponent that turns 2 into \(\tfrac{18}{5}\) — which is exactly what \(\log_{2}\!\left(\tfrac{18}{5}\right)\) names. So \(3x = \log_{2}\!\left(\tfrac{18}{5}\right)\). Same equation, two notations.
“Is line 4 a finished solution?” Yes — it is the exact answer. Dividing by 3 to isolate \(x\) gives \(x = \tfrac{1}{3}\log_{2}\!\left(\tfrac{18}{5}\right)\). Punching this into a calculator gives \(x \approx 0.6155\), but that decimal is an approximation. The log expression is the precise value.
click to advance discussion ▶
Try Saying Between line ___ and line ___, the move was ___ because ___.
Try Saying Line 4 is/is not finished because ___.
The two big takeaways: (1) the (base)(exp) = (constant) structure is the gateway to using a log; (2) a log expression is an exact value — the calculator gives an approximation.

If a student insists line 4 isn't done because "the log isn't simplified," ask: "Is \(\sqrt{2}\) finished, or do we have to write 1.414 to be done?" Same idea — log notation is exact, just like radical notation.
Lens · Today's Math, Today's Story 3 min

Why we're solving \(a \cdot b^{t} = y\) today

The Wrench A union strike works when workers can outlast their employer financially. The math behind that — "how many weeks of leverage do we have?" — is exactly the kind of equation we're solving today: solve for t when t is locked inside an exponent.

The numbers (UAW 2023 Stand-Up Strike)

Quick estimate — no calculator. If every member walked out and got $500/week, roughly how many weeks could the fund last?
3 min. This slide is the bridge: it names the math (\(a \cdot b^{t} = y\), solve for \(t\)) and the story (strike fund leverage clock) at the same time. Take 1–2 noticings on the data packet, then ask the estimate question. Don't resolve — that's the warm-up of Activity 1.

Sample student answer to the estimate: ≈ 11 weeks. Reasoning: 146,000 × $500 = $73M per week; $825M ÷ $73M ≈ 11.3 weeks. (This matches Morningstar's actual 2023 estimate of 6–11 weeks depending on healthcare.) Some students may forget to multiply by membership; that's a productive error to surface.

Trauma-aware: If a student says, "my mom was on strike at X, they got nothing like $500" — affirm the variation: not all unions have a strike fund; some pay less. UAW's was an unusually well-resourced fund.
Activity 1 · Launch 3 min

Meet the Natural Logarithm

The strike fund earns interest while it sits. Public donations also flow in continuously — in 2023 the SAG-AFTRA Foundation pulled in $15M in just 3 weeks, and the Hollywood Entertainment Community Fund's outflows spiked from $75k/week to $400–500k/week. When growth or decay is continuous, the natural base \(e \approx 2.718\) is the right tool. Logs with base \(e\) get a special name and notation:
Math (notation)
\(\ln(x) = \log_{e}(x)\)
We Say
"the natural log of x"
or"l-n of x"
Key idea \(\ln(x)\) means: the exponent that turns \(e\) into \(x\). Same logic as \(\log_{2}\) or \(\log_{10}\) — just a different base.
On your calculator, find the e button and the ln button. Verify: \(\ln(e) = 1\) and \(\ln(10) \approx 2.302585\). What does the second of these mean in plain language?
3 min. Display the MASL pair. Make sure students locate the \(\ln\) and \(e\) keys. Many calculators have \(\ln\) above the natural log key. The disc-prompt answer: "\(e\) raised to the power 2.302585 gives 10." Quick sanity: \(2.7^{2.3} \approx 10\). ✓

Why this matters in the strike-fund context: if the fund earns continuous interest at rate \(r\), it grows as \(F_{0} \cdot e^{r t}\) — that is the only place where \(e\) really earns its keep over base 10.
MASL · Vocabulary
Natural log — the equivalence
Math
\[\begin{aligned} &\text{if } \ln(x) = y \\ &\text{then } e^{y} = x \end{aligned}\]
We Say
"the natural log of x equals y
means
e to the y equals x"
Meaning
\(\ln(x)\) is the exponent that turns the base e into x. Logarithmic and exponential form are the same equation written two ways.
Example \(\ln(18) = y\) is the same equation as \(e^{y} = 18\). On a calculator: \(y \approx 2.890\). Check: \(e^{2.890} \approx 18\). ✓
MASL slide for the natural log equivalence. This pair is the engine for solving any equation of the form \(e^{?} = \text{something}\). Press the point: logarithmic form and exponential form are the same equation. They are not "convert to log to solve" — they're the same statement, with the unknown moved to a different spot.
Activity 1 · Work Time 7 min

