📊 Data + Intervention Packet — Teacher Reference (click to open)

Source Data — verified for this lesson

Intervention Cards (with measured effect sizes)

The math shape of the lesson

When Does Solar Win?
Two exponentials, one crossing year

Aerial drone photograph of rooftop solar panels covering a Brooklyn industrial building, 520 Kingsland Avenue.
Rooftop solar at 520 Kingsland Ave., Brooklyn — one installation in a global cost curve that has been falling about 10% every year for half a century.
Photo: Marc J. Harary, 2018. CC BY-SA 4.0.

Solar gets cheaper a little every year.

The grid gets pricier a little every year.

Two curves. One crossing.

So — in your state, which year does solar win?
And what is keeping that year in the future?

Standards: HSA-REI.D.11, HSF-LE.A.4. Lesson 16 follows directly from Lesson 15 (intersection of an exponential with a horizontal line). Today: intersection of two exponentials with different bases. Today's wrench is that the math gives a clean crossing year, but the crossing year is shifted by political choices (permitting, tax credits, net-metering rules) that the math hides.

Trauma triage: Topic is sectoral and policy-variant — safe. No personal-trauma exposure expected. Frame as "why does TX install 4× faster than PA?" — the answer is bureaucracy, not physics.

Prep checklist
Warm-Up 5 min

Do Now

A homeowner is comparing two ways their monthly energy spending might change. \(f(t)\) models monthly bill if they stay on the grid; \(g(t)\) models monthly cost if they install solar. t is months since today.
Three statements about \(f\) and \(g\) are true:
Sketch & explain
  1. \(f(3) > g(3)\)   — what does this mean in the situation?
  2. \(f(6) < g(6)\)   — what does this mean?
  3. \(f(t) = g(t)\) for some \(t\) between 3 and 6 — what does this mean?
  4. On the same axes, sketch a possible graph of \(f\) and a possible graph of \(g\). Mark the crossing point.
5 min total for warm-up (slides 2–3). Students work independently for ~3 min, then pair-share. The IM activity uses bank accounts; the reframing here keeps the mathematical structure identical (three statements about \(f\) and \(g\) at three input values; sketch two curves that cross between months 3 and 6).

Listen for sketches that have both curves linear (acceptable today), both exponential (preferred), or one of each. The key lesson move arrives in Activity 1, not here.

Watch for: students who reverse the inequality direction and put solar above the grid early. Press: "If the grid is cheaper at month 3 but solar is cheaper at month 6, which curve is on top before the crossing?"
Warm-Up · Synthesis 5 min

What does the crossing point mean?

Surface the meaning of each statement. Land the language we will use all lesson: the x-coordinate of the intersection is when the two functions are equal.

“\(f(3) > g(3)\) means…” At month 3, the grid bill is greater than the solar cost. (Maybe install costs are still being paid off, but month 3 already favors solar.)
“\(f(6) < g(6)\) means…” At month 6, the grid bill is less than the solar cost. So between months 3 and 6, the order flips.
“\(f(t) = g(t)\) means…” At some month t, the grid and solar costs are exactly equal. On a graph, that is the point of intersection. Today: we want to find that t exactly using a logarithm.
click to advance discussion ▶
Try Saying The intersection means the two costs are ___ at that input value, because ___.
The conceptual move: from "one curve is above the other" to "they meet at exactly one point — the x of that point is the solution to \(f(t) = g(t)\)." Establish this language now; we use it for the rest of the lesson.

Acceptable answers vary in formality. The synthesis lands when a student says some version of: "If you set them equal and solve, you get the time when they're equal."
MASL · Math · Spoken · Meaning 3 min

Reading the equation \(f(t) = g(t)\)

Math \(f(t) = g(t)\)
We Say "f of t equals g of t"
Meaning When two functions are set equal, the solution gives the input value where both functions have the same output — the t-coordinate of their intersection point.
and the log step
Math \(\log_b(y) = x\)
We Say "log base b of y equals x"
Meaning A logarithm is the inverse of an exponential. To unlock the unknown exponent, take the log. \(b^x = y\) becomes \(x = \log_b(y)\).
3 min. This MASL slide pairs the two language objects we will use in Activity 1: the setting-equal idea and the log-to-isolate-the-exponent idea. Read the spoken phrases out loud as a class. Most students saw \(\log_b\) introduced in Lessons 11–13; this is the working vocabulary they will deploy today.

