Exponential growth and decay models — half-life, doubling time: Worked examples

This page is a worked-example gym for the parent topic. The parent taught you the machine:

$N (t) = N_{0} e^{k t}, T_{d} = \frac{l n 2}{k}, T_{1/2} = \frac{l n 2}{∣ k ∣} .$

Here we don't re-derive it — we stress-test it against every kind of question an exam can ask. First we lay out the full "scenario matrix" so you can see the whole battlefield; then we fight each cell.

The scenario matrix

Every cell below is a genuinely different situation your algebra must survive. The right column names the example that covers it.

#	Case class	What's unknown / tricky	Covered by
A	Growth, $k > 0$ given → find $N$ at a time	plug in, then verify with base-2 form	Ex 1
B	Decay, half-life given → find age from a ratio	solve for $t$ inside exponent, sign care	Ex 2
C	Find $k$ from two data points (decay)	$k$ comes out negative	Ex 3
D	Find $k$ from two data points (growth)	$k$ comes out positive; then doubling time	Ex 4
E	Degenerate: $k = 0$	rate is zero → constant, NOT exponential	Ex 5
F	Limiting behaviour $t \to \pm \infty$	decay → $0$ but never reaches it; growth → $\infty$	Ex 6
G	Whole-number half-lives (no logs needed)	use $(1/2)^{n}$ shortcut, spot the fraction	Ex 7
H	Real-world word problem with unit conversion	minutes vs hours; read the ratio correctly	Ex 8
I	Exam twist: mixed (given doubling time, asked a tripling time)	fixed-factor logic beyond ×2	Ex 9

Notice the two "sign" axes: is $k$ positive or negative? and is the unknown $N$ , $t$ , or $k$ ? Between them they generate every cell.

Figure — Exponential growth and decay models — half-life, doubling time

Worked examples

Ex 1 — Case A: growth, find the amount

Worked example Bacteria after a set time

A culture starts at $N_{0} = 800$ and obeys $\frac{d N}{d t} = 0.25 N$ (per hour). Find $N$ after $5$ hours.

Forecast: guess — will it more than double? (Doubling time is $ln 2/0.25 \approx 2.77$ h, so in $5$ h it doubles nearly twice → expect roughly $3$ – $4 \times$ .)

Write the model. $N (t) = 800 e^{0.25 t}$ . Why this step? The defining property $\frac{d N}{d t} = k N$ has the unique solution $N_{0} e^{k t}$ ; here $k = 0.25 > 0$ so it's growth.
Substitute $t = 5$ . $N = 800 e^{0.25 \times 5} = 800 e^{1.25}$ . Why this step? The question asks for the amount at a specific time, so we plug that time in — nothing is unknown except the value we want.
Evaluate. $e^{1.25} \approx 3.4903$ , so $N \approx 800 \times 3.4903 \approx 2792$ cells. Why this step? $e^{1.25}$ is the multiplying factor over $5$ hours.

Verify: base-2 form. $N = 800 \cdot 2^{5/2.7726} = 800 \cdot 2^{1.803} \approx 800 \times 3.490 \approx 2792$ ✓. Matches, and it's between $2 \times$ and $4 \times$ as forecast. Units: cells (dimensionless count) ✓.

Ex 2 — Case B: decay, find the age from a ratio

Worked example Dating a sample

A fossil retains $65%$ of its original Carbon-14. C-14 half-life is $5730$ years. How old is it?

Forecast: it's lost a bit more than a third but nowhere near half, so the age should be less than one half-life ( $< 5730$ yr).