Natural Logarithm — Complete the Table

Each row is the same equation written two ways. The first row shows you the pattern. Fill in the missing form. Use a calculator only on the last column.
Exponential form Logarithmic form
given \(e^{0} = 1\) \(\ln(1) = 0\)
a. \(e^{1} = e\) ?
b. \(e^{2} = e^{2}\) ?
c. \(e^{x} = 825\) ?
d. ? \(\ln(46) = x\)
e. ? \(\ln(2) = x\)
Try Writing For row ___, the missing form is ___ because ___.
After you finish
Estimate & check (calculator OK)
  1. Use the \(\ln\) key to estimate the value of \(x\) in row c. Round to 3 decimal places.
  2. Use the \(\ln\) key to estimate the value of \(x\) in row d. Round to 3 decimal places.
  3. Without a calculator, is \(\ln(2)\) bigger than 1 or smaller than 1? Explain.
7 min · Activity 1 Work Time. Leave on screen during work time.

Table answers: a. \(\ln(e) = 1\) · b. \(\ln(e^{2}) = 2\) · c. \(\ln(825) = x\) · d. \(e^{x} = 46\) · e. \(e^{x} = 2\)

Estimates: Q1: \(\ln(825) \approx 6.715\).   Q2: \(\ln(46) \approx 3.829\).   Q3: \(\ln(2) \approx 0.693\) — smaller than 1, because \(e^{1} = 2.718 > 2\), so the exponent that gives 2 must be less than 1.

Watch-for: students who write "\(\ln(e^{2}) = e^{2}\)" — they are confusing the value of the log with the exponent. Press: "What number is the exponent? That's the answer."

The 825 and 46 in the table foreshadow Activity 2: 825 = strike fund balance ($M), 46 = peak number on strike (thousands). The math is intentional.
Amplify · 3 inputs

Order per problem: Text → Math Response (named answer1, answer2) → Note. Problem 3 → Text Response (named answer3).

Format examples: Example: 6.715 · Example: 3.829

Problem 1: Estimate ln(825) to 3 decimal places.
content: when answer1.submitted and answer1.numericValue > 6.71 and answer1.numericValue < 6.72 "Correct! ln(825) ≈ 6.715. So e^(6.715) ≈ 825." when answer1.submitted and answer1.numericValue > 2.91 and answer1.numericValue < 2.92 "Almost — 2.916 is log_10(825), not ln(825). Use the ln button (base e), not the log button (base 10)." when answer1.submitted and not isBlank(answer1.latex) "Almost — make sure you used the ln key (base e). Round to 3 decimal places." otherwise ""
Problem 2: Estimate ln(46) to 3 decimal places.
content: when answer2.submitted and answer2.numericValue > 3.82 and answer2.numericValue < 3.84 "Correct! ln(46) ≈ 3.829. So e^(3.829) ≈ 46." when answer2.submitted and answer2.numericValue > 1.66 and answer2.numericValue < 1.67 "Almost — 1.663 is log_10(46), not ln(46). Use the ln button." when answer2.submitted and not isBlank(answer2.latex) "Almost — use the ln key (natural log, base e). Round to 3 decimal places." otherwise ""
Problem 3: Is ln(2) bigger or smaller than 1? Explain. [Text Response]
content: when answer3.submitted and not isBlank(answer3.content) "Thanks — share your reasoning in the discussion. (Hint: e^1 = 2.718 > 2.)" when answer3.submitted and isBlank(answer3.content) "Please write a short explanation before submitting." otherwise ""
Activity 1 · Synthesis 5 min

Natural Logarithm — Discussion

The natural log isn't a new operation — it's the same logarithm idea you already know, with base \(e\) instead of base 10 or base 2.

