A quantity becomes half as large after four hours. Four hours later, it halves again. How much remains after twelve hours? How long does it take to fall below one-tenth of its starting value?
Those questions illustrate a mathematical idea known as half-life.
The underlying relationship is an exponential decay model, and it connects directly with indices, logarithms, equations, graphs and rates of change.
This is one reason to study Additional Mathematics: students learn to recognise the difference between a constant amount lost and a constant fraction lost.
That distinction matters because the two processes produce very different long-term patterns. A constant subtraction eventually reaches zero after a fixed number of steps. Repeated multiplication by a positive fraction produces an exponential curve that approaches zero without reaching it at a finite time in the ideal continuous model.
Learning to work with both patterns builds mathematical modelling skill, although any application to a real physical or biological process needs suitable evidence and assumptions.
First, define half-life mathematically
In a simple decay model, the half-life is the amount of elapsed time required for the modelled quantity to become half its current value.
Suppose a quantity begins at 160 units and has a half-life of four hours.
After four hours, it is 80 units.
After eight hours, it is 40 units.
After twelve hours, it is 20 units.
After sixteen hours, it is 10 units.
The quantity lost in each four-hour period is not the same.
It decreases by 80, then 40, then 20, then 10 units.
What remains constant is the proportion lost: one-half of the value at the beginning of each four-hour interval.
This is the central feature of exponential decay.
Why the subtraction model gives a different answer
Imagine instead that the quantity loses 80 units every four hours.
The modelled values would be:
160 initially.
80 after four hours.
0 after eight hours.
-80 after twelve hours, if the linear formula were extended without a physical boundary.
This is not the same as halving the current quantity every four hours.
A fixed loss of 80 and a fifty-percent loss are equal only during the first interval in this example.
Afterward they diverge.
This distinction is a common source of misunderstanding when a word problem uses phrases such as “decreases by half” or “decreases by 50 percent”.
Additional Mathematics helps students translate those descriptions into the correct mathematical operation before calculating.
Build a discrete half-life table
Let n count the number of completed four-hour intervals.
Set the initial amount:
A₀ = 160.
The recurring rule is:
A_(n+1) = A_n/2.
Then:
A₁ = 80.
A₂ = 40.
A₃ = 20.
A₄ = 10.
The general expression is:
A_n = 160(1/2)^n,
where n is a non-negative integer.
This is a geometric sequence.
Each successive term is obtained by multiplication by one-half.
A student who understands index laws can calculate a value after any whole number of completed half-lives without listing every intermediate amount.
Translate completed intervals into elapsed time
If each half-life lasts four hours, then n completed intervals correspond to elapsed time t = 4n hours.
For a smooth exponential model, we can write:
A(t) = 160(1/2)^(t/4),
where t is measured in hours.
When t = 4, the exponent is 1.
When t = 8, it is 2.
When t = 12, it is 3.
The formula reproduces the half-life table at those times.
It also defines values between them, such as t = 2. This smooth interpolation is an additional modelling assumption if the original description concerned only discrete four-hour updates.
In many idealised decay models, continuous exponential behaviour is explicitly assumed.
Students should read the question to determine which interpretation is intended.
A general half-life formula
Let A₀ be a positive initial quantity.
Let H be a positive half-life measured in a specified time unit.
Let t be elapsed time measured in the same unit.
The standard ideal exponential decay model is:
A(t) = A₀(1/2)^(t/H).
The ratio t/H counts how many half-lives have elapsed mathematically.
If t = H, then:
A(H) = A₀/2.
If t = 2H, then:
A(2H) = A₀/4.
If t = 3H, then:
A(3H) = A₀/8.
The formula works for real t in the continuous model because the positive base one-half can be raised to any real exponent.
It allows us to study whole and fractional numbers of half-lives consistently.
A half-life is not necessarily the time to reach zero
Consider:
A(t) = A₀(1/2)^(t/H).
Assume A₀ > 0 and H > 0.
