Exponential Decay

Learn how quantities shrink by a constant percentage, use the decay formula y = a(1 - r)^t and solve depreciation, medicine and cooling problems.

Advanced25 min18 problems
Practice Problems

Definition

Exponential decay happens when a quantity decreases by the same percentage in every equal time period. The amount lost gets smaller each time, because each new loss is a percentage of what is left.
y=a(1−r)ty = a(1 - r)^t
Where:
  • aa = initial value (the amount at t=0t = 0)
  • rr = decay rate, the percent lost per period, written as a decimal (15%→r=0.1515\% \to r = 0.15)
  • 1−r1 - r = decay factor bb, the part that is kept each period (1−0.15=0.851 - 0.15 = 0.85)
  • tt = number of time periods
In the general form y=a⋅bty = a \cdot b^t you can tell the type from the base:
  • b>1b > 1: exponential growth
  • 0<b<10 < b < 1: exponential decay
After tt periods, the fraction that remains is (1−r)t(1 - r)^t, no matter how big the starting amount is.

Try it now

Which function shows exponential decay?

Worked Examples

For each function, decide whether it shows growth or decay and find the percent change per period: f(t)=300⋅0.92tf(t) = 300 \cdot 0.92^t and g(t)=50⋅1.03tg(t) = 50 \cdot 1.03^t.
1

Look at the base of ff

b=0.92b = 0.92 and 0<0.92<10 < 0.92 < 1 → ff shows decay
2

Find the decay rate of ff

r=1−b=1−0.92=0.08r = 1 - b = 1 - 0.92 = 0.08 → ff loses 8%8\% per period
3

Look at the base of gg

b=1.03>1b = 1.03 > 1, so r=1.03−1=0.03r = 1.03 - 1 = 0.03 → gg grows 3%3\% per period
4

Read the initial values

f(0)=300⋅0.920=300f(0) = 300 \cdot 0.92^0 = 300 and g(0)=50g(0) = 50 → Starting values 300300 and 5050

Common Mistakes

Using the decay rate as the base: writing 25000⋅0.15t25000 \cdot 0.15^t for a 15%15\% loss

Why it's wrong: 0.15t0.15^t keeps only 15%15\% each year, which means losing 85%85\% per year. The base must be the part that is kept.

Correct: Decay factor =1−r= 1 - r. A 15%15\% decrease gives 25000⋅0.85t25000 \cdot 0.85^t.

Subtracting the same amount every period: '20% of 1000 is 200, so after 5 years the value is 0'

Why it's wrong: That is linear thinking. Each year's loss is 20%20\% of the current value, not of the original value, so the losses shrink.

Correct: Multiply by 0.80.8 each year: 1000→800→640→512→409.6→327.681000 \to 800 \to 640 \to 512 \to 409.6 \to 327.68. After 5 years 327.68327.68 is left.

Computing a⋅1−rta \cdot 1 - r^t instead of a(1−r)ta(1 - r)^t

Why it's wrong: The exponent belongs to the whole decay factor. 1−0.152=0.97751 - 0.15^2 = 0.9775, but (1−0.15)2=0.7225(1 - 0.15)^2 = 0.7225.

Correct: Find 1−r1 - r first, then raise it to the power tt, then multiply by aa.

Believing a decaying quantity eventually reaches exactly zero

Why it's wrong: Multiplying a positive number by a factor like 0.80.8 always gives a positive number.

Correct: In the model y=a(1−r)ty = a(1 - r)^t the values get closer and closer to 00 (the horizontal asymptote y=0y = 0) but never reach it.

Interactive Visual

Function graph

0246810120100200300400tyy = 0

Tap or hover over the graph to read off values.

y = a(1 − r)^tAsymptote

Table of values

t012345
y400320256204.8163.84131.072

Sequence Explorer

100,
80,
64,
51.2,
40.96,
32.768,
26.214,
20.972
, ...

Formula

a_n = 100 × 0.8^(n-1)
12345678
a_1

100

a_5

40.96

a_10

13.422

a_20

1.441

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Expression Calculator

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Practice Problems

18 problems
Problem 1 of 18
Easy
Which function shows exponential decay?

Why It Matters

Exponential decay describes many things that shrink over time:
  • Depreciation: A new car or phone loses a percentage of its value every year, so it loses the most money at the start
  • Medicine: The body removes a fixed percentage of a drug each hour, which decides how often a dose is needed
  • Cooling: A hot drink cools quickly at first and then more and more slowly as it nears room temperature
  • Light and sound: Each meter of water or wall absorbs a percentage of the light or sound passing through
Knowing the decay formula lets you predict values, compare offers (a fixed amount off per year versus a percentage off) and understand why some things never quite reach zero.

