Exponential & Logarithmic Models — Print

Growth and decay that depend on how much is already there, and the logs that solve them.

What the practice questions cover

These questions use 5 ideas from the course. Follow any of them for a definition, where it is taught, and what it leads to.

A cup of coffee left on the desk.

It starts at 80°. After ten minutes it is 53°; after twenty, 37°; after thirty, 28°.

coffee left on the desk room temp 80°53°37°28° equal times it never drops the same degrees — it drops the same share

The drops are 27, then 16, then 9 — shrinking every time, though the ten minutes never change.

What stays steady is the share: each interval closes about 40% of the remaining gap to room temperature.

That is the signature of exponential change. Constant difference means a straight line; constant ratio means an exponential — and the two models make wildly different predictions.

Linear growth against exponential growth adding the same amount, against multiplying by the same amount + 6 each step x 2 each step time amount linear: y = mx + b exponential: y = a(b)^x the exponential starts behind and still overtakes, every single time

Telling the two models apart

Look at what is constant between equal time steps. A constant difference means linear; a constant ratio means exponential.

A table going 3, 6, 12, 24 is exponential — each entry is double the last.

A table going 3, 6, 9, 12 is linear — each entry is three more than the last.

Divide, do not subtract

To test for exponential, divide each term by the one before. If you get the same number every time, you have your growth factor.

Writing the model

The standard form is y = abt, where a is the starting amount and b is the factor per period.

A 5% annual rise gives b = 1.05; a 5% fall gives b = 0.95. The 1 carries the part you keep.

A population of 200 growing 8% a year is y = 200(1.08)t.

The rate is not the factor

A 30% decay has b = 0.70, not 0.30. You keep 70% each period, and it is what you keep that gets multiplied.

Half-life and doubling time

When the period is a half-life, the factor is 12, so y = a(12)t/h with h the half-life.

Dividing t by h counts the halvings rather than the years, which is the only adjustment needed.

Doubling works the same way with a factor of 2 in place of a half.

Count the halvings first

For 22920 years at a half-life of 5730: that is four halvings, so 116 remains. No logs needed when the number divides evenly.

When the unknown is the time

If you know the amount and want the time, the unknown sits in the exponent — so take logs of both sides.

From 200(1.08)t = 500, divide first to get 1.08t = 2.5, then bring t down with the power law.

t = log 2.5log 1.08 ≈ 11.9 years.

Isolate the power first

Take logs only once the exponential term is alone. Taking logs of 200(1.08)t while the 200 is still attached makes unnecessary work.

Reading a log scale

On a log scale, equal distances mean equal multiples, not equal differences.

A magnitude 7 earthquake releases about 32 times the energy of a magnitude 6, not 17% more.

That is why log scales are used for things spanning huge ranges — sound, earthquakes, acidity, star brightness.

Each step multiplies

Going from pH 5 to pH 3 is two steps, so 100 times more acidic. Steps multiply; they never add.

Worked examples

Building the model

A town of 8000 people grows 3% a year. Write the model and find the population after 10 years.

  1. Initial value a = 8000; the factor is 1 + 0.03 = 1.03. The 1 keeps the people already there.
  2. So P = 8000(1.03)t. Standard form abt.
  3. At t = 10: 8000(1.03)10 ≈ 8000(1.344) ≈ 10751. About a third more, not 30% — compounding adds a little extra.

about 10,751

Half-life

A sample has 80 g of an isotope with a half-life of 6 days. How much remains after 24 days?

  1. 24 ÷ 6 = 4 half-lives. Count the halvings rather than the days.
  2. Each halving multiplies by 12, so four gives 116. Not 42 — the halvings compound.
  3. 80 × 116 = 5 g. Halving 80 four times: 40, 20, 10, 5.

5 g

Solving for the time

How long for $1200 at 6% compounded annually to reach $2000?

  1. 1200(1.06)t = 2000, so 1.06t = 20001200 ≈ 1.667. Isolate the power before taking logs.
  2. Take logs: t log 1.06 = log 1.667. The power law brings t down.
  3. t = log 1.667log 1.06 ≈ 8.8 years. Check: doubling at 6% takes about 12 years, so under 9 for a smaller rise is sensible.

about 8.8 years

Words to know

Growth factor
What you multiply by each period. Above 1 is growth, between 0 and 1 is decay.
Half-life
The time for a decaying quantity to fall to half.
Doubling time
The time for a growing quantity to double.
Initial value
The amount at time zero — the a in y = abt.
Continuous growth
Growth compounding at every instant, modelled with A = Pert.