For every finite real t, the exponential factor is positive.
Therefore:
A(t) > 0.
As t increases without bound, the factor approaches zero.
But it never becomes exactly zero for finite t.
So the continuous mathematical model does not predict an ordinary finite time at which the quantity becomes precisely zero.
This is different from a linear model with a fixed rate of subtraction.
In real applications, instruments may have detection limits, quantities may be discrete, and a model may cease to be useful at very small scales.
The mathematical limit and an observed practical threshold should not be confused.
A worked threshold question
Suppose the initial quantity is 160 units and the half-life is four hours.
At what time does the continuous model reach 10 units?
Write:
160(1/2)^(t/4) = 10.
Divide by 160:
(1/2)^(t/4) = 1/16.
Recognise:
1/16 = (1/2)^4.
The bases are equal and positive, so their exponents are equal:
t/4 = 4.
Therefore:
t = 16 hours.
Check the half-life table.
After four half-lives, 160 becomes 80, 40, 20 and then 10.
Four periods of four hours give sixteen hours.
The algebra and intuitive repeated-halving process agree.
Use logarithms when the threshold is not a neat power
Now ask when the same quantity reaches 12 units.
The equation is:
160(1/2)^(t/4) = 12.
Divide:
(1/2)^(t/4) = 12/160 = 0.075.
Take natural logarithms:
(t/4) ln(1/2) = ln(0.075).
Solve:
t = 4 ln(0.075)/ln(1/2).
Numerically, this is approximately 14.95 hours.
The result lies between twelve and sixteen hours, which is reasonable.
At twelve hours, 20 units remain.
At sixteen hours, only 10 remain.
A threshold of 12 must occur between those times in a continuously decreasing model.
This rough bound helps verify the logarithmic calculation.
Why the signs in the logarithmic calculation matter
Both 0.075 and 1/2 lie between zero and one.
Their natural logarithms are negative.
Therefore:
ln(0.075) < 0,
and:
ln(1/2) < 0.
The ratio of these two negative values is positive.
So the elapsed time is positive, as expected.
A student who obtains a negative time may have omitted a negative sign or reversed a logarithmic ratio.
The sign check is useful because the logarithm laws are correct only when applied to the right quantities.
In a model beginning at 160 and decaying towards 12, we expect a positive elapsed time.
Mathematical interpretation provides an independent check on the algebra.
What does a fractional half-life mean?
Suppose the continuous model has a half-life of four hours.
What remains after two hours, or half a half-life?
Use:
A(2) = A₀(1/2)^(2/4).
Therefore:
A(2) = A₀(1/2)^(1/2).
The factor is:
1/√2,
approximately 0.7071.
So about 70.71 percent of the starting quantity remains at two hours under the smooth exponential model.
It is not 75 percent.
That latter number would come from drawing a straight line between 100 percent at time zero and 50 percent at four hours.
Exponential interpolation and linear interpolation give different intermediate results.
This is a useful connection between fractional indices and modelling assumptions.
Compare the two-hour values exactly
For an initial amount of 160:
The continuous exponential model gives:
A(2) = 160/√2 = 80√2,
approximately 113.14 units.
A linear interpolation between 160 and 80 over the first four hours gives:
160 – (80/4)(2) = 120 units.
A step-change model that changes only at four-hour boundaries would remain at 160 before the first update.
All three models agree on the starting value and may agree at the four-hour mark.
They differ between those observations because they describe different mechanisms of change.
Students should not treat one interpolation as an inevitable consequence of the word “half-life” without clarifying what the problem assumes.
Connect half-life with the natural exponential function
The model:
A(t) = A₀(1/2)^(t/H)
can also be written using e.
Recall:
a^x = e^(x ln a)
for positive a.
Therefore:
(1/2)^(t/H) = e^[(t/H)ln(1/2)].
Since ln(1/2) = -ln 2:
A(t) = A₀e^[-(ln 2)t/H].