“Row c: \(e^{x} = 825\). Why is the log form \(\ln(825) = x\)?” Because \(\ln\) is the inverse of \(e^{(\ldots)}\). \(e^{x} = 825\) asks "what exponent on \(e\) gives 825?" The natural-log notation is the name for that exponent: \(\ln(825) = x\). On a calculator, \(x \approx 6.715\).
“Row d: \(\ln(46) = x\). Why is the exponential form \(e^{x} = 46\)?” Same equation, written backwards. \(\ln(46) = x\) says "the exponent that turns \(e\) into 46 is \(x\)." That sentence in symbols is \(e^{x} = 46\). Calculator: \(x \approx 3.829\).
“Without a calculator, why is \(\ln(2) < 1\)?” \(e^{1} = e \approx 2.718\), which is bigger than 2. So to land on 2 we need an exponent smaller than 1. (Calculator confirms \(\ln(2) \approx 0.693\).) Same kind of estimation move you'd use for \(\log_{10}(5)\) being between 0 and 1.
click to advance discussion ▶
Try Saying \(\ln(\,\square\,) = x\) and \(e^{x} = \square\) are the same equation because ___.
The conceptual move: natural log is just \(\log_{e}\) — the underscore-base-letter is just hidden by convention. Once that lands, students can use \(\ln\) the same way they used \(\log_{10}\) in lessons 12–13. The next activity puts it to work on real strike-fund equations.
Activity 2 · Launch 3 min

The Leverage Clock

A strike fund of $825 million. Strike pay $500 / week / striker. Three different strategies. Each one is an equation in t — and t is the number of weeks of leverage.
Scenario A · Full strike
All 146,000 members walk out at once.
Burn: 146,000 × $500 = $73M / week.
Solve: \(73t = 825\).
Scenario B · Stand-Up Strike
Only ~46,000 strike at peak (UAW's actual 2023 plan).
Burn: 46,000 × $500 = $23M / week.
Solve: \(23t = 825\).
Scenario C · Exponential burn
Burn rate grows 9% per week (more plants strike, harder to feed). Fund left after t weeks: \(F(t) = 825 \cdot (0.91)^{t}\).
When does F hit $100M?
Which scenario can you solve without a logarithm? Which one needs one? Why?
3 min. Don't solve any of these yet — just notice the structure.

Sample answers: A and B are linear (\(73t = 825\), \(23t = 825\)) → solve by dividing, no log needed. C has \(t\) locked in an exponent → that's the form that needs a log to extract \(t\). This is exactly the "why we need logs" question from the title slide.

Note on numbers: A and B give 11.3 and 35.9 weeks. UAW's actual 2023 strike was 6.3 weeks (44 days) at well below peak — they spent only ~$58M of the fund. This is the "leverage they didn't need to spend" data point worth flagging.
Reference · How To Activity 2 · Work Time 14 min
Reference · How To — Solving \(a \cdot b^{t} = y\)
Step What to do Example: \(23 \cdot 10^{t} = 825\)
1 Isolate the exponential expression. Get it alone on one side, with a constant on the other. Divide both sides by 23: \(10^{t} = \tfrac{825}{23} \approx 35.87\).
2 Rewrite in logarithmic form. The exponent on the left becomes a log on the right. \(t = \log_{10}\!\left(\tfrac{825}{23}\right)\). For base \(e\): use \(\ln\). For base 10: use \(\log\).
3 Solve for the variable (if there's still arithmetic). This is the exact answer. \(t = \log_{10}\!\left(\tfrac{825}{23}\right)\) — finished.
4 If asked, evaluate with a calculator for an approximation. \(t \approx 1.555\) — the calculator estimate of the exact log expression.
★ If both sides can be written as the same base raised to integer powers, you don't need a log. Set the exponents equal. \(10^{x} = 1{,}000\) → \(10^{x} = 10^{3}\) → \(x = 3\). No log required.
Activity 2 · Work Time
Solve. Use log notation if exact reasoning won't give a clean number. No calculator unless noted.
  1. \(10^{x} = 10{,}000\)
  2. \(10^{x} = 825\)  (strike fund, $M)
  3. \(2 \cdot 10^{x} = 200\)
  4. \(23 t = 825\)  (linear; for comparison)
  5. \(73 t = 825\)  (full-strike weeks)
  6. \(10^{x} = 35.87\)  (calc OK)
  7. \(825 \cdot (0.91)^{t} = 100\)  (calc OK; round to 0.1 wk)
  8. \(e^{x} = 7.389\)
  9. \(e^{2x} = 5\)
  10. \(4 \cdot e^{2x} = 88\)  (calc OK)
14 min · Activity 2 Work Time. Leave this slide on screen during work time.