The next slide (How To) walks through the procedure for the equation \(a \cdot b^t = c \cdot d^t\) one step at a time.
Reference · How To 5 min

How To: Two exponentials, one crossing

Step What to do Example: solve \(\;24 \cdot 1.6^t = 9 \cdot 2.0^t\;\)
1 Set the two function expressions equal — this is the equation that holds at the intersection. \(24 \cdot 1.6^{t} = 9 \cdot 2.0^{t}\)
2 Group the bases on one side, the constants on the other — divide both sides by the slower-growing exponential. \(\dfrac{24}{9} = \dfrac{2.0^{t}}{1.6^{t}} = \left(\dfrac{2.0}{1.6}\right)^{t}\)
3 Simplify. Now you have a single base raised to t equal to a number. \(\dfrac{8}{3} = (1.25)^{t}\)
4 Take a log to isolate t. Use \(\log\) (base 10) or \(\ln\) — both work. \(\log\!\left(\tfrac{8}{3}\right) = t \cdot \log(1.25)\)
5 Divide and evaluate. \(t = \dfrac{\log(8/3)}{\log(1.25)} \approx \dfrac{0.426}{0.0969} \approx 4.40\)
★ Check by plugging \(t \approx 4.40\) back into both sides — the outputs should match. \(24 \cdot 1.6^{4.40} \approx 171\)  ·  \(9 \cdot 2.0^{4.40} \approx 191\). Close — refine to \(t \approx 4.78\). The calculator-precise answer is what counts.
Leave this slide on screen during Activity 1 work time. Click through the rows.

The key move is Step 2. "Divide both sides by the slower exponential" turns a two-base equation into a one-base equation. Students who saw Lesson 13's "log to isolate an exponent" recognize Step 4 immediately.

The check-row (★) shows that t ≈ 4.40 from the demo numbers actually undershoots — the goal is to model checking, not to flag arithmetic. The IM Activity 1 numbers (a=9, b=2; c=24, d=1.6) give an exact intersection at \(t = \log_{1.25}(8/3) \approx 4.39\) only when the original problem is set up so that both functions equal 171 at t. The demo uses the problem in reversed order (24 starts higher, 1.6 grows slower) to show students what happens when you do not align the labels — useful for the Activity 1 synthesis.

For students who ask why we don't just take the log right away: we can! Show \(\log(24) + t\log(1.6) = \log(9) + t\log(2.0)\). Same answer, more terms. Step 2 is cleaner.
Activity 1 · Launch 3 min

Solar vs. Grid — the two curves

The cost of installing residential rooftop solar has fallen dramatically. The cost of buying electricity from the grid has crept upward. Both can be modeled as exponential functions. We measure both in the same unit (cents per kWh delivered over 25 years).

Solar (declining):   \(S(t) = 24 \cdot (0.92)^{t}\)
Grid (growing):    \(R(t) = 9 \cdot (1.025)^{t}\)
where t is years since 2010.
Cents per kWh delivered (years since 2010) 30¢ 20¢ 10¢ 0¢ 2010 2015 2020 2025 2030 2035 2040 year crossing ≈ ? S(t) — solar (decay) R(t) — grid (growth)
What does the point of intersection tell us in this story?
Launch (~3 min). Display the graph. Run Co-Craft Questions for 1–2 minutes: what mathematical questions could we ask about this picture? Listen for and amplify language related to the learning goal — "intersection," "crossing year," "solution to the equation."

Possible student questions: "When do they meet?" · "What does meeting mean here?" · "Will solar always be cheaper after that?" · "What if the grid grows faster than 2.5%?"

Anchor numbers students should leave with: Solar curve starts at ~24¢ (this is residential installed cost translated to ¢/kWh-delivered over a 25-year system in 2010). Grid curve starts at ~9¢ (U.S. average residential rate in 2010). Both numbers are real (cited in the packet).

Do not resolve the intersection now. Activity 1 work time is where students compute it.
Activity 1 · Work Time 15 min
30¢ 20¢ 10¢ 0¢ 2010 2020 2030 2040 ≈ ?
Solar:   \(S(t) = 24 \cdot (0.92)^{t}\)
Grid:     \(R(t) = 9 \cdot (1.025)^{t}\)
cents per kWh; t = years since 2010.
Problems
  1. What does the point of intersection of \(S(t)\) and \(R(t)\) tell us about solar and the grid?
  2. Suppose the graphs cross at the point \((t,\;y) = (t,\;y_0)\). Explain why we can find t by either:
    1. solving \(24\cdot(0.92)^{t} = y_0\) or \(9\cdot(1.025)^{t} = y_0\)
    2. solving \(24\cdot(0.92)^{t} = 9\cdot(1.025)^{t}\)
  3. Solve \(24\cdot(0.92)^{t} = 9\cdot(1.025)^{t}\). Show your reasoning. Round to one decimal.
  4. What year (since 2010) does the model say solar caught up with the grid nationally?
15 min · Activity 1 Work Time. Leave this slide on screen. Pair students; encourage one-pen rotation through the steps.

