EL7 - Exponential growth and decay

Build exponential models for repeated percentage changes, determine their rates from observations and use logarithms to find times. You will distinguish a time within a continuous model from the first complete year or month when a threshold is reached, and interpret growth and decay in context.

Modelling a repeated percentage change

An amount doubles every six hours. How long does it take to become four times as large: twelve hours or twenty-four? Exponential change repeats a multiplier, so equal multiplication factors take equal times. We will return to this after learning how to find the time itself.

You already know

A fixed percentage change multiplies the current amount. The recurrence $Q_{n+1}=aQ_n$ , starting at $Q_0$ , gives $Q_n=Q_0a^n$ . Logarithms solve an unknown exponent; keep intermediate values unrounded.

A $6\%$ increase keeps $106\%$ of the previous amount, giving multiplier $1.06$ . A $6\%$ decrease keeps $94\%$ , giving multiplier $0.94$ . The percentage acts on the changing amount, not on the original amount every time.

Exponential growth and decay models

Q(t)=Q_0a^t

Here $Q_0>0$ is the value at $t=0$ , and $a>0$ is the multiplier per unit of $t$ . Growth has $a>1$ ; decay has $0<a<1$ . For percentage growth $r$ , $a=1+r/100$ ; for percentage loss $r$ , $a=1-r/100$ , with $0<r<100$ .
Not on the formulae sheet: learn it.

The formulae sheet gives the same structure specifically for compound interest. It is useful, but identifying the time unit and retained percentage remains your job.

Compound interest

\text{Total accrued}=P\left(1+\frac r{100}\right)^n

$P$ is the principal, $r$ the percentage interest rate per compounding period, and $n$ the number of times interest is compounded. There are no payments in or out.
On the formulae sheet.

Use whole-number inputs for a process updated only at specified times. A model explicitly assumed to apply continuously may also be evaluated at fractional times; that is an additional modelling assumption, not an automatic consequence of annual interest.

Worked example 1: growth at a fixed percentage

Question. A balance of £3000 earns $6\%$ annual compound interest. Form a model for the balance after $n$ complete years and find its value after three years.

Plan. Multiply by $1.06$ for each completed year.

\begin{aligned} B_n&=3000(1.06)^n && \text{initial value and annual multiplier}\\ B_3&=3000(1.06)^3 && \text{three complete periods}\\ &=3573.048 && \text{evaluate without early rounding} \end{aligned}

Why is the exponent $3$ , not $36$ ? The multiplier is annual, so the input counts years rather than months.

Answer. $B_n=3000(1.06)^n$ ; after three years the balance is £3573.05 to the nearest penny.

Check. The first year's interest is £180, and later interest is larger. The total increase £573.05 is therefore greater than $3\times180=540$ .

Change only the sign of the percentage change, keeping the starting amount and time interval. The multiplier now lies below $1$ .

Worked example 2: decay at a fixed percentage

Question. A machine worth £3000 loses $6\%$ of its current value each year. Form a model for its value after $n$ complete years and find its value after three years.

Plan. Retain $94\%$ of the current value at each annual update.

\begin{aligned} V_n&=3000(0.94)^n && \text{annual retained fraction}\\ V_3&=3000(0.94)^3 && \text{substitute three periods}\\ &=2491.752 && \text{evaluate} \end{aligned}

Why not subtract £180 each year? The $6\%$ loss is calculated from a smaller value each year, so the loss in pounds decreases.

Answer. $V_n=3000(0.94)^n$ ; after three years the value is £2491.75 to the nearest penny.

Check. The model stays positive. Its loss £508.25 is less than three losses of £180, as expected.

A fixed-percentage decrease does not reach zero in a finite number of periods. That can be sensible for an idealised concentration, but a machine's resale value may eventually stop following a smooth percentage model.

Watch out

Wrong: $6\%$ decay gives $Q(t)=Q_0(0.06)^t$ . That keeps only $6\%$ each time. Right: retain $94\%$ , so the multiplier is $0.94$ .

Change the loss percentage and keep the annual units.

Finding an unknown growth or decay rate

Two observations tell you the total multiplier across their time gap. Take the appropriate root to find the multiplier for one period.

For $Q(t)=Q_0a^t$ , dividing the observations gives $Q(t+h)/Q(t)=a^h$ . This removes the initial coefficient even when the observations do not start at time zero. The ratio-and-root method is not on the formulae sheet: learn it.

Worked example 3: recovering a growth rate

Question. A population model $N=ka^t$ , with $t$ in years, has $N=500$ at $t=0$ and $N=720$ at $t=4$ . Determine its annual percentage growth rate.

Plan. Find the four-year multiplier, take its fourth root and convert the excess over $1$ to a percentage.

\begin{aligned} k&=500 && \text{initial observation}\\ 500a^4&=720 && \text{second observation}\\ a&=\left(\frac{720}{500}\right)^{1/4}=1.095445\ldots && \text{one-year multiplier}\\ r&=100(a-1)=9.5445\ldots && \text{percentage increase} \end{aligned}

Why not divide the total $44\%$ growth by $4$ ? Each year's percentage acts on a different population; taking a root reverses repeated multiplication.