Define:
k = ln 2/H.
Then:
A(t) = A₀e^(-kt),
where k is positive.
The two formulas describe exactly the same continuous function.
One emphasises half-life.
The other emphasises a proportional decay-rate coefficient.
Why the decay constant depends on the time unit
If H is measured in hours, k = ln 2/H is measured in inverse hours.
If H is measured in days, k has units of inverse days.
The exponential argument kt must be dimensionless.
Therefore using a value of k per hour with a time t measured in days would require conversion.
A student should not silently switch between units.
For example, a half-life of 24 hours is the same physical interval as one day.
The numerical decay constant expressed per hour differs from the numerical decay constant expressed per day, even though both models describe the same curve.
This shows why parameter values have meanings, not just numerical magnitudes.
Differentiate the half-life model
Take:
A(t) = A₀e^(-kt),
where k > 0.
Differentiate:
A'(t) = -kA₀e^(-kt).
Therefore:
A'(t) = -kA(t).
This says the instantaneous rate of decrease is proportional to the current amount.
The negative sign indicates that the quantity decreases as time increases.
The larger the current amount, the larger the magnitude of its instantaneous decrease for the same k.
As the amount becomes smaller, the absolute rate of loss also becomes smaller.
That explains the shape of the exponential decay graph.
It falls relatively quickly at first in absolute terms and then flattens.
Rate at the beginning and after one half-life
At time zero:
A(0) = A₀.
So:
A'(0) = -kA₀.
At time H:
A(H) = A₀/2.
Therefore:
A'(H) = -kA₀/2.
The magnitude of the instantaneous decay rate has halved after one half-life.
The proportional coefficient k has remained constant.
This is a powerful distinction.
A constant proportional decay rate does not mean a constant number of units is lost each hour.
The absolute loss rate changes because it is calculated from the current remaining amount.
Students who understand this are less likely to confuse exponential and linear models.
Find the decay constant from a measured half-life
Suppose an idealised process has a half-life of six minutes.
Then:
k = ln 2/6,
with k measured per minute.
Numerically:
k ≈ 0.1155 per minute.
The continuous model can be written:
A(t) = A₀e^[-(ln 2)t/6].
At t = 6:
A(6) = A₀e^(-ln 2).
Since e^(-ln 2) = 1/2:
A(6) = A₀/2.
The formula satisfies the half-life definition.
The check is useful because an accidental sign error would instead produce exponential growth.
That would contradict the intended decay model.
Recover an unknown half-life from observations
Suppose an idealised quantity begins at 240 units and after ten hours it is 60 units.
Assume it follows a constant-half-life exponential model.
Then:
60 = 240(1/2)^(10/H).
Divide:
1/4 = (1/2)^(10/H).
Recognise:
1/4 = (1/2)².
Therefore:
10/H = 2.
So:
H = 5 hours.
The quantity experienced two half-lives in ten hours.
This interpretation is simple because the observed ratio is a familiar power of one-half.
The general logarithmic method can handle less convenient ratios.
Use two observations without knowing the initial amount
Suppose the model is:
A(t) = Ce^(-kt),
where C is positive but unknown.
At time t₁ the observed value is A₁.
At time t₂ the observed value is A₂, with t₂ > t₁ and both positive.
Divide the two equations:
A₂/A₁ = e^[-k(t₂ – t₁)].
The unknown initial scale C cancels.
Take natural logarithms:
ln(A₂/A₁) = -k(t₂ – t₁).
Therefore:
k = ln(A₁/A₂)/(t₂ – t₁).
Once k is known, half-life is:
H = ln 2/k.
This is a useful algebraic relationship for a simplified model.
It requires the assumption that one constant exponential decay law describes the interval.
Numerical observations give a check
Suppose at t = 2 hours a quantity is 90 units and at t = 6 hours it is 45 units.
The time difference is four hours.
The ratio is:
45/90 = 1/2.