Strategy ladder (sequence the synthesis in this order):
Whole-power reasoning → Q1, Q3 (and Q5 / Q4 are linear, not exponential).
Definition of log → Q2, Q6, Q8.
Algebraic isolation + log → Q7, Q9, Q10.

Answers: Q1: \(x=4\). Q2: \(x = \log_{10}(825) \approx 2.916\). Q3: \(x=2\). Q4: \(t = \tfrac{825}{23} \approx 35.87\) wk. Q5: \(t = \tfrac{825}{73} \approx 11.30\) wk. Q6: \(x = \log_{10}(35.87) \approx 1.555\). Q7: \(t = \tfrac{\log_{10}(100/825)}{\log_{10}(0.91)} \approx 22.4\) wk. Q8: \(x = \ln(7.389) \approx 2\). Q9: \(x = \tfrac{1}{2}\ln(5) \approx 0.805\). Q10: \(x = \tfrac{1}{2}\ln(22) \approx 1.546\).

Watch-for: Q4/Q5 are linear; some students will try to take a log anyway. Affirm: linear and exponential are different shapes — no log needed when \(t\) is not in the exponent.
Q7 is the wrench question: the fund decays exponentially when burn outpaces inflow. The 22.4-week answer is the leverage clock.
Q9: students may write \(\ln(5)\); it's a half-credit answer because they forgot to divide by 2.
Amplify · 10 inputs

Per problem: Text → Math Response (named answer1 … answer10) → Note with CL.

