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B. Exponential Functions · Exponential Functions

Solving Problems with Exponential Growth: Populations, Bacteria and Interest

B2.3 B2.2About 75 minutes
Learning goals
  • I can solve growth problems by substituting values for the exponent into an exponential equation.
  • I can read an exponential graph to estimate when a quantity reaches a given value.
  • I can find an unknown time by reasoning with powers or by systematic trial (guess, check and improve).
  • I can use A = P(1 + i)ⁿ as an exponential growth model for money earning compound interest.

Words to know

substitute
Replace a variable with a number and evaluate.
interpolate / extrapolate
Estimate a value between known data (interpolate) or beyond them (extrapolate).
systematic trial
Guess a value, check it, and improve the guess in an organized way until it is close enough.
compound interest
Interest that is added to the principal, so later interest is earned on earlier interest too. It grows exponentially.
principal (P)
The amount of money first invested or borrowed.

Planning for a growing town

A town of 8500 people is growing at about 2.4% a year. The town council wants to know: How many people will live here in 10 years? When will we pass 12 000 people and need a second arena? How many lived here 5 years ago?

These are the three kinds of questions you will answer in this lesson: find the amount for a given time, find the time for a given amount, and look backwards. The tools are substitution, reading graphs, and organized guessing.

Strategy 1: Substitute into the equation

If you know the time, put it into the exponent and evaluate. For the town, P = 8500(1.024)ᵗ:

  • In 10 years: P(10) = 8500(1.024)¹⁰ ≈ 10 775 people.
  • Five years ago (t = −5): P(−5) = 8500(1.024)⁻⁵ ≈ 7550 people. Negative t works because the model goes backward, as long as the growth rate really was the same then.

Round people to whole numbers, and money to the cent, at the end of the calculation.

Strategy 2: Read the graph

To find when a quantity reaches a value, draw a horizontal line at that value, find where it meets the curve, and read down to the time axis. For a culture that starts at 200 bacteria and doubles every 4 hours, N = 200(2)^(t/4). When is N = 1600?

t (h)N2468101214164008001200160020002400280032000(12, 1600)N = 200(2)^(t/4)
Reading the graph: go across from N = 1600 to the curve, then down. The culture reaches 1600 at t = 12 h.

Check with powers: 1600 ÷ 200 = 8 = 2³, so there have been 3 doublings, and 3 × 4 = 12 hours. When the numbers work out like this, rewriting as a power of the same base gives the exact answer.

Strategy 3: Systematic trial

When the numbers are not "nice", substitute guesses and improve them. To find when the town passes 12 000:

  • t = 10: 10 775 (too small)
  • t = 14: 11 847 (close, still too small)
  • t = 15: 12 132 (over 12 000)

So the population passes 12 000 during the 15th year (between t = 14 and t = 15). A graph or a graphing tool can give a more precise value if you need one; in later courses (such as MCR3U or MHF4U) you learn logarithms, which solve these equations exactly.

t (years)P24681012141618209 00010 00011 00012 00013 000t = 14: 11 847t = 15: 12 132target 12 0008 000(vertical axis starts at 8 000)
The town's population model with the 12 000 target line. The curve crosses the line between t = 14 and t = 15.
Worked example

Example 1: Find the amount

A small-business owner's online subscribers grow by 8% per month from 2500. (a) Write a model. (b) How many subscribers after 1 year (12 months)? (c) After 18 months?

  1. (a) Growth: b = 1.08. S = 2500(1.08)ᵐ, m in months.
  2. (b) S(12) = 2500(1.08)¹² ≈ 2500 × 2.518 17 ≈ 6295 subscribers.
  3. (c) S(18) = 2500(1.08)¹⁸ ≈ 2500 × 3.996 02 ≈ 9990 subscribers.

About 6295 after 1 year and 9990 after 18 months.

Worked example

Example 2: Find the time using powers

A rumour spreads so that the number of people who have heard it triples every 2 days. Today 5 people know. When will 1215 people know?

  1. Model: N = 5(3)^(t/2).
  2. Set N = 1215: 5(3)^(t/2) = 1215, so 3^(t/2) = 243.
  3. 243 = 3⁵, so t/2 = 5 and t = 10 days.

10 days

Worked example

Example 3: Compound interest

Ava invests $2000 at 4% per year, compounded annually. (a) How much does she have after 6 years? (b) After how many full years will she first have at least $3000?

  1. Compound interest is exponential growth: A = P(1 + i)ⁿ, where i is the interest rate per compounding period and n the number of periods.
  2. (a) A = 2000(1.04)⁶ ≈ $2530.64.
  3. (b) Trial: n = 10 gives 2000(1.04)¹⁰ ≈ $2960.49 (not yet). n = 11 gives ≈ $3078.91. She first has at least $3000 after 11 years.

