A. Exponential and Logarithmic Functions · Exponential and Logarithmic Functions

Solving Growth and Decay Problems

A3.4 A2.4 A3.2About 75 minutes

Learning goals

I can write exponential models for half-life, doubling time and compound interest.
I can solve for the time, rate, half-life or doubling time in a model using logarithms.
I can solve problems about cooling, where the temperature difference decays exponentially.
I can pose a growth or decay problem of my own and solve it.

Words to know

half-life
The time it takes for half of a decaying quantity to remain. Model: A = A₀(1/2)^(t/h), where h is the half-life.

doubling time
The time it takes a growing quantity to double. Model: P = P₀(2)^(t/d), where d is the doubling time.

compound interest
Interest added to the balance so that it earns interest too. A = P(1 + i)ⁿ, where i is the rate per period and n is the number of periods.

initial value
The amount at time t = 0, written A₀ or P₀.

Newton's law of cooling
The difference between an object's temperature and the room's temperature decays exponentially over time.

Now you can solve for time

Remember from Grade 11: in Exponential Functions (MCR3U) you modelled radioactive decay, population growth, cooling and compound interest, and found when something happened by reading a graph or by trial and error. With logarithms you can now solve for an unknown time, rate or half-life exactly, with algebra. These are the same skills used in pharmacology (how long a drug stays in the body), archaeology (carbon dating), finance and environmental science.

The models

Half-life: A = A₀(1/2)^(t/h). After one half-life (t = h), A = A₀/2.
Doubling time: P = P₀(2)^(t/d). After one doubling time, P = 2P₀.
Growth or decay by a percent per period: A = A₀(1 + r)ⁿ, with r > 0 for growth and −1 < r < 0 for decay. Compound interest is this model with r = i, the interest rate per compounding period.

To solve for the time, isolate the power and take logs, as in lesson 12.

Worked example

Example 1: Half-life in medicine

Iodine-131, used in some medical treatments, has a half-life of about 8 days. How long does it take 100 mg to decay to 10 mg?

Model: A = 100(1/2)^(t/8).
Set A = 10: 100(1/2)^(t/8) = 10, so (1/2)^(t/8) = 0.1.
Take logs: (t/8) log 0.5 = log 0.1.
t = 8 × log 0.1 ÷ log 0.5 = 8 × (−1) ÷ (−0.30103) ≈ 26.6.
Check: 26.6 days is between 3 half-lives (12.5 mg left) and 4 half-lives (6.25 mg left) ✓

About 26.6 days

Iodine-131 halves every 8 days: 100 mg, 50 mg, 25 mg, 12.5 mg. The amount reaches 10 mg at t ≈ 26.6 days.

Worked example

Example 2: Carbon dating

Carbon-14 has a half-life of about 5730 years. Charcoal from an ancient campfire has 30% of the carbon-14 that fresh wood has. Estimate its age.

Model: A/A₀ = (1/2)^(t/5730), and A/A₀ = 0.30.
(t/5730) log 0.5 = log 0.3.
t = 5730 × log 0.3 ÷ log 0.5 ≈ 5730 × 1.737 ≈ 9953.

About 10 000 years old. (Real carbon dating adds corrections, so scientists report a range.)

Worked example

Example 3: Finding a half-life from data

A radioactive sample decays from 80 g to 65 g in 10 hours. Find its half-life.

Model: 65 = 80(1/2)^(10/h), so (1/2)^(10/h) = 65/80 = 0.8125.
(10/h) log 0.5 = log 0.8125.
10/h = log 0.8125 ÷ log 0.5 ≈ 0.29956.
h = 10 ÷ 0.29956 ≈ 33.4.

Half-life ≈ 33.4 hours

Worked example

Example 4: Doubling time

A bacteria culture grows from 500 to 3000 cells in 5 hours. Find the doubling time.

3000 = 500(2)^(5/d), so 2^(5/d) = 6.
(5/d) log 2 = log 6.
d = 5 log 2 ÷ log 6 ≈ 1.934.

Doubling time ≈ 1.93 hours (about 1 hour 56 minutes).

Worked example

Example 5: Compound interest compounded monthly

Lina invests $5000 in a GIC at 4.8% per year, compounded monthly. How long until it doubles?

Monthly rate i = 0.048 ÷ 12 = 0.004. Model: A = 5000(1.004)ⁿ, n in months.
5000(1.004)ⁿ = 10 000, so 1.004ⁿ = 2.
n = log 2 ÷ log 1.004 ≈ 173.6 months.
Interest is added monthly, so it first doubles after 174 months = 14.5 years.
Quick check with the "rule of 72": 72 ÷ 4.8 = 15 years. Close ✓

174 months, or 14.5 years

Cooling: a decay that levels off

A hot drink does not cool to 0 °C; it cools to room temperature. Newton's law of cooling says the difference between the drink and the room decays exponentially:

T = Tᵣ + (T₀ − Tᵣ)(1/2)^(t/h), where Tᵣ is the room temperature, T₀ the starting temperature and h the time for the difference to halve. The graph has a horizontal asymptote at T = Tᵣ.