Answer. Approximately $9.54\%$ growth per year.

Check. $500(1.095445\ldots)^4=720$ ; the unrounded multiplier reproduces the observation.

Change only the later observed value from above the starting value to below it. The root now gives a multiplier less than $1$ .

Worked example 4: recovering a decay rate

Question. A quantity is $500$ at $t=0$ and $320$ at $t=4$ , with $t$ in years. Assuming $Q=ka^t$ , determine the annual percentage decrease.

Plan. Take the fourth root of the retained fraction and subtract it from $1$ .

\begin{aligned} a^4&=\frac{320}{500}=0.64 && \text{four-year retained fraction}\\ a&=0.64^{1/4}=0.894427\ldots && \text{one-year retained fraction}\\ r&=100(1-a)=10.5572\ldots && \text{percentage lost each year} \end{aligned}

Why subtract from $1$ ? The multiplier describes what remains, while the requested rate describes what is lost.

Answer. Approximately $10.6\%$ decrease per year.

Check. $500(0.894427\ldots)^4=320$ , and the annual retained fraction is between $0$ and $1$ .

The same calculation can use observations at, say, years $2$ and $6$ : the gap is still four years. Use the gap, not the larger year label.

Watch out

Wrong: $a=720/500=1.44$ is the annual multiplier. That ratio covers four years. Right: $a^4=1.44$ , hence $a=1.44^{1/4}$ .

Keep the decay model, but let the observations begin after the process has already started.

Doubling times and half-lives

A doubling time or half-life depends on the multiplier, not on the amount you start with.

For growth, divide $Q_0a^t=2Q_0$ by $Q_0$ to obtain $a^t=2$ . For decay to half the starting amount, the corresponding equation is $a^t=1/2$ . Logarithms then find $t$ . Use this fractional-time interpretation only when the model is assumed to hold continuously.

Time to reach a specified multiple

Q(t)=fQ_0\quad\Longrightarrow\quad a^t=f \quad\Longrightarrow\quad t=\frac{\log_{10}f}{\log_{10}a}

Here $Q_0>0$ , $a>0$ , $a\ne1$ , $f>0$ . A doubling time uses $f=2$ ; a half-life uses $f=1/2$ . Interpret only non-negative times for a process starting at $t=0$ .
Not on the formulae sheet: learn it.

The initial amount cancels, which explains why a fixed-rate model takes the same time to double from $80$ to $160$ as from $160$ to $320$ .

Worked example 5: doubling under continuous growth

Question. A continuously applicable model is $Q=80(1.12)^t$ , where $t$ is in hours. Find its doubling time.

Plan. Set the amount to $160$ , cancel $80$ and take logarithms.

\begin{aligned} 80(1.12)^t&=160 && \text{twice the starting amount}\\ (1.12)^t&=2 && \text{divide by }80\\ t\log_{10}1.12&=\log_{10}2 && \text{take logs}\\ t&=\frac{\log_{10}2}{\log_{10}1.12}=6.1162\ldots && \text{solve for time} \end{aligned}

Why can the answer be fractional? The model is stated to apply continuously, so a point between complete hours is meaningful.

Answer. The doubling time is approximately $6.12$ hours.

Check. The unrounded time gives $80(1.12)^t=160$ .

Change growth to decay and ask for the matching factor of one half. Both logarithms are now negative, giving a positive time.

Worked example 6: a half-life under continuous decay

Question. A continuously applicable model is $Q=80(0.88)^t$ , with $t$ in hours. Find its half-life.

Plan. Set the amount to $40$ , cancel the initial amount and take logarithms.

\begin{aligned} 80(0.88)^t&=40 && \text{half the starting amount}\\ (0.88)^t&=\frac12 && \text{divide by }80\\ t\log_{10}0.88&=\log_{10}(1/2) && \text{take logs}\\ t&=\frac{\log_{10}(1/2)}{\log_{10}0.88}=5.4222\ldots && \text{divide two negatives} \end{aligned}

Why is this not the same time as in the growth example? A $12\%$ increase and a $12\%$ decrease are not reciprocal multipliers: $1.12\times0.88=0.9856$ , not $1$ .

Answer. The half-life is approximately $5.42$ hours.

Check. At twice the unrounded half-life, the multiplier is $(1/2)^2=1/4$ , giving $Q=20$ .

We can now answer the opening puzzle. A model doubling every six hours doubles again in the next six: after twelve hours its multiplier is $2\times2=4$ . The initial amount does not affect that conclusion.

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Algebra

Enumeration (counting and probability)

Coordinate geometry

Pythagoras' theorem and trigonometry

Calculus

Exponentials and logarithms

Numerical methods

EL7 - Exponential growth and decay

Modelling a repeated percentage change

Finding an unknown growth or decay rate

Doubling times and half-lives

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