How do I know if a real-world situation is exponential growth or just linear growth?

Exponential growth multiplies by the same factor each period (like a population growing by 10% each year), while linear growth adds the same amount each period (like saving $100 per month). If the rate is given as a percentage, it is almost certainly exponential.

Why use e in the formula instead of 2, 10, or some other base?

e \approx 2.71828 is special because A = A_0 e^{kt} represents growth happening at every instant in time, not just at discrete intervals. This continuous model fits natural and financial phenomena better than step-wise growth with other bases. Most textbooks and real applications use e for this reason.

What happens if k = 0?

When k = 0, the exponent becomes 0 \cdot t = 0 regardless of t, so e^0 = 1 and A = A_0 \cdot 1 = A_0 for all time. The amount never changes; there is no growth or decay.

Can t be negative, and what does that mean?

Yes, negative t is valid; it means you are looking backward in time. If t = -3 and k = 0.1, then A = A_0 e^{-0.3}, which tells you the amount 3 time units ago. This is useful when you know the current amount and want to calculate what it was in the past.

Exponential Growth

Use Exponential Growth when a quantity grows by multiplying by a constant factor each time period, like bacteria populations or money earning compound interest.

A = A_0 e^{kt}, \quad k > 0

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What each symbol means

What Exponential Growth takes

Symbol	Meaning
$A$	The amount after time has elapsed; what you measure at time $t$ . Do not confuse with $A_0$ or you will report the starting value as your final answer.
$A_0$	The starting amount at time $t = 0$ , before growth begins; always written with subscript 0. Forgetting the subscript can lead you to use the wrong initial value.
$k$	The growth rate constant (positive for growth), with units of inverse time such as per year or per hour. If $k < 0$ you have decay instead; if $k = 0$ there is no change.
$t$	The elapsed time since $t = 0$ , measured in the same units that define $k$ . If $k$ is per year, $t$ must be in years; mixing units gives wrong answers.

When to use it

Reach for this model when something multiplies by the same percentage each time step, such as populations, investments, or radioactive decay.

Level

Usually taught in: Algebra II

Worked examples

1. Calculate population size after a given time

Problem

A bacteria colony starts with 100 cells and grows exponentially with

k = 0.1

per hour. How many cells are present after 5 hours?

$A = A_0 e^{kt}$
Write the exponential growth formula.
$A = 100 e^{0.1 \cdot 5}$
Substitute $A_0 = 100$ , $k = 0.1$ , and $t = 5$ into the formula.
$A = 100 e^{0.5}$
Simplify the exponent by multiplying $0.1 \cdot 5 = 0.5$ .
$A \approx 100 \cdot 1.6487 = 164.87$
Use a calculator to find $e^{0.5} \approx 1.6487$ , then multiply by 100.

Answer: $A \approx 164.87 \text{ cells}$

We substitute the known values into the exponential growth formula and evaluate. The colony has roughly 165 cells after 5 hours because continuous growth compounds the increase over time.

2. Solve for time when population reaches a target

Problem

A population starts at 5000 people and grows continuously at

k = \ln 2 / 10

per year (doubling every 10 years). How many years until the population reaches 20000?

$20000 = 5000 e^{\frac{\ln 2}{10} \cdot t}$
Set up the equation with $A = 20000$ , $A_0 = 5000$ , and the given $k$ .
$4 = e^{\frac{\ln 2}{10} \cdot t}$
Divide both sides by 5000 to isolate the exponential.
$\ln 4 = \frac{\ln 2}{10} \cdot t$
Take the natural logarithm of both sides to bring $t$ out of the exponent.
$t = \frac{10 \ln 4}{\ln 2} = 10 \cdot \frac{\ln 4}{\ln 2} = 10 \cdot 2 = 20$
Divide both sides by $\ln 2 / 10$ . Since $\ln 4 = \ln 2^2 = 2 \ln 2$ , we get $\frac{\ln 4}{\ln 2} = 2$ .

Answer: $t = 20 \text{ years}$

Starting from 5000, the population quadruples to reach 20000. Since the population doubles every 10 years, it doubles twice in 20 years ( $5000 \to 10000 \to 20000$ ), confirming our answer.