Format examples: Example: 4 · Example: 2.916 · Example: 11.30

Problem 1: 10^x = 10,000.
content: when answer1.submitted and answer1.numericValue = 4 "Correct! 10,000 = 10^4, so x = 4. No log needed — both sides write as a power of 10." when answer1.submitted and answer1.numericValue > 3.99 and answer1.numericValue < 4.01 "Correct!" when answer1.submitted and answer1.numericValue > 999 and answer1.numericValue < 1001 "Almost — 1,000 is the value of 10^3, not the answer. The question asks for x. Try 10^x = 10,000 → x = ?" when answer1.submitted and not isBlank(answer1.latex) "Almost — write 10,000 as a power of 10. (10·10·10·10 = 10,000.) Then read off the exponent." otherwise ""
Problem 2: 10^x = 825.
content: when answer2.submitted and answer2.numericValue > 2.91 and answer2.numericValue < 2.92 "Correct! x = log_10(825) ≈ 2.916. (Or write the exact answer: x = log(825).)" when answer2.submitted and answer2.numericValue > 6.71 and answer2.numericValue < 6.72 "Almost — 6.715 is ln(825), not log_10(825). For 10^x = 825, use base 10 (log button)." when answer2.submitted and not isBlank(answer2.latex) "Almost — rewrite 10^x = 825 in log form: x = log_10(825). What does the calculator give?" otherwise ""
Problem 3: 2 · 10^x = 200.
content: when answer3.submitted and answer3.numericValue = 2 "Correct! Divide by 2: 10^x = 100 = 10^2, so x = 2." when answer3.submitted and answer3.numericValue > 1.99 and answer3.numericValue < 2.01 "Correct!" when answer3.submitted and answer3.numericValue > 99 and answer3.numericValue < 101 "Almost — 100 is what 10^x equals after isolating, not the value of x. Solve 10^x = 100." when answer3.submitted and not isBlank(answer3.latex) "Almost — first divide by 2 to isolate 10^x. Then write the right side as a power of 10." otherwise ""
Problem 4: 23t = 825 (Stand-Up Strike, weeks).
content: when answer4.submitted and answer4.numericValue > 35.86 and answer4.numericValue < 35.88 "Correct! t = 825/23 ≈ 35.87 weeks. With the rolling Stand-Up strategy, the fund stretches almost 9 months." when answer4.submitted and answer4.numericValue > 35.85 and answer4.numericValue < 35.89 "Correct!" when answer4.submitted and answer4.numericValue > 802 and answer4.numericValue < 803 "Almost — 802 is 825 − 23. The equation is 23t = 825, so divide (don't subtract): t = 825 ÷ 23." when answer4.submitted and not isBlank(answer4.latex) "Almost — t is multiplied by 23, so divide both sides by 23. Round to 2 decimal places." otherwise ""
Problem 5: 73t = 825 (full strike, weeks).
content: when answer5.submitted and answer5.numericValue > 11.29 and answer5.numericValue < 11.31 "Correct! t ≈ 11.30 weeks. That's the Morningstar 2023 estimate — about 11 weeks if every UAW member walked out." when answer5.submitted and answer5.numericValue > 752 and answer5.numericValue < 753 "Almost — 752 is 825 − 73. Divide instead: t = 825 ÷ 73." when answer5.submitted and not isBlank(answer5.latex) "Almost — divide both sides by 73. Round to 2 decimal places." otherwise ""
Problem 6: 10^x = 35.87.
content: when answer6.submitted and answer6.numericValue > 1.55 and answer6.numericValue < 1.56 "Correct! x = log_10(35.87) ≈ 1.555. Notice: 10^1 = 10 and 10^2 = 100, so the exponent for 35.87 sits between 1 and 2." when answer6.submitted and answer6.numericValue > 3.57 and answer6.numericValue < 3.59 "Almost — 3.580 is ln(35.87). For base 10, use the log button (not ln)." when answer6.submitted and not isBlank(answer6.latex) "Almost — rewrite 10^x = 35.87 as x = log_10(35.87). Use the log key." otherwise ""
Problem 7: 825 · (0.91)^t = 100.
content: when answer7.submitted and answer7.numericValue > 22.3 and answer7.numericValue < 22.5 "Correct! Divide first: (0.91)^t = 100/825 ≈ 0.1212. Then t = log(0.1212)/log(0.91) ≈ 22.4 weeks. The fund hits $100M after about 5.5 months at this burn rate." when answer7.submitted and answer7.numericValue > -22.5 and answer7.numericValue < -22.3 "Almost — sign error. log(0.1212) is negative AND log(0.91) is negative, so the ratio is positive. Recheck: t = log(0.1212) / log(0.91) ≈ +22.4." when answer7.submitted and answer7.numericValue > 0.12 and answer7.numericValue < 0.13 "Almost — 0.121 is 100/825, the value of (0.91)^t after isolating. You still need to solve for the exponent t. Take log of both sides." when answer7.submitted and answer7.numericValue > 8.04 and answer7.numericValue < 8.06 "Almost — 8.05 looks like log(100/825) without dividing by log(0.91). The setup is t = log(100/825) / log(0.91)." when answer7.submitted and not isBlank(answer7.latex) "Almost — divide by 825 first, then take log of both sides, then divide by log(0.91). Round to 0.1 wk." otherwise ""
Problem 8: e^x = 7.389.
content: when answer8.submitted and answer8.numericValue > 1.99 and answer8.numericValue < 2.01 "Correct! x = ln(7.389) ≈ 2.000. (7.389 ≈ e^2, so the exponent is exactly 2.)" when answer8.submitted and answer8.numericValue > 0.86 and answer8.numericValue < 0.88 "Almost — 0.869 is log_10(7.389). For e^x, use ln (base e), not log (base 10)." when answer8.submitted and not isBlank(answer8.latex) "Almost — rewrite e^x = 7.389 as x = ln(7.389). Use the ln button." otherwise ""
Problem 9: e^(2x) = 5.
content: when answer9.submitted and answer9.numericValue > 0.80 and answer9.numericValue < 0.81 "Correct! Take ln: 2x = ln(5). Divide: x = ½·ln(5) ≈ 0.805." when answer9.submitted and answer9.numericValue > 1.60 and answer9.numericValue < 1.61 "Almost — 1.609 is ln(5), but the exponent is 2x (not x). After ln(5) = 2x, divide both sides by 2: x = ln(5)/2 ≈ 0.805." when answer9.submitted and not isBlank(answer9.latex) "Almost — take ln of both sides: 2x = ln(5). Then x = ln(5)/2." otherwise ""
Problem 10: 4 · e^(2x) = 88.
content: when answer10.submitted and answer10.numericValue > 1.54 and answer10.numericValue < 1.56 "Correct! Divide by 4: e^(2x) = 22. Take ln: 2x = ln(22). x = ½·ln(22) ≈ 1.546." when answer10.submitted and answer10.numericValue > 3.09 and answer10.numericValue < 3.10 "Almost — 3.091 is ln(22) (the exponent 2x, not x). Divide by 2 once more: x = ln(22)/2 ≈ 1.546." when answer10.submitted and answer10.numericValue > 21.99 and answer10.numericValue < 22.01 "Almost — 22 is what e^(2x) equals after dividing by 4. You still need to solve for x. Take ln of both sides." when answer10.submitted and not isBlank(answer10.latex) "Almost — divide by 4 first to get e^(2x) alone, then take ln, then divide by 2." otherwise ""
Activity 2 · Synthesis 5 min