Therefore under the constant-half-life model, H = 4 hours.
We can also use the logarithm formula:
k = ln(90/45)/(6 – 2).
This becomes:
k = ln 2/4.
Then:
H = ln 2/k = 4.
Both approaches agree.
Notice that neither calculation required the original value at t = 0.
The ratio of two positive observations was sufficient because the scale factor cancelled.
That is one reason exponential models are algebraically convenient.
Half-life and doubling time are related but distinct
A growing quantity may double after a fixed interval.
A decaying quantity may halve after a fixed interval.
Both can be described using exponential models.
For growth:
G(t) = G₀2^(t/D),
where D is a positive doubling time.
For decay:
A(t) = A₀(1/2)^(t/H),
where H is a positive half-life.
The signs of their logarithmic growth coefficients differ.
The growth model has derivative greater than zero.
The decay model has derivative less than zero.
These relationships help students organise exponentials as one family while recognising how the base controls the direction of change.
The graph of exponential decay
Consider:
f(t) = 100(1/2)^(t/4),
for t ≥ 0.
At t = 0, the graph begins at 100.
At t = 4, it is 50.
At t = 8, it is 25.
At t = 12, it is 12.5.
The graph stays above the horizontal axis.
It decreases continuously and flattens as time increases.
The horizontal line y = 0 is an asymptote as t tends towards positive infinity.
The graph does not cross into negative outputs in this model.
That is useful for checking numerical results.
A negative amount or a finite-time zero would not be a correct consequence of this strictly positive exponential expression.
An applied interpretation needs limitations
Half-life models can be used in scientific contexts, including certain simplified radioactive-decay models and other processes with approximately proportional decrease.
But real applications may involve multiple interacting processes, replenishment, measurement limits or changing decay behaviour.
The mathematical formula should not be treated as universal for every chemical, biological or environmental situation.
In particular, this educational model is not a basis for making medication, exposure or safety decisions.
Those require appropriate professional standards and evidence beyond a secondary-school exponential example.
Additional Mathematics teaches the structure of the model.
Responsible application requires recognising whether the assumptions fit the actual process.
What if decay does not follow one half-life?
Suppose an observed process falls from 100 to 50 units during its first four hours but from 50 to 40 during its next four hours.
A constant-four-hour-half-life model would have predicted 25 after eight hours, not 40.
The discrepancy indicates that the simple model does not match those observations exactly.
This does not prove which alternative relationship is correct.
It does suggest that more evidence or a different model may be needed.
A student should not force every decreasing data set into one exponential formula simply because one early interval appeared to show halving.
Model checking is part of mathematical reasoning.
The equation’s elegance does not guarantee empirical suitability.
A fixed removal can stop at zero
Consider a quantity initially 100 units that loses 10 units at the end of every hour.
A discrete linear model is:
L_n = 100 – 10n,
for completed periods n within the meaningful range.
It reaches zero after ten hours.
Extending the formula beyond that point would give negative values, which may not describe a physical stock count.
Compare this with:
E_n = 100(0.9)^n.
This model loses ten percent of its current amount each period.
After ten periods it remains positive.
The two models may agree after one hour but diverge afterward.
The difference lies in the description of change, not in arithmetic technique.
Why a percentage threshold needs a careful question
Suppose a half-life model asks:
“When will less than one-eighth of the original amount remain?”
At exactly three half-lives:
(1/2)³ = 1/8.
That is equal to one-eighth, not strictly less.
Therefore the mathematical condition “less than” is satisfied for t > 3H in a continuous decay model.
If the question asks for the first completed half-life period with less than one-eighth remaining, that is the fourth completed half-life.
The difference between ≤ and < affects the endpoint.
A student who simply writes three half-lives may have answered “at most one-eighth” rather than “less than one-eighth”.
Inequality symbols are part of the meaning of the problem.
A full inequality example
Let:
A(t) = 200(1/2)^(t/5),
with t ≥ 0 in hours.
