Growth and decay factors
A phone loses eighteen percent of its value every year. What single number describes how that value changes, year after year?
Where you are 6 in Exponential growth and decay
Why this exists
Many quantities change by the same percent every period. A phone’s resale value falls 18\% a year. A cycling club’s membership grows 4\% a year. After a pond is treated, the algae cover on it shrinks 20\% a day. In each case the amount gained or lost is different every period, because it is a percent of a different amount each time. What stays the same is the number the amount is multiplied by.
The multiplier for a percent change turns one percent change into one multiplication. When the same change repeats period after period, that multiplier describes the whole process, and it gets a name: the growth factor or the decay factor. With it, the value after three years, the change over a week, or the rate per month each takes one calculation.
Definition
A quantity has a constant factor b per period when, at the end of every period, the amount is b times what it was at the start of that period.
- If b > 1, the quantity grows, and b is a growth factor.
- If 0 < b < 1, the quantity shrinks, and b is a decay factor.
- If b = 1, the quantity stays the same.
A percent rate r per period, written as a decimal, gives the factor:
So 4\% growth a year is a factor of 1.04 a year, and 18\% decay a year is a factor of 0.82 a year. Read the other way, a factor of 0.88 is a 12\% decay, because 0.88 = 1 - 0.12, and a factor of 1.035 is a 3.5\% growth, because 1.035 = 1 + 0.035.
Starting from an amount a and multiplying by b once for each of t periods gives
Here a is the amount at the start, when t = 0; b is the factor per period; and t counts periods. A function of this form, with a \neq 0, b > 0 and b \neq 1, is an exponential function.
Intuition
Figure 1 follows the algae. On day 0 it covers 2{,}500 square meters, and each day’s cover is 0.8 of the day before.
The dashed pieces are what each day removes: 500, then 400, then 320, then 256 square meters. The factor is the same every day, but the same percent of a smaller amount is a smaller piece, so the losses shrink. Growth runs the other way: 4\% of a bigger club is more members, so the gains get larger every year. Adding the same amount each period is linear change; multiplying by the same factor is exponential change, and Linear, polynomial and exponential growth compared puts the two side by side.
Figure 2 puts factors on a number line. Everything turns on where b sits compared with 1.
The distance from 1 is the rate. 0.8 is 0.2 below 1, a 20\% decay; 1.6 is 0.6 above 1, a 60\% growth.
Results
Factor and rate. For a growth rate r, b = 1 + r. For a decay rate r, b = 1 - r. Either way, b - 1 is the change per period as a fraction of the amount: positive for growth, negative for decay.
The model. With start a and factor b per period, the amount after t periods is
Money earning a yearly rate r, with the interest added once a year, is the growth case: P dollars become P(1 + r)^t dollars after t years.
A factor over several periods. If b is the factor for one period, the factor for k periods is b^k. A factor of 1.1 a year is 1.1^2 = 1.21 every two years: 21\% growth, not 20\%.
A factor for part of a period. The factor for one twelfth of a period is b^{1/12}, because twelve of them must multiply to b. With t in years,
so b^{1/12} is the factor per month, and 12t counts months.
Equal rates up and down do not cancel. 1.05 \times 0.95 = 0.9975, so 5\% growth followed by 5\% decay ends 0.25\% below the start, as Two percent changes one after the other explains.
Why it is true
One period. Say the amount at the start of a period is A and it grows at the rate r. The new amount is A + rA, and the distributive property writes that as (1 + r)A. So one period multiplies by 1 + r. A decay gives A - rA = (1 - r)A in the same way.
Many periods. After one period the amount is ab. After two it is (ab)b = a(b \cdot b) = ab^2, regrouping by the associative property. After t periods, b has been multiplied in t times, which is ab^t by what an exponent means.
Several periods at once. Over k more periods, starting at any time t, the amount goes from ab^t to ab^{t+k}, and b^{t+k} = b^t \cdot b^k by Multiplying and dividing powers with the same base. So any block of k periods multiplies by b^k, wherever it starts.