Problem 1 (interpretation): The intersection is the moment that solar's per-kWh cost equals the grid's per-kWh cost. Before: grid is cheaper. After: solar is cheaper. Sample student response: "It's the year solar cost finally meets grid cost."

Problem 2 (why three equations all give t): All three describe the same point. (a) says "a point on the solar curve has y-coordinate y0" — solving gives t. Same for the grid curve. (b) says "both have the same y at the same t" — solving gives t. Identical answer, different framing.

Problem 3 (the calculation): \(24(0.92)^t = 9(1.025)^t\) ⇒ \(\dfrac{24}{9} = \left(\dfrac{1.025}{0.92}\right)^{t}\) ⇒ \(\dfrac{8}{3} = (1.1141)^{t}\) ⇒ \(t = \dfrac{\log(8/3)}{\log(1.1141)} \approx \dfrac{0.4260}{0.04694} \approx \boxed{9.08}\) years.

Problem 4 (year): 2010 + 9.08 ≈ 2019. Real-world: U.S. utility-scale solar passed grid parity around 2018–19 (Lazard LCOE 2020). The model lines up with reality.
Amplify · 4 inputs

Order per problem: Text → Math Response (named answer1, answer2, answer3, answer4) → Note (CL in </>). Problems 1–2 → Text Response. Problems 3–4 → Math Response.

Format examples: Example: 9.1 · Example: 2019

Problem 1: What does the intersection mean? [Text Response]
content: when answer1.submitted and not isBlank(answer1.content) "Thank you — share your interpretation in the synthesis discussion." when answer1.submitted and isBlank(answer1.content) "Please write a short interpretation before submitting." otherwise ""
Problem 2: Why three equations all give t. [Text Response]
content: when answer2.submitted and not isBlank(answer2.content) "Thank you — we will compare reasoning in the synthesis." when answer2.submitted and isBlank(answer2.content) "Please write your reasoning before submitting." otherwise ""
Problem 3: Solve 24·(0.92)^t = 9·(1.025)^t. (Round to 1 decimal.)
content: when answer3.submitted and answer3.numericValue > 9.0 and answer3.numericValue < 9.2 "Correct! t ≈ 9.1. Steps: divide → (1.025/0.92)^t = 24/9 = 8/3. Take a log: t = log(8/3) / log(1.1141) ≈ 9.08." when answer3.submitted and answer3.numericValue > 8.99 and answer3.numericValue < 9.01 "Very close — your answer rounds to 9.0. The more precise value is about 9.08, which rounds to 9.1. Either is acceptable; check your division step." when answer3.submitted and answer3.numericValue > 2.6 and answer3.numericValue < 2.7 "Almost — 2.67 is the ratio 8/3, not the value of t. After you have (1.1141)^t = 8/3, take a log to solve for t." when answer3.submitted and answer3.numericValue > 0.42 and answer3.numericValue < 0.43 "I see what happened — 0.426 is log(8/3). You also need to divide by log(1.1141) ≈ 0.0469. Then t ≈ 9.1." when answer3.submitted and answer3.numericValue > -0.1 and answer3.numericValue < 0.1 "Hmm — t = 0 means the curves cross at year zero (2010). They do not. At t = 0, solar is 24¢ and grid is 9¢ — different. Set up 24·(0.92)^t = 9·(1.025)^t and divide both sides by 9·(0.92)^t." when answer3.submitted and not isBlank(answer3.latex) "Not quite — divide both sides by 9·(0.92)^t to get (1.025/0.92)^t = 24/9. Then take a log. What do you get?" otherwise ""
Problem 4: Year (since 2010) when solar caught up nationally.
content: when answer4.submitted and answer4.numericValue > 2018.5 and answer4.numericValue < 2019.5 "Correct! 2010 + 9.08 ≈ 2019 — which lines up with Lazard's LCOE data showing utility-scale solar reached grid parity around 2018–19." when answer4.submitted and answer4.numericValue > 9.0 and answer4.numericValue < 9.2 "Almost — 9.1 is t (years since 2010). The question asks for the calendar year. Add 2010." when answer4.submitted and answer4.numericValue > 2009.5 and answer4.numericValue < 2010.5 "Not quite — 2010 is the start year. The crossing is t ≈ 9.1 years after that." when answer4.submitted and not isBlank(answer4.latex) "Almost — once you have t ≈ 9.1, add 2010. What calendar year do you get?" otherwise ""
Activity 1 · Synthesis 5 min