Half-life form. $N = N_{0} 2^{- t /5730}$ . Why this step? We're given the half-life, not $k$ , so the base-2 form skips a conversion.
Insert the ratio. $0.65 = 2^{- t /5730}$ . Why this step? " $65%$ remains" means $N / N_{0} = 0.65$ ; dividing by $N_{0}$ removes it entirely — this is why half-life never depends on how much you started with.
Take logs to free $t$ . $ln 0.65 = - \frac{t}{5730} ln 2$ . Why this step? $t$ is trapped in the exponent; $ln$ is the one tool that brings an exponent down as a multiplier (Laws of logarithms).
Solve. $t = - 5730 \cdot \frac{ln 0.65}{ln 2} = - 5730 \cdot \frac{- 0.4308}{0.6931} \approx 3562$ years. Why this step? Both $ln$ s are known constants; the two minus signs make the age positive.

Verify: $3562 < 5730$ ✓ (less than one half-life, as forecast). Plug back: $2^{- 3562/5730} = 2^{- 0.6217} \approx 0.650$ ✓.

Ex 3 — Case C: find $k$ from data (decay)

Worked example Drug clearance

Blood concentration drops from $50 mg/L$ to $30 mg/L$ in $2$ hours. Find $k$ and the half-life.

Forecast: concentration falls, so $k$ must be negative. It lost $40%$ in $2$ h — a bit less than half — so half-life is a little over $2$ h.

Model with unknown $k$ . $C (t) = 50 e^{k t}$ , and $C (2) = 30$ . Why this step? Here $k$ is the unknown; the initial value $50$ is known so we keep $e^{k t}$ intact.
Isolate the exponential. $e^{2 k} = \frac{30}{50} = 0.6$ . Why this step? Dividing by $50$ leaves only the factor that time produced.
Log both sides. $2 k = ln 0.6$ , so $k = \frac{1}{2} ln 0.6 \approx - 0.2554 h^{- 1}$ . Why this step? $ln$ undoes $e$ . The result is negative — the sign carries the physics (decay), so we keep it inside $N (t)$ .
Half-life. $T_{1/2} = \frac{ln 2}{∣ k ∣} = \frac{0.6931}{0.2554} \approx 2.71$ h. Why this step? Half-life uses $∣ k ∣$ (a duration is positive); only here do we drop the sign.

Verify: $2.71 > 2$ ✓ (more than the $2$ h it took to lose $40%$ ). Check $50 e^{- 0.2554 \times 2} = 50 e^{- 0.5108} \approx 50 \times 0.600 = 30$ ✓. Units: $k$ in per-hour ✓.

Ex 4 — Case D: find $k$ from data (growth), then doubling time

Worked example Investment growing

An account grows continuously from $£2000$ to $£2600$ in $3$ years. Find $k$ and the doubling time.

Forecast: money grows, so $k > 0$ . It rose $30%$ in $3$ yr, so doubling (a $100%$ rise) takes noticeably longer than $3$ yr — roughly $7$ – $8$ yr.

Model. $A (t) = 2000 e^{k t}$ , with $A (3) = 2600$ . Why this step? Continuous growth (Compound interest in its continuous limit) uses $e^{k t}$ ; $k$ unknown.
Isolate. $e^{3 k} = \frac{2600}{2000} = 1.3$ . Why this step? Removes the known $£2000$ , leaving the growth factor.
Log. $3 k = ln 1.3$ , so $k = \frac{1}{3} ln 1.3 \approx 0.08745 yr^{- 1}$ . Why this step? $k$ positive confirms growth.
Doubling time. $T_{d} = \frac{ln 2}{k} = \frac{0.6931}{0.08745} \approx 7.93$ years. Why this step? $T_{d}$ is the "how long to double" special case — growth uses $ln 2$ (positive $k$ ).

Verify: $7.93$ yr sits in the $7$ – $8$ forecast ✓. Check $2000 e^{0.08745 \times 3} = 2000 e^{0.2624} \approx 2000 \times 1.300 = 2600$ ✓.

Ex 5 — Case E: the degenerate case $k = 0$

Worked example When "exponential" collapses to a flat line

Suppose measurements give $\frac{d N}{d t} = 0 \cdot N$ (i.e. $k = 0$ ), with $N_{0} = 140$ . What is $N (t)$ ? What are the doubling time and half-life?