Mistakes to watch for

Using the rate as the growth factor. A 30% decline has factor 0.70, not 0.30, because 70% is what survives each period. Using 0.30 models a 70% decline instead.

Multiplying the half-life instead of counting halvings. Four half-lives leave 116, not 14. Each halving acts on what the last one left, so they compound.

Treating exponential growth as a fixed amount per year. 8% of 8000 is 640, but next year it is 8% of a bigger number. Adding 640 every year is a straight line, and it drifts further off each year.

Taking logs before isolating the power. In 200(1.08)t = 500, divide by 200 first. Taking logs with the 200 attached gives log 200 + tlog 1.08, which is correct but needlessly messy.

Reading a log scale additively. Magnitude 7 is about 32 times magnitude 6, not one unit more of anything. On a log scale each step multiplies.

Carbon dating is exactly this. Carbon-14 has a half-life of 5730 years, so measuring what is left dates the sample.

Compound interest, virus spread, drug clearance and cooling coffee all share one equation, because all four grow or shrink in proportion to what is present.

The number e ≈ 2.718 turns up wherever growth is continuous rather than in steps — it is what "compounding every instant" converges to.

Problem 1

Is the table 4, 8, 16, 32 linear or exponential?

  1. Exponential
  2. Linear
  3. Neither
  4. Both
Worked answer

Exponential

Each is double the last

A constant ratio means exponential

Problem 2

Is the table 4, 9, 14, 19 linear or exponential?

  1. Linear
  2. Exponential
  3. Neither
  4. Both
Worked answer

Linear

Each adds 5

A constant difference means linear

Problem 3

A quantity grows 6% a year. What is the growth factor?

Worked answer

1.06

1+0.06=1.06

Problem 4

A quantity falls 25% a year. What is the decay factor?

Worked answer

0.75

1-0.25=0.75

Problem 5

For y=100(2)t, find y when t=3.

Worked answer

800

23=8

100 × 8=800

Problem 6

A town of 5000 grows 4% a year. Find the population after 2 years, to the nearest whole number.

Worked answer

5408

1.042=1.0816

5000 × 1.0816=5408

Problem 7

A 64 g sample halves every 3 days. How much remains after 12 days?

Worked answer

4

12 ÷ 3 = 4 halvings

64 → 32 → 16 → 8 → 4

Problem 8

A 200 mg dose halves every 6 hours. How much remains after 18 hours?

Worked answer

25

18 ÷ 6 = 3

200 → 100 → 50 → 25

Problem 9

Write the growth factor for a quantity that triples each period.

Worked answer

3

b=3

Problem 10

$500 at 8% compounded annually. Find the value after 3 years, to the nearest dollar.

Worked answer

630

1.083 ≈ 1.2597

500 × 1.2597 ≈ 630

Problem 11

A population falls 30% a year. Which model is right?

  1. y=a(0.7)t
  2. y=a(0.3)t
  3. y=a(1.3)t
  4. y=a(-0.3)t
Worked answer

y=a(0.7)t

70% survives

So the factor is 0.7

Problem 12

Solve 2t = 40 for t, to two decimal places.

Worked answer

5.32

t=log 40log 2

≈ 5.32

Problem 13

How many years for $1000 at 5% to double? Give the answer to one decimal place.

Worked answer

14.2

t=log 2log 1.05

≈ 14.2

Problem 14

A sample of 80 g has half-life 10 years. How many grams remain after 40 years?

Worked answer

5

80 → 40 → 20 → 10 → 5

Problem 15

A magnitude 8 earthquake releases how many times the energy of a magnitude 6?

  1. About 1000
  2. About 2
  3. About 100
  4. About 32
Worked answer

About 1000

Two steps

32 × 32 ≈ 1000

Problem 16

A culture doubles every 4 hours and now has 1600 cells. How many were there 12 hours ago?

Worked answer

200

12 ÷ 4 = 3 doublings

1600 ÷ 8 = 200

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