What is the difference between the growth rate and the growth factor?

The rate r is the percent change written as a decimal; the factor b is what you multiply by. They are related by b = 1 + r for growth and b = 1 - r for decay. For a 9% increase, the rate is 0.09 and the factor is 1.09. For a 9% decrease, the rate is 0.09 and the factor is 0.91.

How do I tell if an equation shows growth or decay?

Look at the base. If the base is greater than 1, the quantity grows; if it is between 0 and 1, the quantity decays. So y = 40(1.2)^t grows 20% per period and y = 40(0.8)^t decays 20% per period. Fractions count too: y = 40(3/4)^t has base 0.75, so it decays 25% per period.

What if the percent change is given per month but the question asks about years?

Keep t in the unit that matches the rate and convert the time before substituting. With a monthly rate, three years means t = 36. Do not multiply the monthly percent by 12 to get an annual percent — compute (1+r)^12 instead, because the change compounds.

Why can't a decay model ever reach zero?

Each period you multiply by a positive number less than 1, and multiplying a positive number by a positive number never gives zero. The outputs get arbitrarily small and the graph approaches the horizontal asymptote y = 0, but no value of t makes the output exactly zero. In real contexts we usually ask when the amount drops below some small threshold instead.

ALG1-6.4

Exponential Growth & Decay Models

Learn to write, evaluate, and interpret exponential growth and decay models a(1+r)^t and a(1−r)^t, turning percent change into growth or decay factors.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Exponential Growth & Decay Models, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start Exponential Growth & Decay Models free

What this lesson covers

A savings account earning 4% interest, a car losing 15% of its value each year, and a town growing 2% annually all behave the same way mathematically: a quantity is multiplied by the same factor over and over. That repeated multiplication is what makes a model exponential rather than linear.

In this lesson you will turn a sentence like "decreases by 8% per year" into an equation like

y = 250(0.92)^t

, evaluate that equation at a given time, and read an equation backward to say what each number means in the real situation. The whole skill rests on one conversion — percent change into a multiplier — and most mistakes students make trace back to that single step.

From Percent Change to a Growth or Decay Factor

A percent change tells you how much is added or removed relative to the current amount. If a population grows 6% in a year, next year's population is the old population plus 6% of it:

P_{\text{new}} = P + 0.06P = P(1 + 0.06) = 1.06P

The number

1.06

is the growth factor. If a quantity instead loses 6%, you keep 94% of it, so the decay factor is

1 - 0.06 = 0.94

.

Repeat that multiplication

t

times and you get the general models

y = a(1 + r)^t \quad \text{(growth)} \qquad y = a(1 - r)^t \quad \text{(decay)}

where

a

is the starting amount,

r

is the percent rate written as a decimal, and

t

is the number of time periods.

Stated change	$r$ as a decimal	Factor $b$	Model form
grows 5% per year	$0.05$	$1.05$	$a(1.05)^t$
grows 12.5% per year	$0.125$	$1.125$	$a(1.125)^t$
falls 5% per year	$0.05$	$0.95$	$a(0.95)^t$
falls 40% per year	$0.40$	$0.60$	$a(0.60)^t$
grows 100% per year	$1.00$	$2$	$a(2)^t$

Two errors show up constantly. The first is writing the percent as a whole number, giving

y = 500(1+5)^t = 500(6)^t

— that would mean the quantity multiplies by six every year. The second is dropping the 1 and writing

y = 500(0.05)^t

for growth, which actually models losing 95% each period. Always ask: is my factor bigger than 1 (growing) or between 0 and 1 (shrinking)?

Reading a Model in Context

Interpreting is the reverse skill: given

y = a \cdot b^t

, say what the situation is.

The value

a

is the amount when

t = 0

, because

b^0 = 1

. Call it the initial value, starting amount, or original price — whatever fits the story.

The base

b

is the factor. To recover the percent rate, compare

b

to 1. If

b > 1

, then

r = b - 1

and the quantity grows by

r \cdot 100\%

each period. If

0 < b < 1

, then

r = 1 - b

and the quantity decays by

r \cdot 100\%

each period.

For

A(t) = 1{,}400(1.035)^t

where

t

is in years: the starting amount is 1,400, and since

1.035 - 1 = 0.035

, the amount increases 3.5% per year.

For

M(t) = 60(0.78)^t

: the starting amount is 60, and since

1 - 0.78 = 0.22

, the amount decreases 22% per period — not 78%. Saying "it decays 78% per year" is the single most common misreading. The factor tells you what remains; the rate tells you what is lost.

