Lesson 10 — Exponential and Logarithmic Models in the World

Learners use logarithms to solve the two great exponential questions — "how long until it halves (or doubles)?" and "what rate is at work?" — through half-life, doubling time, and radioactive dating. They connect the mathematics to medicine in the body, population, and the age of very old things.

D05 P3: Intellectual & Cognitive Awareness D05.S2 55 minutes Draft

How do I use logarithms to solve for time or rate in real growth and decay — half-life, doubling time, and radioactive dating?

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Two panels. Left: a radioactive decay curve falling toward zero with half-life steps marked, and the equation t equals log of the remaining fraction over log of one half. Right: a growth curve rising with doubling time marked, and the equation t equals log of the growth factor over log of one plus r
Two panels. Left: a radioactive decay curve falling toward zero with half-life steps marked, and the equation t equals log of the remaining fraction over log of one half. Right: a growth curve rising with doubling time marked, and the equation t equals log of the growth factor over log of one plus r

Lesson 10 — Exponential and Logarithmic Models in the World

Summary

Learners use logarithms to answer the two great questions of exponential change: “how long until it halves (or doubles)?” and “what rate is at work?” They solve half-life and doubling-time problems and meet radioactive dating as the tool that reads the age of very old things. The mathematics connects to medicine in the body, population, and the planet.

Objectives

  • Solve growth and decay problems for time or rate using logarithms, and interpret the answers in real contexts such as half-life, doubling time, and radioactive dating. (D05.S2.11.02)

Connection

“Half of this medicine leaves your body every six hours.” “This population doubles every generation.” “This buried charcoal is 12 000 years old.” Each is a question about time in an exponential process, and time is what logarithms unlock: if you know the factor and the fraction, the log tells you how many steps have passed — whether the steps are hours, generations, or millennia.

Materials

  • Growth-and-decay modeling sheet
  • Math journal

Preparation

  • Copy or draw the problem sheet.
  • Retrieval: from Lesson 9, log_b(x) = y means b^y = x; from Grade 10, growth and decay factors and half-life (D05.S1.10.01).
  • Prepare worked examples for half-life, doubling time, and dating.

Facilitator note

This lesson is written to the learner (“you”). The idea to land: to solve b^t = c for t, take logs of both sides: t = log(c)/log(b). For half-life, b = ½; for doubling, b = 2; for growth at rate r, b = 1 + r. Teach the one move — take the log, then divide — with worked examples and guided practice (S-011).

The environment lens: half-life is the honest language of how long pollution and radioactive material persist — carbon-14’s 5 730-year half-life is what lets us date ancient charcoal and climate records. The intellectual lens: one move (log both sides) turns a power equation into a linear one for the unknown time. The critical-thinking lens: learners check every answer by substituting it back into the original. The technology lens: dating labs, medical dose calculators, and climate models all solve these same log equations under the hood. Distinguish the mathematics (exact) from the real-world mess (rates are estimates). Preview: Lesson 11 begins geometry with the unit circle.

Procedure

  1. Recall (5 min). From Lesson 9, log₁₀(1000) = 3 because 10³ = 1000. Today we find the unknown power (time) the same way.
  2. The one move (10 min). To solve b^t = c, take logs of both sides and divide: t = log(c)/log(b). Worked example: a medicine halves every 6 hours (b = ½); after how many steps is 1/16 left? (½)^t = 1/16, so t = log(1/16)/log(½) = 4 steps, or 24 hours.
  3. Half-life and dating (15 min). Carbon-14 has a half-life of about 5 730 years. If a sample has ¼ of its carbon-14 left, two half-lives have passed — about 11 460 years. Worked example: a sample has 30% left. Solve (½)^(t/5730) = 0.30 → t = 5730·log(0.30)/log(0.5) ≈ 9 950 years (S-435). This is how the age of ancient wood and bone is estimated.
  4. Doubling time and the rule of 70 (10 min). A population growing at rate r doubles when (1 + r)^t = 2, so t = log(2)/log(1 + r). Worked example: at 2% a year, t = log(2)/log(1.02) ≈ 35 years. The rule of 70 approximates this: doubling time ≈ 70/(percent rate) ≈ 70/2 = 35 years.
  5. Guided practice (10 min). With a partner: (a) a pollutant halves every 20 years — how long until 1/8 remains? (b) a saving grows 7% a year — when does it double? (c) a sample has 12.5% of its carbon-14 left — how old is it? Check by substitution.
  6. Close (5 min). Say, in one sentence, how a logarithm answers “how long?” for a halving or doubling process.

Differentiation

  • Support: Solve only whole-number-of-steps problems first (½, ¼, ⅛), then move to logs.
  • Extension: Solve for an unknown rate: given a quantity triples in 20 years, find the yearly rate.

Assessment

  • Formative (peer + self): Can the learner solve a growth/decay equation for time with a logarithm and interpret the result in a half-life, doubling-time, or dating context, with a substitution check?
  • Portfolio artifact (unit): The completed modeling sheet, added to the pattern toolkit.

Home connection

Estimate a doubling time near home — how long a plant, a savings amount, or a family tradition’s scale doubles — and check it with the rule of 70.

Resources

  • On worked examples and guided practice: Kirschner, Sweller & Clark (2006), https://doi.org/10.1207/s15326985ep4102_1 (S-011).
  • On carbon-14 dating and log solutions: OpenStax, Algebra and Trigonometry (S-435); on climate-record dating context: IPCC (S-005).