$2530.64; after 11 years

Money gets its own unit

Compound interest is one of the most important uses of exponential growth. In the Financial Mathematics unit of this course you will study it in depth: different compounding periods, present value, comparing simple and compound interest, and annuities. Here, just remember that A = P(1 + i)ⁿ is an exponential function of n with initial value P and growth factor 1 + i.

Just for fun

A legend says the inventor of chess asked a king for 1 grain of rice on the first square of the board, 2 on the second, 4 on the third, doubling every square. The king laughed at such a small reward.

He stopped laughing at square 64, which alone needs 2⁶³ ≈ 9.2 × 10¹⁸ grains, far more rice than the world grows in a year.

Hands-on activity

Grains of rice on a calendar

You need: a calendar page (or a 7-by-5 grid you draw), a pencil, a calculator, and optionally a measuring spoon and some rice.

  1. Put 1 grain on day 1 and double each day. Write the number for each day of a month on the calendar using N = 2^(d − 1).
  2. Optional: count how many grains of rice fill a level teaspoon (about how many?). Use it to estimate how many teaspoons day 20 would need.
  3. Use systematic trial to find the first day the count passes one million.
  4. Change the rule: start with 1 grain and triple each day. On which day does it pass one million now?

What to look for: doubling: day 21 has 2²⁰ = 1 048 576 grains, so day 21 is the first day over one million (day 20 has 524 288). Tripling: 3¹² = 531 441 is not enough and 3¹³ = 1 594 323 is, so day 14 (since N = 3^(d − 1)). A level teaspoon holds roughly 100 to 300 grains, depending on the rice (count yours!), so day 20 alone would need well over a thousand teaspoons.

Your turn

1

P = 8500(1.024)ᵗ. Find the population after 3 years.

2

N = 200(2)^(t/4). Find N after 6 hours.

3

N = 200(2)^(t/4). When is N = 3200?

4

$1500 is invested at 5% per year compounded annually. Find the amount after 8 years.

5

A fish population of 600 grows 10% per year. After how many full years does it first exceed 1000?

6

Use the town model to estimate the population 2 years ago.

7

Explain why the answer to "when does the town pass 12 000?" is "during year 15" rather than "after exactly 15 years".

More practice: warm-up

1

Evaluate 100(2)³.

2

y = 50(3)ˣ. Find y when x = 4.

3

y = 50(3)ˣ. Find x when y = 450.

4

In A = P(1 + i)ⁿ, what does n count?

5

What is the growth factor for 6% interest per year?

More practice: on your own

1

A video had 400 views and its views double every 3 days. Write a model and find the views after 2 weeks (14 days).

2

A colony of 50 ants triples every month. After how many months are there 4050 ants?

3

$5000 is invested at 3.5% per year compounded annually. How much interest is earned in 10 years?

4

A city of 250 000 grows 1.5% per year. Find, by systematic trial, the first full year in which it exceeds 300 000.

5

Bacteria: N = 1000(1.2)ʰ, h in hours. Graphing shows N = 5000 near h = 9. Check by substitution.

Challenge

1

A quick estimate called the "rule of 72" says money at r% per year doubles in about 72 ÷ r years. Test it for 6% by finding the exact doubling time to the nearest year by trial.

2

Two towns: A = 20 000(1.03)ᵗ and B = 26 000(1.01)ᵗ. In which year does town A first become larger?

3

A bacteria count is 3000 at 9:00 a.m. and 24 000 at 3:00 p.m. Assuming exponential growth, find the doubling time and the count at 11:00 a.m.

Check your understanding

Quiz

1.

A town of 5000 grows 3% a year. Which expression gives its population after 7 years?

2.

N = 100(2)^(t/5). When does N reach 800?

3.

$1000 at 6% per year compounded annually for 5 years grows to about:

4.

A town's population is modelled by P = 2000(1.05)ᵗ, t in years from now. What does P(−2) represent?

5.

A quantity is 2900 at t = 9 and 3200 at t = 10. What can you conclude about when it reaches 3000?

6.

Which situation is best modelled by exponential growth?

For parents and teachers

Big idea (B2.3, with B2.2): students solve problems from real applications (population growth, bacteria, compound interest) by substituting values for the exponent into equations and by interpreting graphs.

How to help: Ask three kinds of question about any growth model: "How big after t?", "When does it reach this value?" and "How big was it before?" For "when" questions, encourage an organized table of guesses.

Common difficulties:

  • Rounding too early: keep full calculator values until the last step.
  • Misreading "during year 15" vs "after 15 years".
  • Doubling-time exponent: N = 200(2)^(t/4), not 200(2)^(4t).
  • Thinking compound interest is linear: $1000 at 6% for 5 years is $1338.23, not $1300.

The full study of interest and annuities is in the Financial Mathematics unit.