Real World Applications

Depreciation of Phones and Laptops

Electronics lose a percentage of their value each year. Resale websites and insurers use decay models to estimate what a used device is worth.

Example:

A 1000-dollar laptop that loses 30%30\% per year is worth 1000⋅0.72=4901000 \cdot 0.7^2 = 490 dollars after 2 years.

1Try It Yourself

A new phone costs 1200 dollars and loses 25%25\% of its value each year.

What is the phone worth after 2 years?

Step 1: Write the mathematical expression

Use the decay formula: 1200⋅0.7521200 \cdot 0.75^2

Medicine Dosing

Doctors and pharmacists use exponential decay to know how much of a drug is still in the body and when the next dose should be taken.

Example:

If the body removes 15%15\% of a drug each hour, then after 4 hours 0.854≈0.5220.85^4 \approx 0.522, so about 52.2%52.2\% of the dose remains.

2Try It Yourself

A patient takes 200 mg of a medicine. The body removes 30%30\% of the remaining amount every hour.

How many milligrams are left after 4 hours?

Step 1: Write the mathematical expression

Calculate: 200⋅0.74200 \cdot 0.7^4

Cooling Drinks

Engineers and food scientists model cooling with decay: the temperature difference between an object and the room shrinks by a percentage each minute.

Example:

Coffee at 90°C90°C in a 20°C20°C room: if the difference of 7070 shrinks 10%10\% per minute, after 1 minute it is 6363, so the coffee is 83°C83°C.

3Try It Yourself

A cup of tea is 80°C80°C in a room at 20°C20°C. In a simple model the temperature difference shrinks by 20%20\% every minute.

What is the tea's temperature after 3 minutes?

Step 1: Write the mathematical expression

Calculate: 20+60⋅0.8320 + 60 \cdot 0.8^3

Key Takeaways

  • 1Exponential decay follows y=a(1−r)ty = a(1 - r)^t: the quantity loses the same percentage every period
  • 2The decay rate rr is the percent lost; the decay factor 1−r1 - r is the part kept (a 15%15\% decrease gives 0.850.85)
  • 3In y=a⋅bty = a \cdot b^t, a base 0<b<10 < b < 1 means decay and b>1b > 1 means growth
  • 4After tt periods the fraction remaining is (1−r)t(1 - r)^t; multiply by 100%100\% for the percent remaining
  • 5Linear decrease subtracts the same amount and can reach zero; exponential decrease loses less each period and only approaches zero
  • 6Round only at the end and state the rounding (nearest cent, one decimal place)

Frequently Asked Questions

The decay rate rr is what you lose each period (for example 15%=0.1515\% = 0.15). The decay factor is what you keep: 1−r=0.851 - r = 0.85. The decay factor is the number that goes into the base of the power.
The decay rate rr is what you lose each period (for example 15%=0.1515\% = 0.15). The decay factor is what you keep: 1−r=0.851 - r = 0.85. The decay factor is the number that goes into the base of the power.
Look at the base bb in y=a⋅bty = a \cdot b^t. If b>1b > 1, the quantity grows (for example 1.041.04 means +4%+4\%). If 0<b<10 < b < 1, it decays (for example 0.960.96 means −4%-4\%).
Not in the model. Each period keeps a fixed fraction of a positive number, so the result stays positive. The graph gets closer and closer to the line y=0y = 0, its horizontal asymptote. In real life we stop counting once the amount is too small to matter.
The half-life is the time it takes for exactly half of the quantity to be left. It is a special case of exponential decay with its own formula, covered in the lesson Half-Life Problems.

Glossary

Exponential decay
A decrease by the same percentage over equal time periods, modeled by y=a(1−r)ty = a(1 - r)^t
Decay rate
The percent lost per period, written as a decimal rr (for example 0.150.15 for 15%15\%)
Decay factor
The part kept each period, b=1−rb = 1 - r, with 0<b<10 < b < 1
Initial value
The amount aa at time t=0t = 0
Depreciation
The loss of value of an object, such as a car or a phone, over time
Horizontal asymptote
The line y=0y = 0 that a decay graph approaches but never touches

Formula Card

Exponential Decay Formula

y=a(1−r)ty = a(1 - r)^t

$a$ = initial value, $r$ = decay rate as a decimal, $1 - r$ = decay factor, $t$ = number of periods.

Decay Factor from Percent

b=1−rb = 1 - r

A decrease of $p\%$ per period gives $b = 1 - \frac{p}{100}$. Example: $15\% \to 0.85$.

Percent Remaining

(1−r)t×100%(1 - r)^t \times 100\%

The percent of the starting amount left after $t$ periods, whatever the starting amount is.

Today's challengeOne problem, three tries

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