Exponential growth and decay models — half-life, doubling time: Question bank

Before starting, remind yourself of the one governing rule: a quantity $N$ is exponential when its rate of change $\frac{d N}{d t}$ (how fast it moves at this instant) equals $k N$ — a constant $k$ times the current amount. Solving that (see Solving first-order separable ODEs) gives $N (t) = N_{0} e^{k t}$ , with $N_{0}$ the starting amount at $t = 0$ . Growth means $k > 0$ ; decay means $k < 0$ . Half-life is $T_{1/2} = \frac{l n 2}{∣ k ∣}$ , doubling time is $T_{d} = \frac{l n 2}{k}$ .

True or false — justify

True or false: A radioactive sample twice as large as another has a shorter half-life.

False. Twice the atoms decay twice as fast, so the rate doubles too; the two effects cancel and

T_{1/2} = \frac{l n 2}{∣ k ∣}

contains no

N_{0}

— half-life is fixed for a given substance.

True or false: If a population doubles in 3 years, it will quadruple in 6 years.

True. Two doubling times stack multiplicatively:

\times 2

then

\times 2 = \times 4

. Exponentials multiply by a fixed factor per fixed interval.

True or false: A quantity decaying by 50% each hour reaches zero after 2 hours.

False. Each hour removes half of what remains, giving

\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \dots

— it approaches zero but never arrives. Decay is multiplicative, not additive.

True or false: In

N (t) = N_{0} e^{k t}

you should use

∣ k ∣

so the answer stays positive.

False. Inside

N (t)

the sign of

k

carries the meaning:

k < 0

makes the curve fall. Only the half-life formula uses

∣ k ∣

. Forcing

∣ k ∣

N (t)

turns decay into growth.

True or false: An exponential decay curve eventually crosses the horizontal axis.

False.

N_{0} e^{k t}

with

k < 0

(and

N_{0} > 0

) has the

t

-axis as a horizontal asymptote — it gets arbitrarily close but

e^{k t} > 0

for every real

t

. See Graphs of exponential functions.

True or false: Doubling time and half-life are two different physical mechanisms.

False. They are the same question — "time for a fixed multiplying factor" — asked with factor

2

versus factor

\frac{1}{2}

. Both equal

\frac{l n ( factor )}{k}

True or false: If

N

satisfies

\frac{d N}{d t} = k N

, then plotting

ln N

against

t

gives a straight line.

True (given

N_{0} > 0

N > 0

and

ln N

is defined). Take logs of

N = N_{0} e^{k t}

ln N = ln N_{0} + k t

, a straight line of slope

k

and intercept

ln N_{0}

(uses Laws of logarithms).

True or false: A larger rate constant

∣ k ∣

means a longer half-life.

False.

T_{1/2} = \frac{l n 2}{∣ k ∣}

is inversely proportional to

∣ k ∣

— a faster decay (bigger

∣ k ∣

) halves the amount sooner.

True or false: Discrete compounding

(1 + r)^{t}

can be used for any real time

t

False. The form

(1 + r)^{t}

counts

t

whole compounding periods — each period multiplies by

(1 + r)

. It's literally defined only for integer

t

; between periods nothing changes. To allow any real

t

you must compound continuously and switch to

e^{k t}

(see next question and Compound interest).

Spot the error

"Half-life is 4 years, so after 8 years the sample is completely gone." — find the flaw.

After 8 years = 2 half-lives, so

(\frac{1}{2})^{2} = \frac{1}{4}

remains, not zero. Each half-life halves what's left.

N = N_{0} e^{k t}

, decay with half-life

τ

, so

k = \frac{l n 2}{τ}

." — what's wrong?

The sign. For decay

N

falls, so

k

must be negative:

k = - \frac{l n 2}{τ}

. Dropping the minus makes the model grow.

"Concentration falls from 50 to 30 in 2 h, so it falls 20 more (to 10) in the next 2 h." — spot the error.

That treats decay as linear (constant difference). It's multiplicative: the ratio

\frac{30}{50} = 0.6

recurs, so after 4 h you get

50 \times 0. 6^{2} = 18

, not 10.

"Since

ln \frac{1}{2}

is negative, half-life

T_{1/2} = \frac{l n \frac{1}{2}}{k}

comes out negative — that's impossible." — resolve it.

For decay

k

is also negative, so negative over negative is positive. Or write

T_{1/2} = \frac{l n 2}{∣ k ∣}

and sidestep signs entirely.

"To find how long until 10% remains, solve

0.1 = e^{k t}

and read off

t = ln 0.1

." — what's missing?