Strategy Ladder — When do we need a log?

Three families of equations, three different moves. The wrong move on the wrong family is the most common error of the lesson.

“Q1, Q3: \(10^{x} = 10{,}000\) and \(2 \cdot 10^{x} = 200\). Did you need a log?” No. Once isolated, the right side is a clean integer power of 10 (\(10^{4}\), \(10^{2}\)). Set exponents equal: \(x = 4\), \(x = 2\). Logs work, but they're overkill — same answer, more steps.
“Q4, Q5: \(23t = 825\) and \(73t = 825\). Did you need a log?” No — these are linear. \(t\) is multiplied, not in an exponent. Just divide: \(t = \tfrac{825}{73} \approx 11.30\) weeks (full strike), \(t = \tfrac{825}{23} \approx 35.87\) weeks (Stand-Up). The shape of the equation tells you which tool to grab.
“Q7: \(825 \cdot (0.91)^{t} = 100\). Why did we have to use a log here?” Because \(t\) is locked in the exponent and \((0.91)^{t}\) does not equal a clean integer power of any base. Isolate (\((0.91)^{t} = 0.1212\)), take a log of both sides, and divide: \(t = \tfrac{\log(0.1212)}{\log(0.91)} \approx 22.4\) weeks. Translation: if the burn climbs to 9% per week, the fund shrinks from $825M to $100M in 22.4 weeks.
“Q9: \(e^{2x} = 5\). Why did we use \(\ln\) instead of \(\log_{10}\)?” Because the base is \(e\). \(\ln\) is the inverse for base \(e\), the way \(\log_{10}\) is the inverse for base 10. \(\ln\) cancels \(e\) in one step: \(2x = \ln(5)\), so \(x = \tfrac{1}{2}\ln(5) \approx 0.805\). Could you use \(\log_{10}\)? Yes, but you'd carry around an extra \(\log_{10}(e)\) factor. \(\ln\) is the right wrench for base \(e\).
click to advance discussion ▶
Try Saying A log is needed when ___, but not when ___.
Try Saying For Q___, I would use \(\log_{10}\) / \(\ln\) / no log because the base is ___.
Sequence: clean-power, linear, log-needed, ln-needed. Slow down on the third question — that's the central new skill of the lesson, and it's the same equation shape as every strike-fund/half-life/doubling problem they'll meet for the rest of the unit.
Lesson Synthesis The Wrench 8 min

Order the Moves — Then Move the Rate

Putting it all together: a single equation that holds the whole lesson, then one decision the lesson lets you make.

Part 1 · Order the steps
Solve \(2 \cdot 10^{\,x-5} + 3 = 33\) for \(x\).
“Step 1: which move first?” Subtract 3 from each side: \(2 \cdot 10^{\,x-5} = 30\).
“Step 2?” Divide by 2: \(10^{\,x-5} = 15\).
“Step 3?” Rewrite as a logarithm: \(x - 5 = \log_{10}(15)\).
“Step 4?” Add 5: \(x = 5 + \log_{10}(15) \approx 6.176\).
click to advance discussion ▶
Part 2 · The Wrench — what would buy 6 months?
Math: 6 months = 26 weeks. To last 26 weeks at the full-strike burn ($73M/wk) the fund would need to start at \(73 \cdot 26 = \$1.9\,\text{B}\) — about 2.3× actual.

