How much more does continuous compounding really give?

Not much more, but always more. At 5% for one year, annual compounding multiplies by $1.05$ and continuous compounding by $e^{0.05} \approx 1.0513$. Over long periods and large amounts the difference becomes noticeable.

Is $A = Pe^{-kt}$ the same as $A = Pe^{rt}$ with a negative $r$?

Yes. Some books write decay with a positive decay rate $k$ and a minus sign, others use one formula $Pe^{rt}$ and let $r$ be negative, for example $r = -0.03$. Both give the same result.

How do I find how long it takes to reach a certain amount?

That means solving $A = Pe^{rt}$ for $t$, which needs the natural logarithm $\ln$. You will learn it in the logarithm lessons. For now you can estimate $t$ by trying values in a table.

Teacher Resources

Continuous Growth and Decay

Q: Why do we use $e$ for continuous growth?

Compounding $n$ times a year at rate $r$ gives $\left(1 + \frac{r}{n}\right)^{nt}$. As $n$ gets larger and larger, this approaches $e^{rt}$. So $e$ is what you get when growth happens at every instant.

Use A = Pe^(rt) and A = Pe^(-kt) to model continuous compounding, bacteria growth and radioactive decay, and compare continuous with annual compounding.

Advanced30 minLesson

Printable worksheet

Definition

In the lesson on the natural base

e

you saw that compounding more and more often,

\left(1 + \frac{1}{n}\right)^n

, approaches

e \approx 2.71828

. When a quantity grows continuously (every instant, not once a year), we use

e

as the base.

Continuous growth:

A = P e^{rt}

Continuous decay:

A = P e^{-kt}

where:

$P$ is the initial amount (at $t = 0$ )
$r$ is the continuous growth rate and $k$ the continuous decay rate, both written as decimals ( $5\% = 0.05$ )
$t$ is the time, in the same unit as the rate (years, hours, days)
$A$ is the amount after time $t$

The sign of the exponent tells you the direction: a positive exponent means growth, a negative exponent means decay. You can also write decay as

A = Pe^{rt}

with a negative rate, for example

r = -0.03

In the calculator you can type

e^{x}

directly as e^(x), for example e^0.5 gives

\approx 1.6487

. On a basic calculator without an

e

key, type

2.71828^{x}

instead.

Try it now

A substance decays continuously at a rate of 4% per year. Which formula models the amount

A

after

t

years, starting from

P

Worked Examples

You deposit 1000 dollars in an account that pays 5% interest compounded continuously. How much is in the account after 10 years? Round to the nearest cent.

Identify the values

$P = 1000$ , $r = 0.05$ , $t = 10$ → Values identified

Find the exponent $rt$

$rt = 0.05 \times 10 = 0.5$ → $rt = 0.5$

Find $e^{0.5}$ with the calculator

$2.71828^{0.5} \approx 1.648721$ → $e^{0.5} \approx 1.648721$

Multiply by $P$

$A = 1000 \times 1.648721 \approx 1648.72$ → $A \approx 1648.72$

Answer: After 10 years the account holds about 1648.72 dollars.

Common Mistakes

Writing the rate as a percent in the exponent: $1000 e^{5 \times 10}$

Why it's wrong: The formula needs the rate as a decimal. $e^{50}$ is a gigantic number (about $5 \times 10^{21}$ ), which makes no sense for a savings account.

Correct: Convert first: $5\% = 0.05$ , so $A = 1000 e^{0.05 \times 10} = 1000 e^{0.5}$ .

Forgetting the minus sign for decay: $80 e^{0.03 \times 20}$

Why it's wrong: Without the minus sign the exponent is positive and the amount grows to about 145.77 grams, more than the 80 grams you started with.

Correct: Decay needs a negative exponent: $A = 80 e^{-0.03 \times 20} = 80 e^{-0.6} \approx 43.90$ .

Multiplying before raising to the power: $(1000 \times 2.71828)^{0.5}$

Why it's wrong: The exponent belongs only to $e$ , not to $P$ . Order of operations says powers come before multiplication.

Correct: First compute $e^{rt}$ , then multiply by $P$ . On the calculator: $1000 \times e^{(0.05 \times 10)}$ , with brackets around the exponent.

Using $(1 + r)^t$ when the problem says compounded continuously

Why it's wrong: $(1 + r)^t$ is compounding once per period. Continuous compounding always gives a little more at the same rate.

Correct: Look for the words continuously or continuous rate: they mean $Pe^{rt}$ . Once a year means $P(1 + r)^t$ .

Interactive Visual

Function graph

P = 1000r = 0.05

Tap or hover over the graph to read off values.

