KBAT: Exponential Growth and Decay

Q: Why divide the two models before taking logs?

Dividing collects the two powers into a single base: 1.06t0.96t=(1.060.96)t\frac{1.06^{t}}{0.96^{t}}=\left(\frac{1.06}{0.96}\right)^{t}0.96t1.06t=(0.961.06)t. That leaves one exponential term, which a single logarithm can then solve. Taking logs before combining leaves two ttt terms and makes the algebra much harder.

Q: Why can't the decaying town ever recover in this model?

Because its factor 0.960.960.96 is below 1, so (0.96)t(0.96)^{t}(0.96)t keeps falling towards zero as ttt increases, it never turns upward. The township's factor 1.061.061.06 is above 1, so it grows without bound. Once the growing quantity overtakes the decaying one, it stays ahead permanently.

What makes this a KBAT question

A routine indices question gives you an equation like $2^{x}=32$ and asks for $x$ . A KBAT question describes two changing quantities, a town shrinking each year, a township growing, and expects you to write both as exponential models, combine them with the laws of indices, and use logarithms to find when they meet.

That is Kemahiran Berfikir Aras Tinggi, higher-order thinking: forming $P=P_{0}r^{t}$ , applying $\frac{a^{t}}{b^{t}}=\left(\frac{a}{b}\right)^{t}$ , and taking logs are all Form 4 skills, but deciding to divide the two models, choosing which base to keep, and reading the result as a moment in time are not spelt out. In Add Math this rewards students who understand that a growth factor above 1 climbs without bound while a decay factor below 1 falls towards zero, and who can turn that understanding into a decision.

One worked problem, in the style of Paper 2

This is an original question written in the style of SPM Paper 2. Use a scientific calculator, and try it yourself before reading the solution.

Q1[7 marks]

An old mining town has a population of 90 000 that falls by 4% each year, modelled by $A=90000(0.96)^{t}$ , where $t$ is the number of years from now. A new township nearby starts with 60 000 people and grows by 6% each year, modelled by $B=60000(1.06)^{t}$ .

(a) Find the population of each place after 4 years, to the nearest whole person. (b) Find, to the nearest tenth of a year, the time at which the two populations are equal, and state which place is larger before that time.

(c) A councillor claims that the mining town will eventually recover and overtake the township again. Using the models, explain why this cannot happen.

Show worked solution

Understand. $A$ decays with factor $0.96<1$ ; $B$ grows with factor $1.06>1$ .

Part (a) is direct substitution. Part (b) needs the two models set equal and solved for the exponent $t$ using logarithms.

Part (c) is about the long-run behaviour of decay versus growth.

Plan. Substitute $t=4$ for (a).

For (b), set $A=B$ , divide to collect the powers into a single base using $\frac{1.06^{t}}{0.96^{t}}=\left(\frac{1.06}{0.96}\right)^{t}$ , then take logarithms. For (c), consider what each factor does as $t$ grows large.

Execute and check. (a) At $t=4$ :

A=90000(0.96)^{4}=90000(0.84934656)\approx 76\,441

B=60000(1.06)^{4}=60000(1.26247696)\approx 75\,749

So after 4 years the mining town has about 76 441 people and the township about 75 749.

(b) Set the populations equal and gather the powers:

90000(0.96)^{t}=60000(1.06)^{t} \;\Rightarrow\; \left(\frac{1.06}{0.96}\right)^{t}=\frac{90000}{60000}=\frac{3}{2}

Take logarithms of both sides and make $t$ the subject:

t=\frac{\log 1.5}{\log\!\left(\tfrac{1.06}{0.96}\right)}=\frac{0.17609}{0.04303}\approx 4.09

So the populations are equal at about $t=4.1$ years. Before that time $t<4.1$ , the mining town $A$ is the larger of the two (for example, part (a) shows $A>B$ at $t=4$ ); after it, the township $B$ is larger.