Part of a period. Call the monthly factor m. Twelve months make a year, so m^{12} = b, and the positive number with that property is b^{1/12}, the twelfth root of b (Rational exponents). Then \left(b^{1/12}\right)^{12t} = b^t by Powers of powers, products and quotients, so the monthly form is the same model written in different units, not a new model.
Worked examples
Factors and rates, both ways
(a) Write the factor for each: a phone’s resale value falls 18\% a year; a cycling club’s membership grows 4\% a year; algae cover on a pond shrinks 20\% a day.
(b) Say whether each factor is growth or decay, and give the percent rate: 0.88, 1.035, 2.4.
Solution(a) Write each rate as a decimal. Decay subtracts it from 1 and growth adds it: 1 - 0.18 = 0.82 a year for the phone, 1 + 0.04 = 1.04 a year for the club, and 1 - 0.2 = 0.8 a day for the algae.
(b) Compare each factor with 1. 0.88 < 1, so it is decay, and 1 - 0.88 = 0.12: a 12\% decay. 1.035 > 1, so it is growth, and 1.035 - 1 = 0.035: a 3.5\% growth. 2.4 > 1 and 2.4 - 1 = 1.4: a growth of 140\% per period. A growth rate above 100\% is allowed; it means the amount more than doubles each period.
The phone’s value
A phone is bought for 900 dollars, and its resale value falls 18\% a year. Write its value V(t) after t years, find the value after 3 years, and find the total percent lost.
SolutionThe start is a = 900 and the decay factor is b = 0.82, so V(t) = 900(0.82)^t. Year by year, each value is 0.82 times the one before:
| t (years) | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| V(t) (dollars) | 900 | 738 | 605.16 | 496.23 |
In one step, 0.82^3 = 0.551368 and 900 \times 0.551368 = 496.2312, so after 3 years the phone is worth about 496.23 dollars.
The three-year factor 0.551368 says about 55.1\% of the value remains, so about 44.9\% is lost. That is less than 3 \times 18\% = 54\%, because the second and third years each take 18\% of a smaller value: 162 dollars, then 132.84, then 108.93.
Reading a model
The cycling club’s membership t years after it began is modeled by M(t) = 150(1.04)^t. Say what 150 and 1.04 mean, and estimate the membership after 5 years. Then read g(t) = 320(0.75)^t the same way.
SolutionAt t = 0 the power is 1.04^0 = 1, so M(0) = 150: the club began with 150 members. The base 1.04 is the factor per year, and 1.04 = 1 + 0.04, so membership grows 4\% a year.
After 5 years, 1.04^5 \approx 1.21665, and 150 \times 1.04^5 \approx 182.498. A count is a whole number, so the model predicts about 182 members.
For g, the start is g(0) = 320, and the factor 0.75 is less than 1: a decay of 1 - 0.75 = 0.25, which is 25\% per period.
A factor over a different period
(a) The algae cover shrinks by the factor 0.8 each day. Find the factor for two days and for one week, and say what percent of the cover each removes.
(b) After the treatment, the amount of cleaning chemical in the pond halves every 6 hours, starting from 24 grams. Write the amount C(t) after t hours, and find it after one day and after 9 hours.
Solution(a) Two days multiply by 0.8 twice: 0.8^2 = 0.64. Two days remove 1 - 0.64 = 0.36, which is 36\%, not 40\%. A week is seven days: 0.8^7 = 0.2097152, about 0.21, so a week removes about 79\% of the cover, not 7 \times 20\% = 140\%.