Or: keep the $825M, and lower the weekly burn until \(\tfrac{825}{B} \geq 26\). That gives \(B \leq \$31.7\text{M}/\text{wk}\) — about 63,000 strikers max.

What changes the burn rate?
Stand-Up Strike
UAW 2023: rolling targeted strikes cut burn 3–10×. Actual fund spent ≈ $58M over 44 days vs. ~$320M projected for full strike.
Source: Evercore ISI / Morningstar 2023.
Solidarity donations
SAG-AFTRA Foundation raised $15M in 3 weeks (2023). Adds inflow → effectively shrinks net burn.
Source: Variety, Aug 2023.
Sectoral bargaining
Nordic-style industry-wide funds. Sweden: ~90% workforce coverage vs. ~10% US — bigger pooled funds, longer leverage.
Source: ILO & Wikipedia.
Re-legalize 2° boycotts
PRO Act would restore 1947-era right of other unions to refuse struck goods → multiplies pressure without growing fund.
Source: NLRA §8(b)(4); PRO Act text.
Try Saying (popcorn share) To last 6 months, I would push for ___ because it would change the burn rate from ___ to about ___ — that gives \(\tfrac{825}{\text{burn}} \geq\) 26 weeks.
I Can… (SWBAT)
Use base-10 and natural logarithms to solve exponential equations of the form \(a \cdot b^{t} = y\) for \(t\).
Decide when an equation needs a log and when whole-number reasoning will work.
Read a real leverage clock: identify the burn rate, the starting fund, and what it would take to extend the time horizon.
8 min total. ~3 min on the step-order (left), ~5 min on the wrench (right).

Popcorn rules: 30-second turns, no follow-ups required, frame is on screen. Listen for: students who pick rolling/Stand-Up (the strategy that actually happened) and students who pick public donations (the strategy SAG-AFTRA used). Both are correct — both move the rate. The point is not to land on one answer. The point is that students leave the lesson having proposed a real intervention tied to the math they just did.

Trauma-aware: If a student is dismissive ("strikes don't work"), don't argue — pivot: "the math says 11 weeks at full burn; UAW won in 6 weeks at 1/3 burn. What does that tell us about leverage that wasn't even used?"

SWBAT closes the lesson — it's a confirmation of what they did, not a preview.
Reference · How To Cool-Down 5 min
Reference · How To — Quick recap
Shape Move
\(10^{x} = \) integer power of 10Match exponents. No log.
\(10^{x} = c\), \(c\) not a clean power\(x = \log_{10}(c)\)
\(a \cdot e^{kx} = y\)Divide by \(a\), take \(\ln\), divide by \(k\).
Linear (\(t\) not in exponent)Just divide. No log.
Cool-Down · Solve Some Equations
On your own. Calculator OK on Q3 only.
  1. \(10^{x} = 1{,}000\)
  2. \(4 \cdot e^{2x} = 88\)  (exact form is fine)
  3. \(e^{2x} = 5\) — solve and approximate with a calculator.
Cool-Down · 5 min. Closes the lesson — assesses each move on the strategy ladder (clean-power, ln-isolation, ln + approximate).

Q1 (clean power): \(1{,}000 = 10^{3}\), so \(x = 3\). No log.
Q2 (ln-isolation): divide by 4 → \(e^{2x} = 22\); take ln → \(2x = \ln(22)\); divide → \(x = \tfrac{1}{2}\ln(22) \approx 1.546\).
Q3 (ln + approximate): exact: \(x = \tfrac{1}{2}\ln(5)\). Calculator: \(\approx 0.805\).

Common errors: writing \(\ln(22)\) without dividing by 2 on Q2; using \(\log_{10}\) instead of \(\ln\) on Q2/Q3 (still gives correct value if applied consistently to both sides, but the simpler move is \(\ln\)); reporting just the rounded decimal on Q3 without the exact form.
Amplify · Cool-Down (3 inputs)

Per problem: Text → Math Response (named answer1, answer2, answer3) → Note with CL.

