First — talk for 2 minWhat's your family's electric bill like, roughly? Has anyone in your house thought about solar panels? What stopped them, or pushed them toward it? If no one, Mr. Bennett will share.

Two prices, moving in opposite directions

Every year, rooftop solar panels get a little cheaper. Since 1977, the price of a solar panel has dropped from about \$76.67 per watt to about \$0.11 per watt in 2024 — falling roughly 10% per year for almost fifty years. Engineers call this pattern Swanson's law.

Meanwhile, the price of grid electricity creeps up. The U.S. average residential rate was about 9 cents per kilowatt-hour in 2010 and is around 16.5 cents today — growth of about 2.5% per year. One curve goes down. The other goes up.

Somewhere between those two curves, there is a single year where they cross. Before that year, grid electricity is cheaper; after, rooftop solar is cheaper. So in your state, which year does solar win — and what is keeping that year in the future?

Worked Example #1 — The national crossing year (find t)
ProblemSolar cost \(S(t)\) and grid retail rate \(R(t)\) are measured in cents per kWh, with \(t\) = years since 2010. Find the year both costs are equal.
DataSolar starts at 24¢/kWh and decays at factor 0.92 per year: \(S(t) = 24 \cdot (0.92)^{t}\). Grid starts at 9¢/kWh and grows at factor 1.025 per year: \(R(t) = 9 \cdot (1.025)^{t}\). Find \(t\) where \(S(t) = R(t)\).
Word Bank divide decay logarithm equal isolate multiply growth ratio
Number Bank 9.08 2.667 0.4260 2019 1.1141 0.0469 8/3 15.7
#Reasoning (fill in)Math
1 Set the two costs . This is the equation that holds at the crossing year. \(24 \cdot (0.92)^{t} = 9 \cdot (1.025)^{t}\)
2 both sides by \(9 \cdot (0.92)^{t}\) so the bases are grouped on one side. \(\dfrac{24}{9} = \dfrac{(1.025)^{t}}{(0.92)^{t}} = \left(\dfrac{1.025}{0.92}\right)^{t}\)
3 Simplify the on the left and the base on the right. \(\dfrac{8}{3} \approx\) \(= (\) \()^{t}\)
4 Take a on both sides to free \(t\) from the exponent. \(\log\!\left(\tfrac{8}{3}\right) = t \cdot \log(1.1141)\)
5 Divide to \(t\), then evaluate. Add 2010 to convert \(t\) to a year. \(t = \dfrac{\log(8/3)}{\log(1.1141)} \approx \dfrac{}{}\) \(\approx\) yr ⇒ year
Worked Example #2 — Texas crossing year
ProblemIn Texas, solar installs cheaply and grid rates are low. Find the Texas crossing year.
Data\(S_{\text{TX}}(t) = 22 \cdot (0.92)^{t}\) and \(R_{\text{TX}}(t) = 11 \cdot (1.025)^{t}\), in cents per kWh, with \(t\) = years since 2010. Note \(0.92\) is a decay factor: \(b < 1\).
Word Bank divide log simplify equal substitute isolate add
Number Bank 2 0.301 6.4 2016 1.1141 0.0469 3.2
#Reasoning (fill in)Math
1 Set the two cost functions . \(22 \cdot (0.92)^{t} = 11 \cdot (1.025)^{t}\)
2 both sides by \(11 \cdot (0.92)^{t}\) to group the bases. \(\dfrac{22}{11} = \left(\dfrac{1.025}{0.92}\right)^{t}\)
3 the ratio on the left, then the base on the right. \(= (1.1141)^{t}\)
4 Take a on both sides. Use \(\log_b(\dots)\) or \(\log\) base 10 — same answer. \(\log(2) = t \cdot \log(1.1141)\)
5 Divide to \(t\). Then add 2010 to get the calendar year. \(t = \dfrac{0.301}{0.0469} \approx\) yr ⇒ year
Activity 2 answers — pick one state per pair, then trade
California: \(30 \cdot (0.92)^{t} = 14 \cdot (1.06)^{t}\). Divide: \(\dfrac{30}{14} =\) \( = (1.152)^{t}\). Take a log: \(t \approx\) yr ⇒ year .
Pennsylvania: \(28 \cdot (0.92)^{t} = 11 \cdot (1.022)^{t}\). Divide: \(\dfrac{28}{11} \approx\) \( = (1.111)^{t}\). Take a log: \(t \approx\) yr ⇒ year .
Ranking (earliest crossing first): → → .
Conclude

The math says PA crossed in 2019. So why is rooftop solar still expensive in PA?

Three states. Three crossing years. Same math — different starting values and growth factors. But the math gives the year a system-level average crosses. The year a real household can afford to install is moved by levers the equation hides.

The model says California crosses in ~2015.
Texas crosses in ~2016–2017.
Pennsylvania crosses in ~2019.
Real households face About 65% of a U.S. residential install price is not the panels. It is permitting, inspection, labor, financing, and sales overhead. Germany pays ~\$0.50/W in soft costs. PA pays ~\$1.65/W.

Reflect: Which state would you want to live in if you wanted to install solar today — and what one policy lever (permitting, net metering, community solar) would move your state's crossing year forward the most?

Tier 2 · General Academic Words
Tier 3 · Solar, Grid, & Math Words