A full interpretation also names the time unit. If

t

counts months, then 3.5% is a monthly rate, and the yearly change is not

12 \times 3.5\%

— it is

(1.035)^{12} \approx 1.511

, about a 51.1% increase, because the growth compounds. Never multiply a periodic rate by the number of periods to get the total change in an exponential model.

Evaluating Models and Avoiding Order-of-Operations Traps

To evaluate, substitute the time value and follow the order of operations: exponent first, then multiply. In

y = 24{,}000(0.85)^5

, you must compute

0.85^5 = 0.4437

before multiplying by 24,000. Typing

24000 \times 0.85

and then raising the result to the 5th power gives a wildly wrong answer, and it is worth pressing keys deliberately on a calculator to avoid this.

Time does not have to be a whole number. If

t

is measured in years, then

t = 0.5

means six months and

t = 2.25

means two years and three months. Negative

t

values look backward in time: in

V(t) = 24{,}000(0.85)^t

, the input

t = -1

gives about 28,235, the value one year before the starting point.

When the time unit in the question does not match the model, convert before substituting, not after. If

P(t) = 800(1.02)^t

with

t

in months, then "after 3 years" means

t = 36

, not

t = 3

.

Rounding deserves care too. Keep full precision inside the calculator and round only the final answer, to the precision the context demands — money to the nearest cent or dollar, populations to whole people. Rounding

0.4437053

0.44

early shifts the car's value by almost 90 dollars.

Finally, remember what an exponential model can and cannot do. A decay model approaches zero but never reaches it, so a question asking "when does the value hit zero?" has no solution; the honest answer is that the model predicts the value keeps shrinking without ever equaling zero.

Exponential Versus Linear: Choosing the Right Model

The wording of a problem tells you which model to write. Language about a fixed amount per period is linear; language about a percent of the current amount is exponential.

Wording	Change per period	Model
increases by 40 people per year	constant amount added	$y = a + 40t$
increases by 4% per year	constant factor	$y = a(1.04)^t$
loses 300 dollars per year	constant amount subtracted	$y = a - 300t$
loses 15% of its value per year	constant factor	$y = a(0.85)^t$
doubles every year	factor of 2	$y = a(2)^t$
triples every year	factor of 3	$y = a(3)^t$
halves every year	factor of $\tfrac12$	$y = a(0.5)^t$

You can also spot the difference in a table of values. Compute successive differences: if they are constant, the relationship is linear. Compute successive ratios: if those are constant, it is exponential, and that constant ratio is your factor

b

.

For the table with inputs

0, 1, 2, 3

and outputs

80, 60, 45, 33.75

, the differences are

-20, -15, -11.25

— not constant. The ratios are

60/80 = 0.75

45/60 = 0.75

33.75/45 = 0.75

— constant. So the model is

y = 80(0.75)^t

, a 25% decrease per period.

A related trap: "doubles every 3 years" is not

a(2)^t

when

t

is in years. Because the doubling happens once every three years, the model is

a(2)^{t/3}

. Match the exponent to how many complete periods have passed.

Key terms

Exponential model.: A function of the form $y = a \cdot b^t$ in which the output is multiplied by the constant factor $b$ each time $t$ increases by 1.
Initial value ( $a$ ).: The output when $t = 0$ ; the amount the quantity starts at, since $b^0 = 1$ .
Growth factor.: A base greater than 1, equal to $1 + r$ , that multiplies the quantity each period so it increases.
Decay factor.: A base between 0 and 1, equal to $1 - r$ , that multiplies the quantity each period so it decreases; it represents the fraction that remains.
Rate of change ( $r$ ).: The percent increase or decrease per period, written as a decimal; 7% becomes $r = 0.07$ .
Time period ( $t$ ).: The number of compounding intervals elapsed, measured in the unit the rate is stated in (years, months, hours).
Compounding.: Applying the percent change to the new amount each period rather than to the original, which is why periodic rates cannot simply be multiplied.
Constant ratio.: The fixed quotient between consecutive outputs in an exponential table, equal to the base $b$ .

Worked example

A car is purchased for 24,000 dollars. Its value decreases by 15% each year. Write a model for the value

V

after

t

years, find the value after 5 years to the nearest dollar, and explain what each part of the model means.

Step 1 — Identify the initial value. At

t = 0

the car is worth 24,000 dollars, so

a = 24{,}000

.

Step 2 — Convert the percent to a decimal. A 15% decrease means

r = 0.15

.

Step 3 — Build the factor. Because the value is decreasing, use

1 - r = 1 - 0.15 = 0.85

. The car keeps 85% of its value each year.

Step 4 — Write the model.