Format examples: Example: 3 · Example: 1.546 · Example: 0.805

Problem 1: Solve 10^x = 1,000.
content: when answer1.submitted and answer1.numericValue = 3 "Correct! 1,000 = 10^3, so x = 3. No log needed." when answer1.submitted and answer1.numericValue > 2.99 and answer1.numericValue < 3.01 "Correct!" when answer1.submitted and answer1.numericValue > 999 and answer1.numericValue < 1001 "Almost — 1,000 is the value of 10^x, not the answer. Solve for the exponent: 10^? = 1,000." when answer1.submitted and not isBlank(answer1.latex) "Almost — write 1,000 as a power of 10. (10·10·10 = 1,000.) Then read off the exponent." otherwise ""
Problem 2: Solve 4 · e^(2x) = 88.
content: when answer2.submitted and answer2.numericValue > 1.54 and answer2.numericValue < 1.56 "Correct! Divide by 4: e^(2x) = 22. Take ln: 2x = ln(22). Divide by 2: x = ½·ln(22) ≈ 1.546." when answer2.submitted and answer2.numericValue > 3.09 and answer2.numericValue < 3.10 "Almost — 3.091 is ln(22) (= 2x). Divide by 2 once more: x = ln(22)/2 ≈ 1.546." when answer2.submitted and answer2.numericValue > 21.99 and answer2.numericValue < 22.01 "Almost — 22 is the value of e^(2x) after dividing by 4. You still need to take ln and divide by 2." when answer2.submitted and answer2.numericValue > 1.34 and answer2.numericValue < 1.35 "I see what happened — 1.342 is log_10(22). Use ln (base e) here, not log (base 10). Then divide by 2." when answer2.submitted and not isBlank(answer2.latex) "Almost — divide by 4 first to get e^(2x) = 22. Then take ln of both sides, then divide by 2." otherwise ""
Problem 3: Solve e^(2x) = 5 and approximate.
content: when answer3.submitted and answer3.numericValue > 0.80 and answer3.numericValue < 0.81 "Correct! Exact: x = ½·ln(5). Calculator: x ≈ 0.805." when answer3.submitted and answer3.numericValue > 1.60 and answer3.numericValue < 1.61 "Almost — 1.609 is ln(5) (= 2x). Divide by 2: x = ln(5)/2 ≈ 0.805." when answer3.submitted and answer3.numericValue > 0.69 and answer3.numericValue < 0.70 "Almost — 0.699 is log_10(5). Use ln here (base is e). Then divide by 2." when answer3.submitted and not isBlank(answer3.latex) "Almost — take ln: 2x = ln(5). Then x = ln(5)/2. What does the calculator give?" otherwise ""
Lesson Summary

When the unknown is an exponent — reach for a logarithm

Some exponential equations — like \(10^{x} = 1{,}000\) or \(2^{x} = 32\) — can be solved by spotting the right whole-number power. Others can't. When the value isn't a clean power of the base, the exact answer is written using a logarithm.
Base 10
\(10^{x} = c\)  ⟺  \(x = \log_{10}(c)\)
Calculator key: log.
Base \(e\)
\(e^{x} = c\)  ⟺  \(x = \ln(c)\)
Calculator key: ln.
General \(a \cdot b^{t} = y\)
Divide by \(a\), then take a log, then solve for \(t\).
Lesson Summary The natural log \(\ln\) is the logarithm with base \(e\). It is not a different operation — it is the same idea (the exponent that produces a target value) with a different base. Use \(\log_{10}\) when the base is 10, and \(\ln\) when the base is \(e\). Use neither when both sides can be written as the same integer power.
The Wrench — What you can say now "$825M of strike fund, $73M of weekly burn, gives 11.3 weeks of leverage. UAW spent 6 weeks of it. Cut the burn — by rolling strikes, donations, or sectoral funds — and the same money buys more time. What rate would buy 6 months? Anything under $31.7M/week. The math is the constraint. The strategy is the lever."
Read aloud or have a student read. The yellow box is the general move; the red-teal banner is the takeaway: the math sets the constraint, the strategy moves what's inside it. Tomorrow students will use logs to solve equations involving doubling time and half-life.