Find when A(t) ≤ 25.
Write:
200(1/2)^(t/5) ≤ 25.
Divide by the positive constant 200:
(1/2)^(t/5) ≤ 1/8.
Recognise:
1/8 = (1/2)³.
Because the function (1/2)^u decreases as u increases, the inequality requires:
t/5 ≥ 3.
Therefore:
t ≥ 15 hours.
The equality endpoint is included because the original comparison was “at most.”
A numerical check gives A(15) = 25, confirming the threshold.
Why the inequality direction is easy to reverse incorrectly
For a base between zero and one, larger exponents give smaller positive values.
For example:
(1/2)² = 1/4.
(1/2)³ = 1/8.
The exponent 3 is larger, but the output 1/8 is smaller.
Therefore when comparing two powers with base one-half, the ordering of their exponents runs opposite to the ordering of their positive outputs.
A student who treats (1/2)^u as an increasing function may choose the wrong time interval.
The unit-circle and quadratic topics are not needed here; what matters is understanding exponential monotonicity.
This is another example of Additional Mathematics teaching rules through the structure of the function.
A calculator check using logarithms
For a threshold with an awkward ratio, use:
A(t) = A₀(1/2)^(t/H).
Set A(t) = T, where 0 < T < A₀.
Then:
T/A₀ = (1/2)^(t/H).
Take logarithms:
t = H ln(T/A₀)/ln(1/2).
Both logarithms are negative, so t is positive.
Substitute this value into the original exponential function to check the result.
It is also useful to bound the time between nearby whole half-lives whenever possible.
For example, a threshold between one-quarter and one-eighth of the original amount occurs between two and three half-lives.
This rough estimate can reveal a misplaced logarithm or sign.
What is the difference between an exact answer and an estimate?
An expression such as:
t = 4 ln(0.075)/ln(1/2)
is an exact mathematical answer for the stated continuous model.
A decimal such as 14.95 hours is an approximation.
Rounding should follow the question’s instructions.
If a task instead asks for the first completed four-hour period when the quantity is at or below a target, the final answer may be a whole number of periods rather than a decimal time.
The mathematical expression, the rounding convention and the interpretation should not be confused.
An accurate calculation can still lead to the wrong final statement if the time model is misunderstood.
A five-stage method for decay questions
First, identify the starting amount and the unit of time.
Second, determine whether the problem gives a half-life, a decay factor per period or a continuous decay coefficient.
Third, write a model consistent with that information and state whether time is discrete or continuous.
Fourth, set the model equal to the target or write the required inequality, then solve using powers or logarithms.
Fifth, check the result against familiar halvings, permitted time values and the original question’s wording.
The process is more reliable than inserting every given number into the first exponential formula remembered.
A-Math trains students to explain why the model fits the description.
Four common errors in half-life work
Mistake one: subtracting the initial half every time
The learner calculates 160, 80, 0 instead of 160, 80, 40.
The repair is to multiply the current amount by one-half at each step.
Mistake two: confusing half-life with a fixed rate per hour
The learner treats “halves every four hours” as a constant number of units lost each hour.
The repair is to distinguish constant percentage change from constant absolute change.
Mistake three: using the wrong sign for the exponential coefficient
The learner writes A₀e^(kt) with k positive for decay.
The repair is to check whether the resulting graph increases or decreases.
Mistake four: rounding a threshold without interpreting time
The learner gets 3.2 completed periods and reports three even though the target has not yet been reached.
The repair is to compare nearby complete periods or use the precise continuous threshold as appropriate.
Each mistake requires attention to the model’s meaning, not just calculator practice.
A five-session study plan
Session one: repeated halving
Begin with 160 units and a four-hour half-life.
Write the values after one, two, three and four complete intervals.
Derive the general power-of-one-half formula.
Session two: connect to indices and logarithms
Find when the continuous model reaches 10 units exactly.
Then use logarithms to find when it reaches 12 units.