Worked example

Example 6: Hot chocolate after skating

After skating on a cold January afternoon, you pour hot chocolate at 90 °C in a 20 °C kitchen. The temperature difference halves every 12 minutes. When will it be 50 °C?

Model: T = 20 + 70(1/2)^(t/12).
50 = 20 + 70(1/2)^(t/12), so (1/2)^(t/12) = 30/70 = 3/7.
(t/12) log 0.5 = log (3/7).
t = 12 × log (3/7) ÷ log 0.5 ≈ 14.7.

After about 14.7 minutes.

The hot chocolate cools quickly at first, then more slowly, approaching the room temperature of 20 °C. After 12 minutes the 70° difference has halved to 35°, so T = 55 °C; it reaches 50 °C after about 14.7 minutes.

Pose your own problem

The curriculum asks you to pose problems as well as solve them. Pick a real situation (your savings account, a houseplant's growth, a cooling bowl of soup), choose a model and realistic numbers, and ask a "when will...?" question. Then solve it with logs and check that the answer makes sense.

Just for fun

Why was the radioactive atom so relaxed before the test?

It had already lost half its worries, and in one more half-life it would lose half of the rest.

Hands-on activity

Coin decay

You need: 40 coins (or 40 dried beans with one side marked with a felt pen), a shoebox or large container with a lid, paper and a calculator. If young children are nearby, keep the coins or beans away from them (choking risk).

Put all 40 coins in the box, close it and shake. Open it and remove every coin showing heads. Count what is left.
Record the toss number n and the number of coins left. Repeat until no coins are left.
Each coin has a 1/2 chance of being removed each toss, so the model is C = 40(1/2)ⁿ. What is the "half-life" in tosses?
Use logs to predict after how many tosses fewer than 3 coins should remain. Compare with your data.
Repeat the experiment and average the two runs.

What to look for: the half-life is 1 toss. 40(1/2)ⁿ < 3 means (1/2)ⁿ < 0.075, so n > log 0.075 ÷ log 0.5 ≈ 3.74: after about 4 tosses fewer than 3 coins should remain (the model gives 2.5). Real counts wobble around the model, especially when few coins are left; averaging runs brings the results closer, just as real radioactive samples with billions of atoms follow the model very closely.

Your turn

A substance has a half-life of 30 years. How long until only 25% remains?

How long until 20% of a carbon-14 sample has decayed (80% remains)? Use h = 5730 years.

$1000 is invested at 3% per year, compounded annually. When will it reach $1500?

A population of 2000 grows at 3.5% a year. When will it triple?

A sample decays from 50 g to 30 g in 3 days. Find the half-life.

A culture grows from 200 to 900 cells in 6 hours. Find the doubling time.

Soup at 90 °C is in a 20 °C room; the difference halves every 12 minutes. When will it be 35 °C?

More practice: warm-up

In A = A₀(1/2)^(t/h), what does h represent?

A drug has a half-life of 6 hours. What fraction remains after 12 hours?

A population doubles every 10 years. By what factor does it grow in 30 years?

Write the model for $800 at 5% per year compounded annually.

What is the monthly rate for 6% per year compounded monthly?

More practice: on your own

How long does it take money to double at 6% per year compounded monthly?

Strontium-90 has a half-life of about 28.8 years. How long until 5% of a sample remains?

A car worth $32 000 loses 15% of its value each year. When will it be worth $10 000?

A town grew from 8000 to 11 000 people in 6 years. Find the annual growth rate, assuming exponential growth.

Pose and solve a problem: your $300 savings earns 2.4% per year compounded monthly. How long until you have $400?

Challenge

Show that a quantity with half-life h can be written A = A₀(0.5)^(t/h) = A₀ × bᵗ, and find b for h = 8 days.

Two investments: $2000 at 5% compounded annually and $2500 at 3% compounded annually. When will they be equal?

A drug has a half-life of 4 hours. A patient takes 200 mg. The drug stops working when less than 30 mg is left. How long does a dose work? If a second 200 mg dose is taken at that moment, how much is in the body right after?

Check your understanding

Quiz

A substance has a half-life of 12 years. What fraction remains after 36 years?

Which equation gives the time for $600 at 4% per year (compounded annually) to reach $900?

A sample decays from 100 g to 70 g in 5 days. Which is the half-life (to one decimal)?

In T = 20 + 70(1/2)^(t/12), what does the 20 represent?

A population doubles every 15 years. Which model fits, with P₀ the starting population?

$4000 at 6% per year compounded monthly. Which equation finds when it reaches $6000?

For parents and teachers

Big idea (A3.4, A2.4): half-life, doubling time, compound interest and cooling are all exponential models. To find an unknown time or rate, isolate the power and take logarithms.

How to help: Use real numbers: look at a savings or GIC rate together and ask your teen how long it would take to double money at that rate, then check with the rule of 72. Ask them to explain what each number in their model means.

Common difficulties:

Using the annual rate for monthly compounding (use i = annual rate ÷ 12 and count n in months).
Writing (1/2)^(t·h) instead of (1/2)^(t/h).
Forgetting that the cooling model approaches room temperature, not 0 °C.
Not rounding up to a whole number of compounding periods when money is credited only at the end of each period.