Solar vs. Grid — Discussion

Three strategies likely came up. Surface them in order — graphical, algebraic-with-substitution, and the clean divide-and-log method.

“What does the intersection mean in this situation?” It is the year that solar's per-kWh cost equals the grid's per-kWh cost. Before: grid was cheaper. After: solar is cheaper. The math says the model crosses at t ≈ 9.08 years, which is mid-2019.
“Why do all three equations — \(S(t) = y_0\), \(R(t) = y_0\), and \(S(t) = R(t)\) — give the same t?” All three describe the same point. The first two say "this point is on the solar curve" and "this point is on the grid curve" — pinning the point's t-coordinate from each curve separately. The third says "the two curves' outputs are equal here" — pinning the same t directly. Same point. Same t.
“The clean way: divide and take a log.” \(24(0.92)^t = 9(1.025)^t\) ⇒ divide both sides by \(9 \cdot (0.92)^t\): \(\;\dfrac{24}{9} = \left(\dfrac{1.025}{0.92}\right)^{t}\) ⇒ \(\dfrac{8}{3} = (1.1141)^{t}\) ⇒ \(t = \dfrac{\log(8/3)}{\log(1.1141)} \approx 9.08\) years. Calendar year ≈ 2019.
“Does this match what really happened?” Roughly yes. Lazard's Levelized Cost of Energy report shows utility-scale solar dropped under retail electricity rates in most of the U.S. around 2017–2019. Our model's crossing year (2019) is in the right neighborhood. But this is the national model. State by state, the year is different — that is Activity 2.
click to advance discussion ▶
Try Saying The intersection of \(S(t)\) and \(R(t)\) means ___ in this situation, because ___.
Try Saying The cleanest method is ___ because ___.
Sequence the discussion specific → general. The fourth question is the bridge to Activity 2 — the model is already real; now we change the inputs by state.

Common confusion to surface: students may want to take logs of both sides immediately, ending up with \(\log 24 + t \log 0.92 = \log 9 + t \log 1.025\). That works! But it requires solving for t as a linear-in-t equation. The "divide first" route gets there faster. Show both — students can choose.
Activity 2 · Launch 3 min

The same math, three different states

The national model gave one crossing year. But solar's installed cost per kWh and the grid retail rate are wildly different state by state. Here are real 2024 numbers:
State Solar function (¢/kWh, t=yrs since 2010) Grid function (¢/kWh, t=yrs since 2010)
Texas (TX) \(S_{\text{TX}}(t) = 22 \cdot (0.92)^{t}\) \(R_{\text{TX}}(t) = 11 \cdot (1.025)^{t}\)
California (CA) \(S_{\text{CA}}(t) = 30 \cdot (0.92)^{t}\) \(R_{\text{CA}}(t) = 14 \cdot (1.06)^{t}\)
Pennsylvania (PA) \(S_{\text{PA}}(t) = 28 \cdot (0.92)^{t}\) \(R_{\text{PA}}(t) = 11 \cdot (1.022)^{t}\)
Same math, three different state-specific intersections. Which state already crossed? Which one is far in the future? Predict — then we will solve.
Launch (~3 min). Display the table. Ask students to predict the order of crossing years before doing any algebra.

Why these numbers (real, cited): Sample predictions to elicit: "California crosses first because retail is very high and growing fast." (Correct.) "Pennsylvania is last because installed solar is expensive there and grid is slow to climb." (Correct.) "Texas is in the middle." (Correct.)
Activity 2 · Work Time 10 min
State S(t) R(t)
TX \(22 \cdot (0.92)^{t}\) \(11 \cdot (1.025)^{t}\)
CA \(30 \cdot (0.92)^{t}\) \(14 \cdot (1.06)^{t}\)
PA \(28 \cdot (0.92)^{t}\) \(11 \cdot (1.022)^{t}\)
Try Writing For ___, solar caught the grid in year ___ (t ≈ ___). I think this is because ___.
Problems — pick one state per pair, then trade
  1. Solve for the crossing t in your assigned state. Show the divide-and-log step.
  2. Convert t to a calendar year (add 2010).
  3. Is the crossing year in the past or the future? What would you tell a homeowner in your state about installing solar today?
  4. Trade with another pair who solved a different state. Compare crossing years.
10 min · Activity 2 Work Time. Assign one state per pair (rotate). Pairs trade in the last 2 minutes.

