🌱 Growth and Decay

OCR FSMQ Additional Maths · Exponentials and Logarithms (EL7)

Level 3 · Ages 15–16

← Back to topic overview
1 The Model
EL7: use exponentials and logarithms in problems involving exponential growth and decay. This is where the whole topic pays off — the model describes savings, populations, radioactivity, depreciation and cooling.
The standard model
$A = A_0 \times r^t$
$A_0$ = the initial amount  ·  $r$ = the multiplier per period  ·  $t$ = number of periods
Getting $r$ right
Increase of $p\%$:  $r = 1 + \dfrac{p}{100}$
Decrease of $p\%$:  $r = 1 - \dfrac{p}{100}$
In wordsMultiplier $r$Growth or decay?
Increases by $6\%$$1.06$Growth
Increases by $50\%$$1.5$Growth
Decreases by $15\%$$0.85$Decay
Loses $3\%$ a year$0.97$Decay
Halves each period$0.5$Decay
Triples each period$3$Growth
The multiplier is not the rate. A $6\%$ increase means multiplying by $1.06$, not by $0.06$. And a $15\%$ decrease means multiplying by $0.85$, not subtracting $0.15$ each time.
Why repeated percentage change is exponential. Applying the same multiplier $t$ times gives $r^t$, which puts $t$ in the index. That is what makes compound interest fundamentally different from simple interest.
2 Finding the Amount
Worked Example 1 — Compound interest

£$8000$ is invested at $3.5\%$ compound interest per year. Find its value after $7$ years.

①$r = 1.035$, $A_0 = 8000$, $t = 7$
②$A = 8000 \times 1.035^7$
③$1.035^7 = 1.272279$
④$A = £10\,178.23$ (to the nearest penny)
⑤Compare simple interest: $8000 + 7(0.035)(8000) = £9960$ — compounding is worth about £$218$ more over seven years.
Worked Example 2 — Depreciation

A car worth £$18\,000$ loses $18\%$ of its value each year. Find its value after $4$ years.

①Losing $18\%$ leaves $82\%$, so $r = 0.82$.
②$A = 18\,000 \times 0.82^4 = 18\,000 \times 0.45212$
③$A = £8138.19$ (to the nearest penny)
④The car has lost about $55\%$ of its value in four years, even though it loses only $18\%$ each year.
Percentage losses do not add up. Four years of $18\%$ loss is not $72\%$ — each loss applies to a smaller amount, so the total loss is less than the naive sum.
3 Finding the Time
Solving for $t$
$A_0 r^t = A \;\Longrightarrow\; r^t = \dfrac{A}{A_0} \;\Longrightarrow\; t = \dfrac{\log\left(A/A_0\right)}{\log r}$
Worked Example 3 — Reaching a target

£$5000$ is invested at $4.5\%$ per year. After how many complete years does it first exceed £$8000$?

①$5000 \times 1.045^t > 8000$
②$1.045^t > 1.6$
③$t\log 1.045 > \log 1.6$
④$t > \dfrac{0.204120}{0.019116} = 10.678$
⑤So $t = 11$ complete years — rounding up, since after $10$ years the target has not yet been reached.
⑥Check: $5000 \times 1.045^{10} = £7764$ (under) and $5000 \times 1.045^{11} = £8113$ (over) ✓
Round up for a growth target, down for a decay target. If you want a value to exceed something, you need the next whole period. Always verify with the two neighbouring integers.
Worked Example 4 — Half-life

A radioactive substance decays at $7\%$ per day. Find its half-life.

①$r = 0.93$. Half-life means the amount falls to half, whatever the starting amount.
②$0.93^t = 0.5$
③$t\log 0.93 = \log 0.5$
④$t = \dfrac{-0.301030}{-0.031517} = 9.551$ days
⑤The half-life is $9.55$ days (3 s.f.).
The half-life is independent of the starting amount. Dividing by $A_0$ removes it from the equation entirely, which is why half-life is a property of the substance alone.
Worked Example 5 — Finding the rate

A population grows from $4000$ to $9000$ in $6$ years. Find the annual percentage growth rate.

①$4000 r^6 = 9000$, so $r^6 = 2.25$.
②$r = 2.25^{1/6} = \sqrt[6]{2.25}$
③$r = 1.14471$
④So the growth rate is $14.5\%$ per year (3 s.f.).
⑤Check: $4000 \times 1.14471^6 = 9000$ ✓
Take the root, do not divide. The population grew by $125\%$ over six years, but the annual rate is not $125 \div 6 = 20.8\%$. Compounding means the annual rate is much lower.
4 Models with a Limiting Value
The cooling model
$T = T_{\text{room}} + \left(T_0 - T_{\text{room}}\right) r^t$
It is the excess temperature above the surroundings that decays exponentially.
Worked Example 6 — A cooling drink

A drink at $80$ °C is left in a room at $20$ °C. After $10$ minutes it is $56$ °C. Find its temperature after $30$ minutes.

①Initial excess: $80 - 20 = 60$ °C. Excess after $10$ min: $56 - 20 = 36$ °C.
②$60 r^{10} = 36$, so $r^{10} = 0.6$.
③$r = 0.6^{1/10} = 0.95020$
④After $30$ minutes the excess is $60 \times r^{30} = 60 \times \left(r^{10}\right)^3 = 60 \times 0.6^3 = 60 \times 0.216 = 12.96$ °C.
⑤$T = 20 + 12.96 = 32.96$, so about $33.0$ °C.
⑥Note the neat shortcut: the excess falls to $0.6$ of its value every $10$ minutes, so after $30$ minutes it is $0.6^3$ of the original.
Never apply the exponential to the temperature itself. A drink cooling in a $20$ °C room approaches $20$ °C, not $0$ °C. Subtract the room temperature first, model the excess, then add it back.
5 Quick Reference

The model

$A = A_0 r^t$.

