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Exponential Growth And Decay

Exponential Growth And Decay

Finding the number of years (or days)

Finding The Number Of Years Extended 6 min read

On this page
  1. In this note
  2. Formula bank
  3. Searching for n
  4. Ways they test it
  5. Mistakes to avoid
  6. Summary
  7. Practice
  8. Can I do this?
  9. Quiz

In this note

Syllabus 1.17: exponential change - finding the time
  • 1.17Find the number of complete years (hours, days ...) for a quantity to reach, pass or fall below a value.
  • 1.17Use trial and improvement with the multiplier, showing the values either side of the target.
  • 1.17Find doubling and halving times.

Before you start: exponential growth and decay. No logarithms are needed.

Formula bank

The condition to testLearn it\(\text{start}\times m^{n}\ \ge\ \text{target}\) (or \(<\) for decay)try whole numbers n.Use it when "how many complete years …".
Doubling timeLearn itsmallest n with \(m^{n}\ge2\)the starting amount does not matter.Use it when "double".
Halving timeLearn itsmallest n with \(m^{n}\le0.5\)Use it when "half".

Searching for n

To solve \(3.2\times1.028^n=4\) exactly you'd need logarithms, which are not in this syllabus. Instead try values of n and watch the value pass the target. The answer is the first whole number of periods that does the job.

General methodHow to find the number of complete periods

  1. Write the model: start × multiplierⁿ.
  2. Estimate n: total % change needed ÷ rate (a first guess - growth gets there a bit sooner).
  3. Work out the value for your guess; move n up or down.
  4. Write the LAST value before the target and the FIRST value after it - both earn marks.
  5. Answer: the first n that meets the condition (reach / pass / first below).

Example A population of 3.2 million grows at 2.8% per year. After how many complete years will it first reach 4 million?

Estimate 25% ÷ 2.8% ≈ 9, then test either side.

  1. \(n=8\): \(3.2\times1.028^{8}=3.991\) (not yet)
  2. \(n=9\): \(4.103\) ✓

Answer 9 years

Example How many complete years does it take for an amount to double at 6% per year?

Test \(1.06^{n}\) against 2.

  1. \(1.06^{11}=1.898\)
  2. \(1.06^{12}=2.012\) ✓

Answer 12 years

Write the values for BOTH trial years. 'n = 9' alone may score only the final mark.

Try it yourself

1
100 grows 10% a year. First whole year it is more than 200?
Hint
1.1⁷ = 1.95, 1.1⁸ = 2.14.
Show answer

Answer 8

2
Decay 20% per hour. First whole hour below half?
Show answer

Answer 4

3
50 g decays at 12% per day. First whole day below 5 g?
Show answer

Answer 19

Ways they test it

Type 1 Complete years to reach a value

"Calculate the number of complete years it takes to reach 8 million." (3 marks)

Spot it "complete years", a target value.

  1. A trial with start × mⁿ (M1).
  2. Values either side (M1).
  3. n (A1).

For example 5.4 million, +3.6%, target 8 million

  1. \(n=11\): 7.968
  2. \(n=12\): 8.255

Answer 12 years

Type 2 First below a value

"Find the number of whole days until the mass is first less than 1 g."

Spot it decay and "first less than".

  1. Same search with < .

Type 3 Doubling or halving

"After how many complete years is the value first half or less?"

Spot it no starting value.

  1. Test mⁿ against 2 or 0.5.

Mistakes to avoid

  • ✗ 11 years (value 7.97 million)✓ 12 yearsIt must REACH 8.
  • ✗ Only the final n written✓ show both valuesThose are the method marks.
  • ✗ 5% for 10 years = half✓ 14 yearsExponential, not linear.

Summary

  • No logs: try whole numbers of periods.
  • Show the value before and after the target.
  • Answer with the first n that meets the condition.
  • Doubling / halving: compare multiplierⁿ with 2 or 0.5.

If you see …

  • "complete years"a whole number - the first one that works.
  • "double" / "half"the start value cancels.

Practice

Practice: easy, medium, hard

Easy

1
Work out \(3.2\times1.028^{2}\) to 3 d.p.
Show answer

Answer 3.382

2
$6000 grows 5% a year. First whole year it is above $8000?
Show working
  1. 5 years: 7657.69
  2. 6 years: 8040.57

Answer 6

3
Decay 20% per hour. Is 0.8³ above or below 0.5? Give 0.8³.
Show answer

Answer 0.512

Medium

1
A town of 36 000 people decreases by 3% a year. After how many complete years is it first below 30 000?
Guide me

What is it asking? n

What do we know? \(36\,000\times0.97^{n}<30\,000\)

Plan try 5 and 6

Show working
  1. 5 years: 30 914
  2. 6 years: 29 987

Answer 6

2
20 g of a substance decays at 10% per day. Whole days until it is first less than 1 g?
Guide me

What is it asking? n

What do we know? \(20\times0.9^{n}<1\)

Plan try near 28

Show working
  1. 28 days: 1.047 g
  2. 29 days: 0.942 g

Answer 29

3
A machine's value decreases exponentially at 5% per year. Complete years until it is first half or less of its original value?
Guide me

What is it asking? n

What do we know? \(0.95^{n}\le0.5\)

Plan try 13 and 14

Show working
  1. \(0.95^{13}=0.513\)
  2. \(0.95^{14}=0.488\)

Answer 14

Hard

1
A population of 5.4 million increases exponentially at 3.6% per year. Complete years until it reaches 8 million?
Guide me

What is it asking? n

What do we know? \(5.4\times1.036^{n}\ge8\)

Plan try 11 and 12

Show working
  1. 11 years: 7.968
  2. 12 years: 8.255

Answer 12

PartAnswer / workingMarksGuidance
trial seenM1
7.97 and 8.26M1
12A1
Total: 3 marks
2
A town of 54 000 decreases exponentially at 2% per year. Complete years until it is first below 45 000?
Guide me

What is it asking? n

What do we know? \(54\,000\times0.98^{n}<45\,000\)

Plan try 9 and 10

Show working
  1. 9 years: 45 022
  2. 10 years: 44 122

Answer 10

PartAnswer / workingMarksGuidance
\(54\,000\times0.98^n\) trialM1
45 022 and 44 122M1
10A1
Total: 3 marks
3
An investment grows at 6% a year. Complete years until it has doubled?
Guide me

What is it asking? n

What do we know? \(1.06^{n}\ge2\)

Plan start cancels

Show working
  1. \(1.06^{11}=1.898\)
  2. \(1.06^{12}=2.012\)

Answer 12

PartAnswer / workingMarksGuidance
both valuesM1
12A1
Total: 2 marks

Can I do this?

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Quiz

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