Forecast: if the rate of change is zero, nothing ever changes — so it should be a horizontal line, and "doubling" should be impossible.

Solve the model. $N (t) = 140 e^{0 \cdot t} = 140 e^{0} = 140$ for all $t$ . Why this step? $e^{0} = 1$ , so the multiplying factor is always $1$ — a constant. This is the boundary between growth and decay.
Try doubling time. $T_{d} = \frac{ln 2}{k} = \frac{ln 2}{0}$ — undefined (division by zero). Why this step? The formula screams that doubling never happens: to reach $2 \times$ you'd wait forever, so no finite $T_{d}$ exists.
Try half-life. Same: $\frac{ln 2}{∣0∣}$ undefined — it never halves either. Why this step? A flat quantity neither doubles nor halves, exactly as the algebra warns.

Verify: the graph is a horizontal line at $140$ ; $N (1000) - N (0) = 0$ ✓. This is the one case where the doubling/half-life formulas deliberately have no answer — a healthy sanity check, not an error.

Ex 6 — Case F: limiting behaviour, $t \to \pm \infty$

Worked example Where does the curve end up?

For decay $N (t) = 200 e^{- 0.5 t}$ , describe $N$ as $t \to + \infty$ and as $t \to - \infty$ . Does it ever reach $0$ ?

Forecast: decay should sink toward the horizontal axis but (you may recall the graph) never touch it; going backward in time it should blow up.

As $t \to + \infty$ . $e^{- 0.5 t} \to 0^{+}$ , so $N \to 0^{+}$ — approaches zero from above but is always positive. Why this step? $e^{(large negative)}$ is a tiny positive number; it can be as small as you like but never exactly $0$ . So the $t$ -axis is a horizontal asymptote (Graphs of exponential functions).
Concrete points. $N (10) = 200 e^{- 5} \approx 1.35$ ; $N (20) = 200 e^{- 10} \approx 0.0091$ . Still positive. Why this step? Numbers make "never reaches zero" tangible — this is exactly why "half of what remains" never empties the pile.
As $t \to - \infty$ . $e^{- 0.5 t} \to + \infty$ , so $N \to + \infty$ . Why this step? Running the exponential backwards undoes the shrinking, growing without bound.

Verify: $N (10) = 200 e^{- 5} \approx 1.348 > 0$ ✓ and $N (20) \approx 0.00908 > 0$ ✓ — both strictly positive, confirming zero is never attained.

Ex 7 — Case G: whole-number half-lives (no logs)

Ex 8 — Case H: real-world word problem with unit conversion

Worked example Cooling coffee (rate given per minute, question in hours)

A hot object's excess temperature decays with $k = - 0.03 min^{- 1}$ starting at $60^{\circ} C$ above room. What is the excess temperature after $1.5$ hours? What is the half-life in minutes?

Forecast: $1.5$ h is $90$ min; the half-life will be $ln 2/0.03 \approx 23$ min, so $90$ min is nearly four half-lives → expect roughly $60/16 \approx 4^{\circ}$ C.

Convert units first. $1.5 h = 90 min$ , because $k$ is per minute — mixing units is the classic trap. Why this step? $t$ and $k$ must share the same time unit or $k t$ is meaningless.
Model. $T (t) = 60 e^{- 0.03 t}$ with $t$ in minutes. Why this step? Excess temperature obeys the same $\frac{d N}{d t} = k N$ law (Newton's cooling).
Substitute $t = 90$ . $T = 60 e^{- 0.03 \times 90} = 60 e^{- 2.7} \approx 60 \times 0.06721 \approx 4.03^{\circ} C$ . Why this step? Plug the converted time in.
Half-life. $T_{1/2} = \frac{ln 2}{0.03} \approx 23.1$ min. Why this step? Uses $∣ k ∣$ ; answers "how long to lose half the excess heat."