(c) As $t$ grows without bound, $(0.96)^{t}\to 0$ , so $A\to 0$ ; meanwhile $(1.06)^{t}\to\infty$ , so $B$ increases without limit. A decaying quantity can never turn upward again under this model, so once $B$ overtakes $A$ at $t\approx 4.1$ , $B$ stays larger forever.

The councillor's claim is therefore incorrect.

Check. At $t=5$ , $A=90000(0.96)^{5}\approx 73\,384$ and $B=60000(1.06)^{5}\approx 80\,294$ , so $B>A$ , consistent with the crossover at $t\approx 4.1$ and with $A$ continuing to fall.

Finding a sensible first step

When two exponential quantities must be compared, the reliable first step is to write each as $P_{0}r^{t}$ , a starting value times a yearly factor, and read the factor from the percentage. A 4% fall means multiply by $0.96$ each year; a 6% rise means multiply by $1.06$ .

Getting those factors right is the whole battle. To find when the two are equal, do not expand anything: set the models equal and divide, so the two powers combine into one base using $\frac{a^{t}}{b^{t}}=\left(\frac{a}{b}\right)^{t}$ .

Only then take logarithms, because a logarithm is the tool that brings an exponent down to ground level where you can solve for it. Naming the factors and collecting the powers before touching logs keeps the working short and the method marks safe.

What markers reward

Marking is analytic, so method marks are awarded line by line. On an exponential growth-and-decay question a marker looks for:

Each yearly factor read correctly from the percentage, $0.96$ for a 4% fall, $1.06$ for a 6% rise.
Correct substitution for part (a), with answers rounded sensibly to whole people.
The two models set equal and the powers combined, $\left(\tfrac{1.06}{0.96}\right)^{t}=\tfrac{3}{2}$ .
Logarithms applied correctly to isolate $t$ , giving $t=\dfrac{\log 1.5}{\log(1.06/0.96)}$ .
A numerical answer to the required accuracy, $t\approx 4.1$ years, with a statement of which place is larger before it.
A long-run argument for part (c), decay tends to zero, growth increases without bound, so no recovery is possible.

How a teacher helps

Growth-and-decay questions improve fastest when a student is fluent with the factor idea and confident taking logs. In a one-to-one lesson our teachers ask you to write the yearly factor from the percentage before anything else, to combine the powers with index laws rather than expanding, and to say in words what a growth factor above 1 does over time.

We practise the calculator steps for logarithms so the numbers are dependable, and we press on the interpretation in part (c), because that reasoning earns the KBAT marks. Teachers at spmaddmath.com.my are experienced, and lessons are online and taught in English.

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Frequently asked questions

How do I turn a percentage change into a factor?

Start from 100, add the percentage for growth or subtract it for decay, then divide by 100. A 6% rise gives a factor of $1.06$ ; a 4% fall gives $0.96$ .

Each year you multiply by this factor, so after $t$ years the quantity is $P_{0}r^{t}$ , the heart of every growth-and-decay model.

Why divide the two models before taking logs?

Dividing collects the two powers into a single base: $\frac{1.06^{t}}{0.96^{t}}=\left(\frac{1.06}{0.96}\right)^{t}$ . That leaves one exponential term, which a single logarithm can then solve.

Taking logs before combining leaves two $t$ terms and makes the algebra much harder.

Which base of logarithm should I use?

Either base works, as long as you use it on both sides. Common logarithm (base 10) or natural logarithm (base $e$ ) both give $t=\dfrac{\log 1.5}{\log(1.06/0.96)}\approx 4.1$ .

A non-programmable scientific calculator has both keys, so pick whichever you are comfortable with.

Why can't the decaying town ever recover in this model?

Because its factor $0.96$ is below 1, so $(0.96)^{t}$ keeps falling towards zero as $t$ increases, it never turns upward. The township's factor $1.06$ is above 1, so it grows without bound.

Once the growing quantity overtakes the decaying one, it stays ahead permanently.

Source:SRC-DSKP-EN SRC-FORMAT

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026