You forgot

k

and

N_{0}

. Correctly

\frac{N}{N _{0}} = 0.1 = e^{k t} \Rightarrow t = \frac{l n 0.1}{k}

; dividing by

k

is the whole point.

\frac{d N}{d t} = k N

means the graph of

N

is a straight line because the rate is constant." — find the mistake.

The rate

k N

is not constant — it grows with

N

. A constant rate would give a line; a rate proportional to

N

gives the ever-steepening exponential curve.

"The population grows exponentially, so it will exceed any number — real populations must do the same." — critique.

The model is unbounded, but reality adds limits (food, space). Exponential growth is an early-stage approximation, not an eternal law.

Why questions

Why does the doubling time not depend on

N_{0}

Because

N = N_{0} e^{k t}

hits

2 N_{0}

when

e^{k t} = 2

— the

N_{0}

cancels on both sides, leaving

k t = ln 2

. The condition is about the ratio, never the size.

Why do we separate variables to solve

\frac{d N}{d t} = k N

To get all

N

-terms on one side and all

t

-terms on the other, so each side can be integrated independently — you can't integrate a mixed

N

-and-

t

expression directly.

Why does

e

(rather than

2

10

) appear naturally in

N = N_{0} e^{k t}

Because

\frac{d}{d t} e^{k t} = k e^{k t}

—

e

is the base whose exponential is its own derivative, exactly matching "rate proportional to amount". See Natural logarithm and e.

Why must we take a logarithm to solve for

t

inside an exponent?

The unknown sits up in the exponent of

e^{k t}

. The logarithm is the inverse of exponentiation — it's the only tool that brings that

t

back down to the ground (uses Laws of logarithms).

Why is a constant ratio per equal time-step the fingerprint of exponential behaviour?

Multiplying by the same factor

e^{k Δ t}

over each equal

Δ t

is precisely what

N_{0} e^{k t}

does — linear data instead shows a constant difference, not ratio.

Why can the same formula

\frac{l n ( factor )}{k}

describe both a bank account and a decaying atom?

Both obey

\frac{d N}{d t} = k N

; the mathematics doesn't know the physics. Only the sign and size of

k

differ (Radioactive decay (Physics) vs Compound interest).

How does continuous compounding turn the discrete

(1 + r / n)^{n t}

into

e^{k t}

Compound

n

times per period at rate

r / n

each: the balance is

(1 + \frac{r}{n})^{n t}

. As

n \to \infty

(compounding every instant), this limit equals

e^{r t}

— so the continuous rate constant is

k = r

. That's exactly why

e

shows up in Compound interest. The bare

(1 + r)^{t}

is just the

n = 1

case restricted to integer

t

Edge cases

What does the model predict if

k = 0

N (t) = N_{0} e^{0} = N_{0}

— the amount never changes. Zero rate constant means neither growth nor decay; a flat horizontal line.

What is

N

t = 0

, and why must that be so?

N (0) = N_{0} e^{0} = N_{0}

, the initial amount. The constant of integration is fixed precisely to make this true — it's what makes the solution specific.

What happens to

N (t) = N_{0} e^{k t}

t \to \infty

for decay (

k < 0

N \to 0^{+}

: it decreases toward zero without ever reaching it, hugging the

t

-axis as an asymptote.

What happens as

t \to - \infty

(looking into the past) for a decay model?

N \to + \infty

: run the decay backwards and the amount blows up — physically it means the sample was larger earlier, valid only as far back as the process existed.

N_{0} = 0

, what does the model give and is it meaningful?

N (t) = 0

for all

t

— nothing there means nothing to grow or decay. It's the trivial (degenerate) solution; a valid but uninteresting fixed point. (Note

ln N

is undefined here, which is why we normally require

N_{0} > 0

After how many half-lives does less than 1% remain?

Solve

(\frac{1}{2})^{n} < 0.01

: since

(\frac{1}{2})^{7} \approx 0.0078 < 0.01

, it takes 7 half-lives. (Six give

\approx 1.6%

, still too much.)

Can

k

be complex or must it be real for these models?

For growth/decay

k

is real. A complex exponent would introduce oscillation (rotation), which describes waves, not monotone growth or decay.

Recall One-line self-test

Cover every answer above and re-derive the reason, not just the verdict. If your justification uses the word "ratio", "asymptote", or "sign of $k$ " correctly, you've internalised the topic.

Connections

Exponential growth and decay models — half-life, doubling time (index 3.2.5) — parent note with full derivations.
Natural logarithm and e · Laws of logarithms — the tools these traps hinge on.
Solving first-order separable ODEs — where $N = N_{0} e^{k t}$ comes from.
Graphs of exponential functions — the asymptote and shape questions.
Compound interest · Radioactive decay (Physics) — the two worlds one formula spans.