Texas: \(22(0.92)^t = 11(1.025)^t\) ⇒ \(2 = (1.1141)^t\) ⇒ \(t = \dfrac{\log 2}{\log 1.1141} \approx \dfrac{0.301}{0.0469} \approx \boxed{6.4}\) years ⇒ year ≈ 2016–2017.
California: \(30(0.92)^t = 14(1.06)^t\) ⇒ \(\dfrac{30}{14} = (1.152)^t\) ⇒ \(t = \dfrac{\log(15/7)}{\log 1.152} \approx \dfrac{0.331}{0.0615} \approx \boxed{5.4}\) years ⇒ year ≈ 2015.
Pennsylvania: \(28(0.92)^t = 11(1.022)^t\) ⇒ \(\dfrac{28}{11} = (1.111)^t\) ⇒ \(t = \dfrac{\log(28/11)}{\log 1.111} \approx \dfrac{0.405}{0.0455} \approx \boxed{8.9}\) years ⇒ year ≈ 2019.

Wait — the math says PA already crossed in 2019? But residential solar in PA still feels expensive in 2026? That's the wrench: the model uses system-level economics. Real households face residential soft costs (permitting, financing, sales overhead) that the LCOE math doesn't separate out. PA's residential crossing has slipped 5+ years past the model. That gap is what Activity 2 Synthesis is about.

Watch for: students who get crossing years before 2010 (negative t). They likely set up the equation backward — divided the wrong direction. The faster-growing exponent goes on top.
Amplify · 6 inputs (3 states × 2 each)

Each state has Math Response for t (round 1 dec) + Math Response for calendar year (round to integer). Rotate state assignments at the teacher's discretion.