Check that the time lies between twelve and sixteen hours.
Session three: compare time interpretations
Calculate the halfway value under a continuous exponential interpolation and a linear interpolation.
Explain why the two answers differ.
Session four: differentiate the model
Rewrite the half-life formula as A₀e^(-kt).
Derive k = ln 2/H and A'(t) = -kA(t).
Interpret the negative sign and proportional rate.
Session five: solve a threshold inequality
Find when a quantity is at most a specified fraction of its starting value.
Compare strict and non-strict inequalities and decide whether the final answer is a real-valued time or a completed-period count.
The aim is to connect formulas with interpretation.
How parents can ask useful questions
Ask whether the amount lost is the same in every interval or whether the proportion lost remains the same.
Ask why a quantity that has halved four times is one-sixteenth of the original.
Ask what the exponent t/H represents.
Ask why a decay model has a negative derivative.
Ask whether an answer such as 2.7 half-lives describes a smooth model or a count of completed events.
These questions help students reason about the model before doing the arithmetic.
A learner who can explain the difference between repeated halving and fixed subtraction has understood the central idea.
Singapore’s G3 Additional Mathematics connection
The official 2027 Singapore-Cambridge Secondary Education Certificate G3 Additional Mathematics syllabus, subject K341, includes exponential and logarithmic functions as models, together with differentiation and rates of change.
The half-life examples here connect those assessed ideas. Some continuous modelling interpretations are explanatory extensions, not predictions of specific examination questions.
For current requirements, consult the SEAB 2027 G3 syllabus directory.
The worked quantities are illustrative. The article does not claim that any particular real substance or process follows the assumed half-life.
How small-group teaching can help
One learner may know exponent laws but build an additive model for repeated percentage loss.
Another may choose the correct exponential equation but reverse the logarithmic inequality.
A third may calculate a correct fractional threshold time yet misinterpret a question asking for completed periods.
These students need different explanations.
In a carefully managed three-student Additional Mathematics tutorial, a tutor can inspect the model choice, algebra and final interpretation separately.
The immutable Secondary 1 Mathematics small-group tutorial in Clementi illustrates why secure percentages, indices and mathematical language are useful foundations before advanced exponential models.
Students who are already progressing independently may not need tuition. Additional support should address a clearly identified learning difficulty.
Frequently asked questions
What is a half-life?
In a constant-half-life model, it is the fixed time required for the quantity to fall to half its current amount.
Does the same number of units disappear every half-life?
No. The same fraction disappears. The absolute amount lost becomes smaller as the remaining quantity decreases.
Why does the formula use t/H?
It represents how many half-life intervals have elapsed, using consistent time units.
Does the continuous model reach zero after enough time?
No. A positive exponential decay model approaches zero asymptotically and remains positive for every finite time.
How is the half-life related to the decay coefficient?
For A(t) = A₀e^(-kt), where k > 0, the half-life is H = ln 2/k.
Why might a fractional time and a whole-period answer differ?
A smooth model allows intermediate time values. A process updated only after completed periods may require the first qualifying whole period.
Are half-life calculations guarantees about real processes?
No. They are consequences of a model that assumes a constant halving interval. Real situations require evidence that the model is appropriate.
The lasting reason to understand exponential decay
A half-life is a compact way to describe repeated proportional decrease.
It connects the arithmetic of halving with powers, logarithms and the derivative of an exponential function.
Additional Mathematics helps students derive the model, calculate thresholds and check the meaning of the result.
It also teaches them that a mathematical process may be smooth, discrete or only approximately representative of real observations.
Those distinctions build judgement as well as algebraic fluency.
A student who understands why a quantity halves rather than simply memorising the formula has gained a transferable way of thinking.
For related reading, explore Why Study Additional Mathematics | Use Logarithms to Find How Long Growth Takes, Why Study Additional Mathematics | Compare Discrete Changes with Continuous Rates and the eduKateSG Additional Mathematics project library.