Verify: $4.03^{\circ}$ C matches the $\approx 4$ forecast ✓. Cross-check via half-lives: $90/23.1 \approx 3.90$ half-lives, $60 \times 2^{- 3.90} \approx 60 \times 0.0671 \approx 4.03$ ✓. Units: temperature in °C, half-life in min ✓.

Ex 9 — Case I: exam twist — a tripling time

Worked example Beyond doubling

A population has doubling time $T_{d} = 5$ years. How long until it triples? (Don't just say $7.5$ years.)

Forecast: tripling is more than doubling, so it takes more than $5$ yr — but less than two doublings ( $10$ yr, which gives $\times 4$ ). Expect $7$ – $9$ yr.

Recover $k$ from the doubling time. $k = \frac{ln 2}{T _{d}} = \frac{ln 2}{5} \approx 0.1386 yr^{- 1}$ . Why this step? We only need the multiplying rate; the doubling time hands it to us.
Set up the tripling condition. $3 N_{0} = N_{0} e^{k T_{3}} \Rightarrow e^{k T_{3}} = 3$ . Why this step? Same "fixed-factor" idea as doubling, but the factor is $3$ not $2$ — this is the general rule "same ratio, same time" generalised to any ratio.
Solve. $T_{3} = \frac{ln 3}{k} = \frac{ln 3}{ln 2/5} = 5 \cdot \frac{ln 3}{ln 2} = 5 \times \frac{1.0986}{0.6931} \approx 7.92$ years. Why this step? The generic fixed-factor time is $ln (factor) / k$ ; tripling just swaps $ln 2$ for $ln 3$ .

Verify: $7.92$ yr lies between $5$ (double) and $10$ (quadruple) ✓ as forecast, and closer to $10$ since $3$ is nearer $4$ than $2$ . Check: $2^{7.92/5} = 2^{1.584} \approx 3.00$ ✓.

Active Recall

Which case has no finite doubling time, and why?

k = 0

— the amount is constant, so

ln 2/ k

divides by zero; it never doubles.

A quantity keeps

70%

after time

t

; is

t

more or less than one half-life?

Less — it hasn't yet lost half.

Why must you convert hours to minutes before using

k

in per-minute?

k t

needs a single consistent time unit or the exponent is meaningless.

Time to triple when doubling time is

T_{d}

T_{3} = T_{d} \cdot \frac{l n 3}{l n 2}

t \to \infty

for decay, what is the limit and is it ever reached?

Tends to

0

from above, never exactly reached (horizontal asymptote).

Fraction remaining after

n

whole half-lives?

(1/2)^{n}

Connections

Exponential growth and decay models — half-life, doubling time (index 3.2.5) — the parent this drills.
Solving first-order separable ODEs — where $N_{0} e^{k t}$ comes from.
Laws of logarithms — used every time we free $t$ from an exponent.
Graphs of exponential functions — the asymptote behaviour in Ex 6.
Compound interest — the continuous-growth setting of Ex 4.
Radioactive decay (Physics) — Ex 2 and Ex 7 in the wild.
Natural logarithm and e — why $ln$ and $e$ pair up in all solving steps.

Worked examples — Exponential growth and decay models — half-life, doubling time

The scenario matrix

Worked examples

Ex 1 — Case A: growth, find the amount

Ex 2 — Case B: decay, find the age from a ratio

Ex 3 — Case C: find $k$ from data (decay)

Ex 4 — Case D: find $k$ from data (growth), then doubling time

Ex 5 — Case E: the degenerate case $k = 0$

Ex 6 — Case F: limiting behaviour, $t \to \pm \infty$

Ex 7 — Case G: whole-number half-lives (no logs)

Ex 8 — Case H: real-world word problem with unit conversion

Ex 9 — Case I: exam twist — a tripling time

Active Recall

Connections

Case Map