(b) The factor is 0.5 per six-hour period. Here t is measured in hours, so t/6 counts how many six-hour periods have passed: after 12 hours, 12/6 = 2 periods. Multiplying by 0.5 once per period gives
One day is 24 hours, which is 4 periods: the factor is 0.5^4 = 0.0625, and C(24) = 24 \times 0.0625 = 1.5 grams. After 9 hours, 9/6 = 1.5 periods have passed, and a calculator gives 0.5^{1.5} \approx 0.35355, so C(9) \approx 8.49 grams. That sits between the 12 grams left at 6 hours and the 6 grams left at 12 hours, as it should.
From a yearly factor to a monthly one
A quantity grows 12\% a year, so it is modeled by Q(t) = a(1.12)^t with t in years. Rewrite the model with a monthly factor, and find the monthly growth rate.
SolutionA year is twelve months, so the monthly factor m must satisfy m^{12} = 1.12, which makes m = 1.12^{1/12}. By Powers of powers, products and quotients,
A calculator gives 1.12^{1/12} \approx 1.009489, which rounds to 1.0095. With 12t counting months, the quantity grows by a factor of about 1.0095 a month, a monthly rate of about 0.95\%.
It is not 12\% \div 12 = 1\%. A monthly 1\% would give 1.01^{12} \approx 1.1268 in a year, about 12.7\%, which is too much: each month’s 1\% is taken of an amount that has already grown.
How long until the cover is small
The algae cover starts at 2{,}500 square meters and shrinks 20\% a day. After how many whole days is it first below 500 square meters?
SolutionThe cover after d days is 2500(0.8)^d, and the question asks for the first d with 2500(0.8)^d < 500. There is no need to solve for the exponent: keep multiplying by 0.8. Days 1 to 6 give 2{,}000, 1{,}600, 1{,}280, 1{,}024, 819.2 and 655.36 square meters, and then
| day d | 6 | 7 | 8 |
|---|---|---|---|
| cover (square meters) | 655.36 | 524.288 | 419.4304 |
On day 7 the cover is still above 500; on day 8 it is below. The cover first drops below 500 square meters after 8 days.
The start does not matter as much as it seems. 500 is a fifth of 2{,}500, so the question is the first d with 0.8^d < 0.2, and 0.8^7 \approx 0.2097 while 0.8^8 \approx 0.1678. Any amount shrinking 20\% a day takes 8 days to fall below a fifth of where it started.
Boundaries
The factor must be positive. An amount cannot lose more than all of itself, so a decay rate is below 100\% and 0 < b < 1. Growth rates have no ceiling: 150\% growth is the factor 2.5.
A factor belongs to its period. “0.8” means nothing until you know it is per day. Changing the period raises the factor to a power; it never multiplies or divides the rate. Two days of 20\% decay remove 36\%, and 12\% a year is about 0.95\% a month.
A rounded monthly factor is approximate. 1.0095^{12} \approx 1.1201, close to 1.12 but not equal. Keep more decimal places when more accuracy is needed.
b = 1 is neither. a \cdot 1^t = a for every t, a constant.
Decay never reaches zero in the model. Multiplying a positive amount by 0.8 leaves a positive amount, so 2500(0.8)^d > 0 for every d. A real pond may be cleared; the model cannot say when.
A model is only as good as its constant factor. Real phones, clubs and ponds do not change by exactly the same percent every period, and a model’s output may need rounding to make sense, as with 182.498 members.
Where this goes next
Plotting y = a \cdot b^t gives a curve that climbs ever faster or flattens toward zero, studied in Graphs of exponential functions. The time a factor takes to double or halve an amount is Doubling time and half life. Working backwards, two values of a quantity give its start and its factor: see Building an exponential model from two points and Writing an exponential function from a table or graph. The amounts one period apart, a, ab, ab^2, \dots, form a geometric sequence whose common ratio is the factor (Geometric sequences and the explicit formula). Interest added several times a year is Compound interest and compounding periods.
Exercises
Factors and rates
Write the factor per period.
- A quantity grows 7\% a year.
- A quantity falls 30\% a day.
- A quantity grows 150\% a decade.
- A quantity falls 0.5\% a month.