Growth

$r = 1 + \dfrac{p}{100} > 1$.

Decay

$r = 1 - \dfrac{p}{100} < 1$.

Finding $t$

$t = \dfrac{\log(A/A_0)}{\log r}$.

Whole periods

Round up for growth targets; check both integers.

Finding $r$

Take the $t$th root, do not divide.

Half-life

Solve $r^t = 0.5$; independent of $A_0$.

Doubling time

Solve $r^t = 2$; also independent of $A_0$.

Cooling

Model the excess above the surroundings.

Interpret

State units, and comment on the model's limits.

6 Practice Questions
Question 1

Write down the multiplier for (a) a $7\%$ increase, (b) a $22\%$ decrease.

▶ Show solution

(a) $1.07$

(b) $0.78$

Question 2

£$3000$ is invested at $5\%$ per year. Find its value after $8$ years.

▶ Show solution

$3000 \times 1.05^8 = 3000 \times 1.477455$

$= £4432.37$ (to the nearest penny)

Question 3

A machine worth £$25\,000$ depreciates at $12\%$ per year. Find its value after $5$ years.

▶ Show solution

$r = 0.88$

$25\,000 \times 0.88^5 = 25\,000 \times 0.527732 = £13\,193.30$

Question 4

A population of $1200$ grows at $3\%$ per year. How long until it reaches $2000$?

▶ Show solution

$1.03^t = \dfrac{2000}{1200} = 1.66667$

$t = \dfrac{\log 1.66667}{\log 1.03} = \dfrac{0.221849}{0.012837} = 17.3$ years (3 s.f.)

Question 5

A substance decays at $10\%$ per hour. Find its half-life.

▶ Show solution

$0.9^t = 0.5$

$t = \dfrac{\log 0.5}{\log 0.9} = \dfrac{-0.301030}{-0.045757} = 6.58$ hours (3 s.f.)

Question 6

£$4000$ is invested at $6\%$ per year. After how many complete years does it first exceed £$6000$?

▶ Show solution

$1.06^t > 1.5$

$t > \dfrac{0.176091}{0.025306} = 6.959$

So $t = 7$ complete years.

Check: $4000 \times 1.06^6 = £5674$ (under) and $4000 \times 1.06^7 = £6015$ (over) ✓

Question 7

A town's population grew from $20\,000$ to $28\,000$ in $5$ years. Find the annual growth rate.

▶ Show solution

$r^5 = \dfrac{28\,000}{20\,000} = 1.4$

$r = 1.4^{1/5} = 1.06961$

The growth rate is $6.96\%$ per year (3 s.f.).

Question 8

A quantity halves every $4$ days. Find its daily percentage decay rate.

▶ Show solution

$r^4 = 0.5$, so $r = 0.5^{1/4} = 0.840896$.

The daily multiplier is $0.8409$, so the quantity loses $1 - 0.8409 = 0.1591$, i.e. about $15.9\%$ per day.

Question 9

Explain why a quantity decaying at $50\%$ per year does not disappear after two years.

▶ Show solution

Each year the quantity is multiplied by $0.5$, so it halves what remains rather than removing a fixed amount.

After one year $50\%$ is left; after two years $25\%$; after three, $12.5\%$.

Since $0.5^t$ is positive for every $t$, the quantity approaches zero but never reaches it. The percentage losses do not add to $100\%$ because each applies to a progressively smaller base.

Question 10

A cup of coffee at $90$ °C is left in a room at $22$ °C. After $5$ minutes its temperature is $73$ °C.

(a) Find the multiplier $r$ for the excess temperature, per minute.   (b) Find the temperature after $20$ minutes.   (c) Find when the coffee reaches $40$ °C, to the nearest minute.   (d) Explain why the model predicts the coffee never reaches room temperature, and whether this is a reasonable feature.

▶ Show solution

(a) Initial excess: $90 - 22 = 68$ °C. Excess after $5$ min: $73 - 22 = 51$ °C.

$68 r^5 = 51 \;\Rightarrow\; r^5 = \dfrac{51}{68} = 0.75$

$r = 0.75^{1/5} = 0.944088$, so $r = 0.9441$ (4 d.p.).

(b) Excess after $20$ min $= 68 \times r^{20} = 68 \times \left(r^5\right)^4 = 68 \times 0.75^4$

$= 68 \times 0.316406 = 21.52$ °C

$T = 22 + 21.52 = \mathbf{43.5}$ °C (3 s.f.)

(c) At $40$ °C the excess is $40 - 22 = 18$ °C:

$68 r^t = 18 \;\Rightarrow\; r^t = \dfrac{18}{68} = 0.264706$

$t = \dfrac{\log 0.264706}{\log 0.944088} = \dfrac{-0.577005}{-0.024988} = 23.1$ minutes

So about $\mathbf{23}$ minutes.

Check: $22 + 68(0.944088)^{23} = 22 + 18.1 = 40.1$ °C ✓

(d) The excess is $68 r^t$, and since $r > 0$ this is positive for every $t$, however large. So the model gives a temperature always slightly above $22$ °C, approaching it asymptotically but never equalling it.

This is actually a reasonable feature of the model, and it matches the physics: heat flows because of a temperature difference, so as the difference shrinks the cooling slows, and the approach to room temperature genuinely does take longer and longer.

In practice the coffee becomes indistinguishable from room temperature once the excess falls below the accuracy of any thermometer — here the excess falls below $0.5$ °C after about $85$ minutes. The asymptote is a mathematical idealisation of a real and observable slowing, not a flaw.

Growth and Decay (EL7) · OCR FSMQ Additional Maths · Created with MathJax