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Asymptote

Table of values

t	0	5	10	15	20
A = P·e^(rt)	1000	1284.03	1648.72	2117	2718.28
A = P(1 + r)^t	1000	1276.28	1628.89	2078.93	2653.3

Interest Calculator

Principal ($1,000.00)

Interest Rate (5%)

Time Period (10 years)

Compounding

Formula

Compound Interest:A = Pe^(rt)A = 1000 · e^(0.05 · 10) = $1,648.72

Compound Interest

Interest Earned:$648.72

Total Amount:$1,648.72

Same money, different compounding

Compounding	n	Total Amount	Interest Earned
Annually	1	$1,628.89	$628.89
Monthly	12	$1,647.01	$647.01
Daily	365	$1,648.66	$648.66
Continuously	∞	$1,648.72	$648.72

Growth Over Time

Interactive Sandbox

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History

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Practice Problems

18 problems

Problem 1 of 18

Easy

A substance decays continuously at a rate of 4% per year. Which formula models the amount $A$ after $t$ years, starting from $P$ ?

Practice Problems: Continuous Growth and Decay

1. A substance decays continuously at a rate of 4% per year. Which formula models the amount A after t years, starting from P?
2. In the model A = 250e^{0.07t}, what is the initial amount?
3. Use A = Pe^{rt} with P = 1000, r = 0.04 and t = 5. Use e^{0.2} \approx 1.2214. What is A? Round to one decimal place.
4. A population follows N = 300e^{0.12t}, with t in years. What is the continuous growth rate as a percent? Write only the number, without the % sign.
5. You invest 1500 dollars at 3% compounded continuously for 10 years. Use e^{0.3} \approx 1.3499. How much do you have at the end? Round to the nearest dollar.
6. A culture of 400 bacteria grows continuously at 25% per hour. How many bacteria are there after 4 hours? Use e ≈ 2.71828 and round to the nearest whole number.
7. What does the model N = 50e^{-0.3t} describe?
8. Two accounts start with the same amount, the same interest rate and the same time. One compounds annually, the other continuously. Which one ends with more money?
9. 200 grams of a radioactive substance decay continuously at a rate of 5% per year. Use e^{-0.5} \approx 0.6065. How many grams are left after 10 years? Round to one decimal place.
10. How much must you deposit now, at 5% compounded continuously, to have 10000 dollars after 10 years? Use e^{0.5} \approx 1.6487 and round to the nearest dollar.
11. A patient receives 60 mg of a medicine. The body removes it continuously at a rate of 20% per hour. Use e^{-0.6} \approx 0.5488. How many mg have been removed after 3 hours? Round to one decimal place.
12. You invest 3000 dollars at 4% for 5 years. Use 1.04^{5} \approx 1.2167 and e^{0.2} \approx 1.2214. Compare annual and continuous compounding. Round each amount to the nearest dollar.
13. How much must you deposit today, at 4% compounded continuously, to have 6000 dollars after 5 years? Use the calculator (e ≈ 2.71828) and round the final answer to the nearest dollar.
14. Offer A: 6% compounded annually. Offer B: 5.85% compounded continuously. Use e^{0.0585} \approx 1.0602. Which offer gives more money after one year?
15. 5000 dollars are invested at 8% for 10 years. Use 1.08^{10} \approx 2.158925 and e^{0.8} \approx 2.225541. How many dollars more does continuous compounding give than annual compounding? Round to the nearest dollar.
16. You want to invest 2000 dollars for 10 years. Bank A pays 5% compounded annually, Bank B pays 4.9% compounded continuously. Use 1.05^{10} \approx 1.6289 and e^{0.49} \approx 1.6323. Round each amount to the nearest dollar. How many dollars more does the better bank give?
17. A sample of 500 grams of a radioactive substance decays continuously at 2% per year. Use e⁻¹ ≈ 0.3679. How many grams are left after 50 years? Round to the nearest gram.
18. A culture starts with 1000 bacteria and grows continuously at 30% per hour. Use e^{0.6} \approx 1.8221 and e^{1.2} \approx 3.3201. By what factor does the culture grow from hour 2 to hour 4? First find the number of bacteria after 2 hours and after 4 hours (nearest whole number), then give the factor to two decimal places.

Why It Matters

Many processes do not wait for the end of the year to change: they change at every moment.

Finance: some savings accounts and loans compound interest continuously, and $Pe^{rt}$ is the upper limit of what any compounding schedule can give at the same rate
Biology: bacteria in a lab divide all the time, not on a timetable, so their growth is modeled with $e$
Physics and medicine: radioactive atoms decay at random moments, and a drug leaves the bloodstream gradually, so both follow $Pe^{-kt}$
Science in general: almost every growth or decay formula in chemistry, physics and economics is written with base $e$

Once you can read

Pe^{rt}

, you can compare bank offers, predict lab results and understand how scientists date old objects.

