5.03a, 6.02a - Exponential growth models

5.03a, 6.02a - Exponential growth models

Turn repeated percentage change into a formula for any whole number of periods. Use it to find amounts, original values and rates, then decide when a prediction crosses a target and whether the model's assumptions are reasonable.

This lesson is for Higher tier.

Building a growth formula

An account starts with £2000 and grows by 7% each year. Will it reach £2500 after three yearly interest payments or four? A formula can compress all those successive multiplications, but you still need to interpret the number of payments correctly.

Begin with a smaller example. £100 growing by 10% becomes £110 after one period, £121 after two and £133.10 after three. Each step multiplies by 1.10. A power counts those identical factors:

100×1.10×1.10×1.10=100(1.10)3100\times1.10\times1.10\times1.10=100(1.10)^3

The initial amount corresponds to zero completed periods: 100(1.10)0=100100(1.10)^0=100.

Exponential growth

A=P(1+r100)n=Pbn,b=1+r100A=P\left(1+\frac{r}{100}\right)^n=Pb^n,\qquad b=1+\frac r{100}

For growth, P>0P>0, r>0r>0 and b>1b>1. The rate rr is a percentage per period, and nn counts completed periods. With annual payments, n=0,1,2,…n=0,1,2,\ldots. The compound-interest formula is supplied on both 2027 formula sheets.

If you instead write A=P(1+r)nA=P(1+r)^n, then rr must be the decimal rate, such as 0.07 for 7%. The sheet uses the percentage-rate version above. Decide which meaning your letter has before substituting.

Worked example 1: writing and using a growth formula

Question. £800 is invested at 8% per year compound interest. Write a formula for its value AA after nn years, then calculate its value after 2 years. Do not round intermediate values.

Plan. Form the annual multiplier, use a power to count years, then substitute n=2n=2.

b=1+8100=1.08annual multiplierA=800(1.08)none factor per yearA=800(1.08)2=933.12two years\begin{aligned} b&=1+\frac8{100}=1.08 &&\text{annual multiplier}\\ A&=800(1.08)^n &&\text{one factor per year}\\ A&=800(1.08)^2=933.12 &&\text{two years} \end{aligned}

Answer. A=800(1.08)nA=800(1.08)^n; after 2 years, £933.12.

Check. Year one gives 800+64=864800+64=864. Year two adds 864×0.08=69.12864\times0.08=69.12, giving £933.12.

Keep the principal and the two-year duration, but change the unknown to the annual rate. Divide out the principal before reversing the power.

Worked example 2: finding a growth rate

Question. £800 grows to £1250 after 2 years at a constant annual compound-interest rate. Find the percentage rate.

Plan. Find the two-year multiplier, take its positive square root, then convert the yearly multiplier to a percentage increase.

1250=800b2two equal yearly factorsb2=1250÷800=1.5625divide by the principalb=1.5625=1.25positive growth multiplierr=(1.25−1)×100=25percentage above 1\begin{aligned} 1250&=800b^2 &&\text{two equal yearly factors}\\ b^2&=1250\div800=1.5625 &&\text{divide by the principal}\\ b&=\sqrt{1.5625}=1.25 &&\text{positive growth multiplier}\\ r&=(1.25-1)\times100=25 &&\text{percentage above 1} \end{aligned}

Answer. 25% per year.

Check. 800×1.25=1000800\times1.25=1000, then 1000×1.25=12501000\times1.25=1250. The negative square root is unsuitable: an ordinary savings balance is not multiplied by a negative factor each year.

For nn equal periods, a positive multiplier can be recovered with b=A/Pnb=\sqrt[n]{A/P}, then r=100(b−1)r=100(b-1). This needs a positive principal, positive final amount and a positive whole number of periods. For example, taking a cube root undoes three yearly multiplications. The case n=0n=0 cannot reveal a rate, because every positive multiplier raised to zero gives the same initial amount.

Watch out

The line A=800(1+8)nA=800(1+8)^n interprets 8% as a decimal rate of 8, giving a multiplier of 9. Use 1+8/100=1.081+8/100=1.08. Also, the exponent counts periods, not the percentage rate.

Reverse the repeated factor with a new investment.

Building a decay formula

Repeated decay has the same power structure as growth, but its positive multiplier is less than one.

If a value loses 15% each year, it retains 85%, so its yearly multiplier is 0.85. After two years it retains 0.852=0.72250.85^2=0.7225 of its original value: 72.25%, not 70%. Later losses are percentages of a smaller amount.

Exponential decay

A=P(1−d100)n=Pbn,b=1−d100A=P\left(1-\frac d{100}\right)^n=Pb^n,\qquad b=1-\frac d{100}

For ordinary percentage decay, P>0P>0, 0<d<1000<d<100 and 0<b<10<b<1. The compound-interest formula supplied on the 2027 sheets has the same repeated-factor structure; for decay, construct the retaining multiplier from the decrease.

A rate of 0% gives a constant amount, rather than growth or decay. A 100% loss makes the value zero after the first period and cannot be reversed by division by its multiplier. The model with 0<b<10<b<1 remains positive after every finite number of periods, although it can become very small. Real equipment can still be scrapped or have a minimum resale value; those are limits of the model.

Worked example 3: writing and using a decay formula

Question. A vehicle is initially worth £24000 and depreciates by 15% of its current value each year. Write a formula for its value VV after nn years, then calculate its value after 2 years.

Plan. Use the fraction retained each year, raised to the number of years.

b=1−0.15=0.85yearly retaining multiplierV=24000(0.85)none factor per yearV=24000(0.85)2=17340two years\begin{aligned} b&=1-0.15=0.85 &&\text{yearly retaining multiplier}\\ V&=24000(0.85)^n &&\text{one factor per year}\\ V&=24000(0.85)^2=17340 &&\text{two years} \end{aligned}

Answer. V=24000(0.85)nV=24000(0.85)^n; after 2 years, £17340.

Check. The first loss is £3600, leaving £20400. The second loss is 20400×0.15=£306020400\times0.15=£3060, leaving £17340.

Keep the same rate and duration, but change the unknown to the initial value. Reverse the whole retaining factor.

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