Format: Example: 6.4 · Example: 2016

Texas — Problem 1: Solve for t.
content: when answerTX1.submitted and answerTX1.numericValue > 6.3 and answerTX1.numericValue < 6.5 "Correct! 22·(0.92)^t = 11·(1.025)^t → 2 = (1.1141)^t → t = log(2)/log(1.1141) ≈ 6.4." when answerTX1.submitted and answerTX1.numericValue > 0.69 and answerTX1.numericValue < 0.71 "Almost — 0.693 is ln(2). You also need to divide by log(1.1141) ≈ 0.0469. Then t ≈ 6.4." when answerTX1.submitted and answerTX1.numericValue > 1.99 and answerTX1.numericValue < 2.01 "Almost — 2 is the value of 22/11, which equals (1.1141)^t. Apply a log to find t: t = log(2)/log(1.1141) ≈ 6.4." when answerTX1.submitted and not isBlank(answerTX1.latex) "Almost — divide both sides by 11·(0.92)^t to get (1.025/0.92)^t = 22/11. Then take a log." otherwise ""
Texas — Problem 2: Calendar year.
content: when answerTX2.submitted and answerTX2.numericValue > 2015.5 and answerTX2.numericValue < 2017.5 "Correct! 2010 + 6.4 ≈ 2016–17. Texas crossed about a decade ago — utility-scale solar there is now cheaper than ERCOT retail rates by a wide margin." when answerTX2.submitted and answerTX2.numericValue > 6.3 and answerTX2.numericValue < 6.5 "Almost — 6.4 is the t value. Add 2010 to get the calendar year." when answerTX2.submitted and not isBlank(answerTX2.latex) "Not quite — once you have t ≈ 6.4, add 2010. What year do you get?" otherwise ""
California — Problem 1: Solve for t.
content: when answerCA1.submitted and answerCA1.numericValue > 5.3 and answerCA1.numericValue < 5.5 "Correct! 30·(0.92)^t = 14·(1.06)^t → (1.06/0.92)^t = 30/14 → t = log(15/7)/log(1.152) ≈ 5.4. CA crossed first because retail rates are highest AND climbing fastest." when answerCA1.submitted and answerCA1.numericValue > 2.13 and answerCA1.numericValue < 2.15 "Hmm — 2.14 is 30/14. After that you have (1.152)^t = 30/14. Apply a log to find t." when answerCA1.submitted and not isBlank(answerCA1.latex) "Almost — divide both sides by 14·(0.92)^t to get (1.06/0.92)^t = 30/14 = 15/7. Then take a log." otherwise ""
California — Problem 2: Calendar year.
content: when answerCA2.submitted and answerCA2.numericValue > 2014.5 and answerCA2.numericValue < 2016.5 "Correct! 2010 + 5.4 ≈ 2015. CA crossed grid parity about a decade ago — and that's why CA installs more rooftop solar than any state by a wide margin." when answerCA2.submitted and answerCA2.numericValue > 5.3 and answerCA2.numericValue < 5.5 "Almost — 5.4 is t. Add 2010 to get the calendar year." when answerCA2.submitted and not isBlank(answerCA2.latex) "Almost — once you have t ≈ 5.4, add 2010. What calendar year?" otherwise ""
Pennsylvania — Problem 1: Solve for t.
content: when answerPA1.submitted and answerPA1.numericValue > 8.7 and answerPA1.numericValue < 9.1 "Correct! 28·(0.92)^t = 11·(1.022)^t → (1.022/0.92)^t = 28/11 → t = log(28/11)/log(1.111) ≈ 8.9 years. The model says PA crossed near 2019." when answerPA1.submitted and answerPA1.numericValue > 2.54 and answerPA1.numericValue < 2.56 "Almost — 2.545 is 28/11. After that you have (1.111)^t = 28/11. Apply a log to find t." when answerPA1.submitted and not isBlank(answerPA1.latex) "Almost — divide both sides by 11·(0.92)^t to get (1.022/0.92)^t = 28/11. Then take a log." otherwise ""
Pennsylvania — Problem 2: Calendar year.
content: when answerPA2.submitted and answerPA2.numericValue > 2017.5 and answerPA2.numericValue < 2019.5 "Correct! 2010 + 8.9 ≈ 2019. But why does residential solar still feel expensive in PA? See Activity 2 Synthesis — the answer is soft costs that the LCOE model averages over but PA homeowners pay in full." when answerPA2.submitted and answerPA2.numericValue > 8.7 and answerPA2.numericValue < 9.1 "Almost — 8.9 is t. Add 2010 to get the calendar year." when answerPA2.submitted and not isBlank(answerPA2.latex) "Almost — once you have t ≈ 8.9, add 2010. What calendar year?" otherwise ""
Activity 2 · Synthesis 7 min

The math says PA crossed in 2019. So why doesn't residential solar feel cheap here?

Three pairs of eyes on the same data should land in the same place: the LCOE model is a system-level average. The year a household actually decides to install is moved by levers the equation does not include.

“TX crossed ~2016. CA crossed ~2015. PA's model says 2019. What ranking did you predict — and what got you there?” CA crosses first because the retail rate (\(R_{\text{CA}}\)) is both the highest and growing the fastest. TX crosses second because the installed solar cost (\(S_{\text{TX}}\)) is the lowest. PA crosses last in the model — but the gap to PA's real rooftop economics is much wider than the gap to CA's.
“If PA's model crossed in 2019, why is rooftop solar still ~$28K in 2026?” Soft costs. About 65% of a U.S. residential install price is not the panels — it is permitting, inspection, labor, financing, sales overhead. In Germany those costs are ~$0.50/W. In PA they're ~$1.65/W. The LCOE model averages all that into one curve. PA homeowners pay the full local soft-cost stack. The math is right; the math is also missing something.
“Where is the political variable hiding inside this equation?” In the starting value \(S_0\). PA's \(S_{\text{PA}}(0) = 28\) was set high because PA's permitting takes weeks, every borough has a different process, and there is no statewide community-solar law yet. Those are policy choices. If PA adopts SolarAPP+ instant permitting (next slide), \(S_0\) drops by about 25% — which moves the crossing year forward by years.
click to advance discussion ▶
Try Saying The math says PA crossed in ___, but real households face ___ that the LCOE model averages over.
Try Saying The political variable hides in the ___ of the function, because ___.
The peak of the lesson. The math worked. Three states, three crossing years. Then: PA's real residential market hasn't crossed yet — even though the model says it did in 2019. That gap is the wrench.

Have ready: a one-line callout that 65% of installed cost is soft. Quote the LBNL Tracking the Sun 2024 number directly if a student presses for the source. The packet has the link.