Real World Applications

Banking: Continuously Compounded Savings

Banks compare offers using the most generous possible compounding, which is continuous. It shows the ceiling of what a given rate can earn.

Example:

At 5% compounded continuously, 1 dollar grows to $e^{0.05} \approx 1.0513$ dollars in a year, so the effective yearly rate is about 5.13%, a bit more than 5%.

1Try It Yourself

You put 4000 dollars into an account paying 4.5% compounded continuously and leave it for 4 years.

How much is in the account after 4 years? Round to the nearest dollar.

Step 1: Write the mathematical expression

Calculate $4000 \times e^{0.045 \times 4}$ :

Biology: Bacteria in the Lab

Microbiologists model a bacteria culture that has plenty of food with $N(t) = N_0 e^{rt}$, where $r$ is the continuous growth rate per hour.

Example:

A culture of 1000 bacteria growing at a continuous rate of 30% per hour has $1000 e^{0.6} \approx 1822$ bacteria after 2 hours.

2Try It Yourself

A lab culture starts with 2000 bacteria and grows continuously at 40% per hour.

How many bacteria are there after 5 hours? Round to the nearest whole number.

Step 1: Write the mathematical expression

Calculate $2000 \times e^{0.4 \times 5}$ :

Medicine: Radioactive Tracers

Hospitals use radioactive substances such as iodine-131 for diagnosis and therapy. The amount left in a sample follows $A = Pe^{-kt}$, so doctors know how much remains on any day.

Example:

Iodine-131 decays continuously at about 8.6% per day. After 1 day, $e^{-0.086} \approx 0.918$ , so about 91.8% of the sample is left.

3Try It Yourself

A hospital receives 100 mg of iodine-131, which decays continuously at a rate of 8.6% per day.

How much is left after 10 days? Round to one decimal place.

Step 1: Write the mathematical expression

Calculate $100 \times e^{-0.086 \times 10}$ :

Key Takeaways

1Continuous growth: $A = Pe^{rt}$ ; continuous decay: $A = Pe^{-kt}$ , with rates written as decimals
2A positive exponent means growth, a negative exponent means decay
3In the calculator, type $e^{x}$ as e^(x) (or $2.71828^{x}$ without an $e$ key) and put brackets around the exponent
4At the same rate and time, continuous compounding always gives a little more than annual compounding
5To find the initial amount, divide: $P = \frac{A}{e^{rt}}$
6Solving for the time $t$ needs logarithms, which come in later lessons

Frequently Asked Questions

Compounding

n

times a year at rate

r

gives

\left(1 + \frac{r}{n}\right)^{nt}

. As

n

gets larger and larger, this approaches

e^{rt}

. So

e

is what you get when growth happens at every instant.

Glossary

Continuous growth: Growth that happens at every instant, modeled by $A = Pe^{rt}$ with $r > 0$
Continuous decay: Decrease that happens at every instant, modeled by $A = Pe^{-kt}$ with $k > 0$
Continuous compounding: Interest added at every instant; the limit of compounding more and more often
Continuous rate: The rate $r$ or $k$ in the exponent, written as a decimal ( $4\% = 0.04$ )
Euler's number $e$: The constant $e \approx 2.71828$ , the natural base for continuous growth and decay
Initial amount: The starting value $P$ at time $t = 0$

Formula Card

Continuous growth

A = P e^{rt}

$P$ = initial amount, $r$ = continuous growth rate (decimal), $t$ = time, $A$ = amount after time $t$.

Continuous decay

A = P e^{-kt}

$k$ = continuous decay rate (decimal, positive). The minus sign makes the amount shrink.

Annual compounding (for comparison)

A = P(1 + r)^t

Interest added once per year. Gives slightly less than $Pe^{rt}$ at the same rate.

Initial amount

P = \frac{A}{e^{rt}}

Divide the final amount by the growth factor $e^{rt}$.

Continuous Growth and Decay

Definition

Worked Examples

Common Mistakes

Interactive Visual

Function graph

Table of values

Interest Calculator

Compound Interest

Same money, different compounding

Growth Over Time

Interactive Sandbox

Expression Calculator

History

Practice Problems

Why It Matters

Real World Applications

Banking: Continuously Compounded Savings

Biology: Bacteria in the Lab

Medicine: Radioactive Tracers

Key Takeaways

Frequently Asked Questions

Glossary

Formula Card

Continuous growth

Continuous decay

Annual compounding (for comparison)

Initial amount

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Exponential Growth

Graphing Exponential Functions

Exponential Decay