V(t) = 24{,}000(0.85)^t

Step 5 — Evaluate at $t = 5$ . Exponent first:

0.85^5 = 0.4437053125

Then multiply:

24{,}000 \times 0.4437053125 = 10{,}648.9275

After 5 years the car is worth about 10,649 dollars.

Step 6 — Interpret. The 24,000 is the purchase price. The 0.85 is the decay factor: each year the car retains 85% of the previous year's value, which is a 15% annual loss. The exponent

t

counts years since purchase. Notice the car did not lose

5 \times 15\% = 75\%

of its value; it lost about 55.6%, because each year's 15% is taken from a smaller amount than the year before.

Check for reasonableness. After one year the value is

24{,}000(0.85) = 20{,}400

, and after two years

20{,}400(0.85) = 17{,}340

. Continuing that pattern three more years lands near 10,649, so the answer is consistent.

Practice questions

A town has 1,200 residents and its population grows 3% each year. Which equation models the population

P

after

t

years?

$P = 1200(3)^t$
$P = 1200(1.03)^t$
$P = 1200(0.03)^t$
$P = 1200 + 0.03t$

Answer: $P = 1200(1.03)^t$

A 3% increase means the town keeps 100% of its people and adds 3% more, so the factor is

1 + 0.03 = 1.03

. The choice with base 3 would triple the town each year. The choice with base 0.03 would shrink the town to 3% of its size each year. The last choice is linear — it adds a fixed 0.03 people per year instead of 3% of the current population.

A medicine leaves the bloodstream so that the amount remaining is modeled by

A(t) = 320(0.86)^t

, where

A

is in milligrams and

t

is in hours. Interpret the numbers 320 and 0.86 in context, then find how much remains after 6 hours.

Answer: 320 mg is the initial dose at $t = 0$ ; 0.86 means 86% of the medicine remains each hour, so the amount decreases 14% per hour. After 6 hours, $A(6) = 320(0.86)^6 \approx 129.5$ mg.

The base tells you what remains, and

1 - 0.86 = 0.14

tells you what is lost, so the hourly decay rate is 14% — not 86%. To evaluate, raise first:

0.86^6 \approx 0.404567

, then multiply by 320 to get about 129.5 mg. A quick sanity check: the amount should be less than half of 320 after 6 hours, since losing 14% per hour for six hours removes more than half.

An investment is modeled by

V(t) = 2500(1.045)^t

, with

t

in years. What is the annual percent increase, and why is the total increase after 10 years more than 45%?

Answer: The annual increase is 4.5%. After 10 years the value is $2500(1.045)^{10} \approx 3882.42$ , an increase of about 55.3%, more than 45% because each year's interest is computed on a larger balance.

Subtract 1 from the base:

1.045 - 1 = 0.045 = 4.5\%

. Multiplying

4.5\% \times 10 = 45\%

would be correct only if the same fixed dollar amount were added each year (a linear model). In an exponential model the growth compounds — year two's 4.5% is taken from the already-grown balance — so the total exceeds the simple product. Computing

1.045^{10} \approx 1.5530

shows the value grows to about 155.3% of the original.

FAQ

What is the difference between the growth rate and the growth factor?: The rate $r$ is the percent change written as a decimal; the factor $b$ is what you multiply by. They are related by $b = 1 + r$ for growth and $b = 1 - r$ for decay. For a 9% increase, the rate is 0.09 and the factor is 1.09. For a 9% decrease, the rate is 0.09 and the factor is 0.91.
How do I tell if an equation shows growth or decay?: Look at the base. If the base is greater than 1, the quantity grows; if it is between 0 and 1, the quantity decays. So $y = 40(1.2)^t$ grows 20% per period and $y = 40(0.8)^t$ decays 20% per period. Fractions count too: $y = 40\left(\frac{3}{4}\right)^t$ has base 0.75, so it decays 25% per period.
What if the percent change is given per month but the question asks about years?: Keep $t$ in the unit that matches the rate and convert the time before substituting. With a monthly rate, three years means $t = 36$ . Do not multiply the monthly percent by 12 to get an annual percent — compute $(1+r)^{12}$ instead, because the change compounds.
Why can't a decay model ever reach zero?: Each period you multiply by a positive number less than 1, and multiplying a positive number by a positive number never gives zero. The outputs get arbitrarily small and the graph approaches the horizontal asymptote $y = 0$ , but no value of $t$ makes the output exactly zero. In real contexts we usually ask when the amount drops below some small threshold instead.

Learn this with a teacher, not a page

The Crimsora tutor teaches Exponential Growth & Decay Models live — explaining on a whiteboard, asking you questions, and adapting to where you get stuck.

Try a live session free