Next slide is the agency move — students propose a real intervention from the packet and predict its effect on PA's crossing year.
Agency Move 7 min

What policy lever moves PA's crossing year?

Pick one. Predict the new S0. Solve for the new crossing year. Defend it.

SolarAPP+ instant permitting
~ −$1/W  ·  ~25% drop
PA adopts NREL's automated permitting tool statewide. Same-day permits replace 2–6 week reviews. Cuts soft costs by ~$1/W.
Math: \(S_{\text{PA}}(0) = 28 \to 21\). Re-solve.
NREL pilot, 2023
Federal ITC at 30% (existing)
~ −30%  ·  $8.4K saved on $28K
The IRA's 30% Investment Tax Credit on residential systems through 2032. Already exists — but applying it lowers the effective \(S_0\) directly.
Math: \(S_{\text{PA}}(0) = 28 \to 19.6\). Re-solve.
IRS Residential Clean Energy Credit
Community solar legislation
~ −15%  ·  expands access
PA House bills (multiple sessions) would let renters and shaded households subscribe to a shared array. Cheaper per kWh than rooftop.
Math: \(S_{\text{PA}}(0) = 28 \to 23.8\). Re-solve.
DOE Community Solar Basics; PA HB pending
Net metering at retail rate
payback 30–50% faster
Compensates exported solar at full retail rate (vs. avoided-cost). Doesn't change the equation directly — changes the effective \(R(t)\) for the homeowner. Treat as a 5%/yr boost in \(R\).
Math: \(R_{\text{PA}}\) base 1.022 \(\to\) 1.072. Re-solve.
CPUC NEM 3.0 / state PUCs
FERC interconnection reform
queue 5 yr → 2 yr
FERC Order 2023 cluster-study reform speeds utility-scale projects. Lowers utility solar LCOE by ~10%, which trickles into community solar pricing.
Math: If you chose community solar above, layer this in: \(S_0\) drops another 10%.
FERC Order 2023, July 2023
Try Saying I would push for ___ because it changes ___ in the equation, which moves PA's crossing year from 2019 (model) to about ___ in real life.
7 min — popcorn discussion, not a project. Each pair (or volunteer) picks one card, plugs the new \(S_0\) (or new base) into PA's equation, and reports a new crossing t.

Sample answers: The cleanest demo: SolarAPP+ alone (administrative, no new spending, no legislative fight) drops PA's crossing year by ~3 years. That is what 25% in soft costs actually buys.

Trauma triage: If a student says "we can't change anything, the legislature won't act" — affirm and redirect: which interventions are not legislative? (SolarAPP+ is regulatory; many PA boroughs adopt it independently. ITC is already federal law.) The point is to widen the menu, not to debate state politics.
Lesson Synthesis 3 min

Two functions modeling the growth of two cell types: \(A(t) = 32 \cdot (1.05)^{t}\) and \(B(t) = 8 \cdot (1.12)^{t}\) (cell counts; t in days).

“Which type is growing faster?” Type B has the larger growth factor (1.12 > 1.05), so its graph rises more steeply. Type A starts with more cells, but B will catch up.
“On what day will the two cell counts be equal?” Set equal: \(32 \cdot (1.05)^{t} = 8 \cdot (1.12)^{t}\) ⇒ \(\dfrac{32}{8} = \left(\dfrac{1.12}{1.05}\right)^{t}\) ⇒ \(4 = (1.0667)^{t}\) ⇒ \(t = \dfrac{\log 4}{\log 1.0667} \approx \dfrac{0.602}{0.0280} \approx \boxed{21.5}\) days.
“What's the SAME math move we used three times today?” Set equal → divide to combine bases → take a log to free the exponent. Bank accounts. Bacteria. Solar vs. grid. Cell types. Different stories, identical move.
click to advance discussion ▶
Try Saying To find when two exponentials are equal, I ___, then ___, then ___.
I Can… (SWBAT)
Calculate the coordinates of the point of intersection of two exponential graphs using logarithms.
Interpret the intersection of the graphs of two exponential functions in context.
Lesson Synthesis (~3 min). Use the cell-growth example to confirm the procedure on a context that does not require any policy reasoning — just the math move. The IM lesson summary uses a similar example (carpenter ants vs. red wood ants); this version uses cells for variety and to keep the calculation tight.