3. Calculate investment return with continuous compounding

Problem

You deposit

\$1000

into a savings account earning 8% annual interest, compounded continuously. Using

A = A_0 e^{kt}

with

k = 0.08

, how much money will you have after 5 years? Round to the nearest cent.

$A = A_0 e^{kt}$
Write the continuous-compound-interest formula, which is the exponential growth formula.
$A = 1000 e^{0.08 \cdot 5}$
Substitute $A_0 = 1000$ dollars, $k = 0.08$ per year, and $t = 5$ years.
$A = 1000 e^{0.4}$
Simplify the exponent: $0.08 \cdot 5 = 0.4$ .
$A \approx 1000 \cdot 1.4918 = 1491.82$
Calculate $e^{0.4} \approx 1.4918$ with a calculator, then multiply by 1000 and round to the nearest cent.

Answer: $A \approx \$1491.82$

Continuous compounding means interest is earned on interest at every instant, not just annually. Over 5 years, your $1000 grows to approximately $1491.82, an increase of about 49%, because the 8% rate is applied continuously.

Common mistakes

Where Exponential Growth usually goes wrong

The mistake	Why it is wrong	The fix
Writing $A = A_0 e^k$ (forgetting the $t$ in the exponent)	This formula gives a constant amount no matter how much time passes, so it fails to model growth.	Always include $t$ : $A = A_0 e^{kt}$ . The product $kt$ is the exponent, which changes as time increases.
Using a negative rate like $k = -0.05$ to represent 5% growth	A negative $k$ produces exponential decay (the amount shrinks), not growth. Negative rates describe radioactive decay or population decline.	Use $k = 0.05$ for 5% growth. Reserve negative $k$ for decay problems, where you label it explicitly as decay.
Writing $k = 5$ when the growth rate is 5% per year	A percentage is not a decimal rate. The percentage 5% converts to the decimal $0.05$ ; using $k = 5$ treats growth as 500% per year, which is incorrect.	Convert percentage to decimal: $5\% = 0.05$ , so $k = 0.05$ . For any percentage $p\%$ , use $k = p / 100$ .

Tips and when to use something else

Check your units carefully: if $k$ is per year, $t$ must be in years. Mixing different time units will give nonsensical answers.
For discrete growth (like doubling every month), convert to continuous rate using $k = \frac{\ln(\text{factor})}{\text{period}}$ ; for doubling, that is $k = \ln 2 \approx 0.693$ .
When you need to solve for $t$ , take the natural logarithm of both sides to free $t$ from the exponent: $\ln(e^{kt}) = kt$ .
Do not confuse this with Linear Growth ( $A = A_0 + kt$ ), which adds a fixed amount each period and grows much slower than exponential growth.

Related formulas

Natural LogarithmNatural logarithm finds the power to which e must be raised to get a number, and it's essential for solving equations involving exponential functions.Euler's NumberEuler's Number e is a fundamental constant approximately equal to 2.71828 that models continuous exponential growth and decay.Product Rule for ExponentsThe Product Rule for Exponents simplifies multiplication of powers with the same base by adding the exponents into a single power.

Frequently asked questions

How do I know if a real-world situation is exponential growth or just linear growth?: Exponential growth multiplies by the same factor each period (like a population growing by 10% each year), while linear growth adds the same amount each period (like saving $100 per month). If the rate is given as a percentage, it is almost certainly exponential.
Why use $e$ in the formula instead of 2, 10, or some other base?: $e \approx 2.71828$ is special because $A = A_0 e^{kt}$ represents growth happening at every instant in time, not just at discrete intervals. This continuous model fits natural and financial phenomena better than step-wise growth with other bases. Most textbooks and real applications use $e$ for this reason.
What happens if $k = 0$ ?: When $k = 0$ , the exponent becomes $0 \cdot t = 0$ regardless of $t$ , so $e^0 = 1$ and $A = A_0 \cdot 1 = A_0$ for all time. The amount never changes; there is no growth or decay.
Can $t$ be negative, and what does that mean?: Yes, negative $t$ is valid; it means you are looking backward in time. If $t = -3$ and $k = 0.1$ , then $A = A_0 e^{-0.3}$ , which tells you the amount 3 time units ago. This is useful when you know the current amount and want to calculate what it was in the past.

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Reviewed 2026-09-18