Say whether each factor is growth or decay, and give the percent rate.
- 0.65
- 1.009
Reading a model
- For P(t) = 4800(0.93)^t, give the starting amount, the factor, whether it is growth or decay, and the percent rate per period.
- For A(t) = 75(1.6)^t, give the starting amount, the factor, whether it is growth or decay, and the percent rate per period.
- For f(t) = 12\left(\tfrac{3}{4}\right)^t, give the factor and the percent rate, then find f(4).
Using a model
- Another cycling club has 240 members and grows 4\% a year. Estimate its membership after 3 years.
- A phone bought for 650 dollars loses 18\% of its resale value a year. Write its value V(t) after t years and find its value after 2 years.
- Algae covering 1{,}800 square meters shrinks 20\% a day. Find the cover after 5 days, to the nearest square meter.
A different period
- A quantity grows by the factor 1.06 a year. Find its factor for 2 years and for 5 years, and the percent growth over each.
- A pond holds 40 grams of a cleaning chemical, and the amount halves every 6 hours. Find the amount after 18 hours, and the factor per hour to four decimal places.
- A phone’s resale value falls 18\% a year. Find the monthly factor to four decimal places, and the monthly percent decay.
How long until
- The phone in Example 2 is worth 900(0.82)^t dollars after t years. After how many whole years is it first worth less than 300 dollars? Show the values you compare.
Reasoning
- Conall’s club grows 5\% one year and shrinks 5\% the next. He says it is back to its old size. Is he right? Find the single factor for the two years and explain.
- Uriel says algae shrinking 20\% a day is gone after 5 days, because 5 \times 20\% = 100\%. Explain what is wrong, and find what fraction of the cover is left after 5 days.
Answers
- 1.07. Growth adds the rate: 1 + 0.07.
- 0.7. Decay subtracts the rate: 1 - 0.3.
- 2.5. 1 + 1.5 = 2.5; the amount more than doubles each decade.
- 0.995. 1 - 0.005.
- Decay of 35\%. 0.65 < 1 and 1 - 0.65 = 0.35.
- Growth of 0.9\%. 1.009 > 1 and 1.009 - 1 = 0.009.
- Start 4{,}800, factor 0.93, decay of 7\% per period, since 1 - 0.93 = 0.07.
- Start 75, factor 1.6, growth of 60\% per period, since 1.6 - 1 = 0.6.
- Factor 0.75, a decay of 25\%; f(4) = 12 \times \tfrac{81}{256} = 3.796875, about 3.80.
- About 270 members. 240(1.04)^3 = 240 \times 1.124864 \approx 269.97.
- V(t) = 650(0.82)^t, and V(2) = 650 \times 0.6724 = 437.06, so 437.06 dollars.
- About 590 square meters. 1800(0.8)^5 = 1800 \times 0.32768 = 589.824.
- 1.06^2 = 1.1236, growth of 12.36\% over 2 years; 1.06^5 \approx 1.3382, growth of about 33.8\% over 5 years.
- 5 grams after 18 hours, since 18 hours is 3 halvings: 40 \times 0.5^3 = 5. The hourly factor is 0.5^{1/6} \approx 0.8909, about 10.9\% lost each hour.
- 0.82^{1/12} \approx 0.9836, a decay of about 1.64\% a month.
- After 6 years. 900(0.82)^5 \approx 333.67 dollars, still at least 300, and 900(0.82)^6 \approx 273.61 dollars, below 300.
- No. The single factor is 1.05 \times 0.95 = 0.9975, so the club ends 0.25\% smaller. The 5\% loss is taken of the larger, grown club, so it removes more members than the 5\% gain added.
- Each day removes 20\% of what is left, a smaller amount each time, so the rates cannot be added. After 5 days the factor is 0.8^5 = 0.32768, so about 32.8\% of the cover is left. Multiplying a positive amount by 0.8 always leaves a positive amount, so in this model no number of days removes all of it.