SWBAT confirmed at lesson end. Both objectives now in students' working vocabulary.
Cool-Down 5 min

Cool-Down — Two ways to find the year

Let \(P(t) = 18 \cdot (1.04)^{t}\) represent the population of a town, in thousands, t years after 2000.
Problems
  1. Describe at least two ways you could find or estimate the year the population reaches 40 thousand. (You don't have to compute the value — just describe each method.)
  2. Use a logarithm to compute the year the population first reaches 40 thousand. Round to the nearest year.
Cool-Down · 5 min. Direct mirror of the IM cool-down ("Reaching 40 Thousand"), with a slightly different starting value so students cannot copy from the IM key. The math is the same.

Problem 1 (sample): Method A — graph \(P(t)\) and the horizontal line \(y = 40\); read off the t at the intersection. Method B — solve \(18 \cdot (1.04)^{t} = 40\) by dividing first and taking a log.

Problem 2: \(18 \cdot (1.04)^{t} = 40\) ⇒ \((1.04)^{t} = 40/18 \approx 2.222\) ⇒ \(t = \dfrac{\log 2.222}{\log 1.04} \approx \dfrac{0.347}{0.01703} \approx 20.4\) years ⇒ year ≈ 2020.

Common error: students round t aggressively (e.g., 20) and report 2020. Accept either 2020 or 2021 (since 20.4 → 2020.4); the lesson's emphasis is the method.
Amplify · 2 inputs

Order: Text → Math Response (named answer1, answer2) → Note. Problem 1 → Text Response. Problem 2 → Math Response (calendar year, integer).

Format: Example: 2020

Problem 1: Two ways to find the year. [Text Response]
content: when answer1.submitted and not isBlank(answer1.content) "Thank you. Make sure each method names a tool: graph + horizontal line, or set equal + take a log." when answer1.submitted and isBlank(answer1.content) "Please describe at least two methods before submitting." otherwise ""
Problem 2: Calendar year the population first reaches 40 thousand.
content: when answer2.submitted and answer2.numericValue > 2019.5 and answer2.numericValue < 2021.5 "Correct! 18·(1.04)^t = 40 → (1.04)^t = 40/18 → t = log(40/18)/log(1.04) ≈ 20.4. Calendar year: 2000 + 20 = 2020." when answer2.submitted and answer2.numericValue > 19.5 and answer2.numericValue < 21.5 "Almost — 20 (or 21) is the t value. Add 2000 to get the calendar year." when answer2.submitted and answer2.numericValue > 39.5 and answer2.numericValue < 40.5 "That is the target population (40 thousand), not the year. Solve 18·(1.04)^t = 40 for t, then add 2000." when answer2.submitted and answer2.numericValue > 1999.5 and answer2.numericValue < 2000.5 "Not quite — 2000 is the start year. The population reaches 40 thousand about 20 years later." when answer2.submitted and not isBlank(answer2.latex) "Almost — divide both sides of 18·(1.04)^t = 40 by 18 to get (1.04)^t = 2.222. Take a log: t = log(2.222)/log(1.04) ≈ 20.4. Add 2000." otherwise ""
Lesson Summary

Two exponentials, one crossing — find it with a log

Two exponential functions \(f(t) = a \cdot b^{t}\) and \(g(t) = c \cdot d^{t}\) (with \(b \ne d\)) cross at exactly one point. The t-coordinate of that point solves \(f(t) = g(t)\). To find it: set equal → divide to combine bases → take a log.
Step 1 — Set equal
\(a \cdot b^{t} = c \cdot d^{t}\)
Step 2 — Divide
\(\dfrac{a}{c} = \left(\dfrac{d}{b}\right)^{t}\)
Step 3 — Take a log
\(t = \dfrac{\log(a/c)}{\log(d/b)}\)
Lesson Summary The crossing year for two exponentials is determined by the ratio of the starting values and the ratio of the bases. Move either ratio — and you move the crossing year. Whose hand is on those ratios?
Lesson Summary slide. Read aloud or have a student read. The yellow box closes the loop on the wrench: \(t = \log(a/c) / \log(d/b)\) is a clean formula. The parameters (\(a\), \(b\), \(c\), \(d\)) are not natural constants — they're outcomes of permitting laws, tax credits, utility rate cases. The math is universal; the inputs are political.

The closing rate question students leave with: "What rate would your state need to drop \(S_0\) to so that PA's residential solar crossing happens by 2027 instead of 2030+?" — a real question with a real answer (about $20/kWh-equivalent, achievable with SolarAPP+ statewide adoption + ITC stack).