Algebra 2 B (California) → Exponential Growth and Decay, Rewritten

By the end of this lesson I can…

rewrite exponential expressions to show growth or decay rates over different time periods, and interpret growth factors, percent rates and half-lives.

Instruction

An exponential function has the form f(t) = a · b^t with a > 0 and b > 0, b ≠ 1. The initial value a is f(0), and the growth factor b is what the output is multiplied by each time t increases by 1. If b = 1 + r, then r is the percent rate of change: b = 1.05 means 5% growth per unit of time, and b = 0.8 means 20% decay.

The same growth, measured differently. An account grows 5% per year: V(t) = 1,000(1.05)^t, with t in years. What is the monthly growth rate? Because t years is 12t months, rewrite using the power rule (A-SSE.2):

1.05^t = (1.05^1/12)^12t ≈ 1.004074^12t.

So the value grows about 0.407% per month, not 5/12 ≈ 0.417%. Dividing the annual rate by 12 slightly overstates the monthly rate, because monthly growth compounds. Per decade: 1.05^t = (1.05¹⁰)^t/10 ≈ 1.6289^t/10, so the value grows about 62.9% per decade. Each rewritten form is the same function (F-IF.8); it simply puts a different, useful rate in plain sight. Treat 1.05^1/12 as a single number (A-SSE.1.b): it is the monthly growth factor.

Half-life. Carbon-14 decays with a half-life of about 5,730 years: every 5,730 years, half of what remains decays. The fraction remaining after t years is

F(t) = (1/2)^t/5730.

The exponent t/5730 counts half-lives. Rewritten, (1/2)^t/5730 = ((1/2)^1/5730)^t ≈ 0.999879^t, a decay of about 0.0121% per year. After 1,000 years, F(1000) ≈ 0.886, so about 88.6% remains. The half-life form is easier to reason with; the annual form is easier to compare with other rates.

Rate of change is not constant (F-IF.6). For V(t) = 1,000(1.05)^t, the average rate of change from year 0 to year 10 is (1,628.89 − 1,000)/10 ≈ $62.89 per year, while from year 10 to year 20 it is (2,653.30 − 1,628.89)/10 ≈ $102.44 per year. For an exponential function, the rate of change is proportional to the current value. That is the defining feature of exponential growth, and it is why a constant percentage produces ever-larger absolute increases.

Reading a form (MP.7). Look at 200 · 3^t/4. The initial value is 200, and the output triples every 4 units of time. Rewritten as 200(3^1/4)^t ≈ 200(1.316)^t, it grows about 31.6% per unit of time. Before computing, ask what question you want the expression to answer, then choose the form that answers it.

Formative check

Work through these before moving on. They are not graded — they tell you, and your teacher, whether the standard below has landed yet.

+50 XP

V(t) = 1,000(1.05)^t, t in years. Which expression shows the monthly growth factor?

Fill in the blank

With a half-life of 5,730 years, the fraction of carbon-14 left after 11,460 years is , and after 1,000 years it is about .

Myth or Fact?

For V(t) = 1,000(1.05)^t, the average rate of change is the same over every 10-year interval.

A town’s population is P(t) = 12,000(1.024)^t, t in years. Rewrite P to show the growth factor per decade and the growth factor per month. State each rate as a percent and explain why the monthly rate is not 2.4% divided by 12.

Quick self-check

How confident are you that you can rewrite exponential expressions to reveal rates over different time periods?

Not yetVery confident

Practice

Work these on paper or in your notebook, then open Check your answer. Aim for all of Fluency and Application; try at least one Challenge.

Printable version: this unit’s practice workbook (PDF)

Fluency

Build speed and accuracy with the core skill.

A quantity starts at 8,000 and grows by 2% each year. Write a model and find its value after 15 years, to the nearest whole number.
Check your answer
Answer: A = 8,000(1.02)^t; after 15 years ≈ 10,767
The growth factor is 1 + 0.02 = 1.02.
A quantity starts at 1,200 and grows by 2% each year. Write a model and find its value after 5 years, to the nearest whole number.
Check your answer
Answer: A = 1,200(1.02)^t; after 5 years ≈ 1,325
The growth factor is 1 + 0.02 = 1.02.
A quantity starts at 2,500 and grows by 3% each year. Write a model and find its value after 12 years, to the nearest whole number.
Check your answer
Answer: A = 2,500(1.03)^t; after 12 years ≈ 3,564
The growth factor is 1 + 0.03 = 1.03.
A quantity starts at 2,500 and decreases by 2% each year. Write a model and find its value after 11 years, to the nearest whole number.
Check your answer
Answer: A = 2,500(0.98)^t; after 11 years ≈ 2,002
The growth factor is 1 − 0.02 = 0.98.
A quantity starts at 2,500 and decreases by 12% each year. Write a model and find its value after 12 years, to the nearest whole number.
Check your answer
Answer: A = 2,500(0.88)^t; after 12 years ≈ 539
The growth factor is 1 − 0.12 = 0.88.
A quantity starts at 8,000 and decreases by 6% each year. Write a model and find its value after 7 years, to the nearest whole number.
Check your answer
Answer: A = 8,000(0.94)^t; after 7 years ≈ 5,188
The growth factor is 1 − 0.06 = 0.94.
V(t) = 500(1.06)^t, t in years. Rewrite to show the monthly growth factor and the per-decade growth factor.
Check your answer
Answer: 500(1.004868)^12t and 500(1.7908)^t/10: about 0.487% a month, 79.08% a decade
Use (b^1/12)^12t = b^t.
A medication (illustrative) has a half-life of 6 hours. What fraction remains after 24 hours, and what is the hourly decay factor?
Check your answer
Answer: 1/16; (1/2)^1/6 ≈ 0.8909, about 10.91% lost per hour
24 hours is 4 half-lives.
Rewrite 200 · 3^t/4 to show the growth factor per unit of time.
Check your answer
Answer: 200(1.3161)^t, about 31.6% per unit

Application

Use the skill in context. Show your reasoning.

California grew about 0.6% a year in the 2010s. What growth factor per decade is that?
Check your answer
Answer: 1.006¹⁰ ≈ 1.0616, about 6.2% a decade
Carbon-14 has a half-life of about 5,730 years. What fraction of the original carbon-14 remains in a 3,000-year-old sample?
Check your answer
Answer: About 0.6957 (roughly 70%)
(1/2)^3000/5730.
Which is better for a saver: 5% a year compounded yearly, or 0.4% a month compounded monthly? Compare annual factors.
Check your answer
Answer: 5% yearly: 1.05 is larger than 1.004¹² ≈ 1.04907

Challenge

Stretch problems. Expect to think before you write.

Which grows faster: a quantity that doubles every 7 years or one that grows 10% a year? Compare annual factors.
Check your answer
Answer: Doubling every 7 years: 2^1/7 ≈ 1.1041 > 1.10

Review

Keep earlier skills sharp.

Simplify: (−5 − i)(5 − 6i)
Check your answer
Answer: −31 + 25i
Use i² = −1.
Evaluate exactly: 1000^−2/3
Check your answer
Answer: 1/100
Take the 3th root of 1000 first, then raise to the 2 power; a negative exponent means reciprocal.