When should I use exponential decay vs. the half-life formula?

Exponential decay A = A_0 e^{-kt} is the general formula and works when you know (or can calculate) the decay constant k. The half-life formula A = A_0 (\tfrac{1}{2})^{t/t_{1/2}} is a convenient shortcut when you know how long it takes for the amount to halve. Both describe the same process; choose whichever matches your given information.

Why is k always positive in the formula A = A_0 e^{-kt}?

The negative sign in front of k is what makes the exponent negative, causing the quantity to shrink. If k itself were negative, the formula would become A = A_0 e^{kt}, which is growth. By convention, k is defined as a positive rate, and the decay formula incorporates the negative sign.

How do I find k if I only know the half-life?

If the half-life is h (the time for the amount to halve), then at time t = h, you have A = \tfrac{1}{2} A_0. Substituting into the formula: \tfrac{1}{2} A_0 = A_0 e^{-kh}, which simplifies to 0.5 = e^{-kh}. Taking \ln: \ln(0.5) = -kh, so k = -\ln(0.5)/h = \ln(2)/h \approx 0.693/h.

Can I use exponential decay for real populations?

Yes, but only for declining populations where the number of deaths is proportional to the current population (like radioactive atoms decaying). For real human or animal populations, births must also be accounted for, making the model more complex. Exponential decay works best for declining populations under environmental stress with no reproduction, such as bacteria under starvation.

Exponential Decay

Exponential Decay models quantities that shrink at a rate proportional to their current amount, used for radioactive decay, cooling, and depreciation.

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

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What each symbol means

What Exponential Decay takes

Symbol	Meaning
$A$	The amount remaining at time $t$ , measured in the same units as $A_0$ (grams, liters, organisms, etc.); if you use $A$ as the starting amount instead of $A_0$ , your predictions will be backwards.
$A_0$	The starting quantity at time $t = 0$ ; must be positive and measured in the same units as $A$ ; confusing $A_0$ with $A$ reverses the direction of decay entirely.
$k$	The decay constant determining how fast the quantity shrinks; always positive, with units of inverse time (per day, per year, etc.); if $k$ is negative, the formula becomes growth instead of decay.
$t$	The time elapsed from the start ( $t = 0$ ), measured in the same time units as $k$ 's rate (hours if $k$ is per hour); must be non-negative; using negative time values has no real-world meaning and produces incorrect results.

When to use it

Use exponential decay when a quantity decreases over time in proportion to how much remains.

Level

Usually taught in: Algebra II

Worked examples

1. Find remaining amount with given decay constant

Problem

A radioactive sample has an initial mass of 100 grams. If

k = 0.05

per day, how much will remain after 3 days?

$A = 100 e^{-0.05 \cdot 3}$
Substitute $A_0 = 100$ , $k = 0.05$ , and $t = 3$ into the exponential decay formula.
$A = 100 e^{-0.15}$
Simplify the exponent by multiplying $0.05 \cdot 3 = 0.15$ .
$A \approx 100 \cdot 0.8607$
Use a calculator to evaluate $e^{-0.15} \approx 0.8607$ .
$A \approx 86.07 \text{ grams}$
Multiply to obtain the final answer.

Answer: $A \approx 86.07 \text{ grams}$

We substitute the given values directly into the formula, simplify the exponent, and use a calculator to evaluate the exponential. This is the straightforward approach when you know the decay constant and want to find the remaining amount after a specific time.

2. Find the decay constant from measurements

Problem

A sample of 500 mg decays to 150 mg over 20 hours. Find the decay constant

k

$150 = 500 e^{-20k}$
Substitute $A = 150$ , $A_0 = 500$ , and $t = 20$ into the exponential decay formula.
$\frac{150}{500} = e^{-20k}$
Divide both sides by 500.
$0.3 = e^{-20k}$
Simplify the fraction.
$\ln(0.3) = -20k$
Take the natural logarithm of both sides to isolate the exponent.
$-1.2040 \approx -20k$
Use a calculator to evaluate $\ln(0.3) \approx -1.2040$ .
$k \approx 0.0602 \text{ per hour}$
Divide both sides by -20 to solve for $k$ .

Answer: $k \approx 0.0602 \text{ per hour}$

When given measurements at different times, use them to set up an equation and solve for the unknown decay constant using logarithms. This is essential for making predictions about future decay, since the rate of decay must be determined from the data you have.

3. Predict amount after two driving legs using measured decay

Problem

On a road trip, you start with 480 mL of fuel additive in your tank. After a 4-hour highway leg, the level drops to 360 mL. If additive depletes exponentially due to engine consumption, how much will remain after driving an additional 6 hours on mountain roads?

$360 = 480 e^{-4k}$
Using data from the first leg, substitute $A = 360$ , $A_0 = 480$ , and $t = 4$ to find $k$ .
$\frac{360}{480} = e^{-4k}$
Divide both sides by 480.
$0.75 = e^{-4k}$
Simplify the fraction.
$\ln(0.75) = -4k$
Take the natural logarithm of both sides.
$-0.2877 \approx -4k$
Use a calculator to evaluate $\ln(0.75) \approx -0.2877$ .
$k \approx 0.0719 \text{ per hour}$
Divide both sides by -4.
$A = 480 e^{-0.0719 \cdot 10}$
For both legs combined ( $4 + 6 = 10$ hours total), find $A$ using the calculated $k$ .
$A = 480 e^{-0.719}$
Simplify the exponent: $0.0719 \cdot 10 = 0.719$ .
$A \approx 480 \cdot 0.4874$
Use a calculator to evaluate $e^{-0.719} \approx 0.4874$ .
$A \approx 233.95 \text{ mL}$
Multiply to obtain the final answer.

Answer: $A \approx 233.95 \text{ mL}$

Real-world decay problems often require two steps: first determine the decay constant from observed measurements, then use it to predict the amount at a future time. This problem spans two driving legs to illustrate how the same decay constant applies throughout.

Common mistakes

Where Exponential Decay usually goes wrong

The mistake	Why it is wrong	The fix
$A = A_0 e^{kt}$	This formula has a positive exponent, so as $t$ increases, the exponent becomes more positive and $e^{kt}$ grows, making $A$ larger — this is exponential growth, not decay.	Use $A = A_0 e^{-kt}$ with a negative sign in front of $kt$ to ensure the exponent is negative and $A$ shrinks over time.
When solving $0.5 = e^{-20k}$ , students write $\ln(0.5) = 20k$ instead of $\ln(0.5) = -20k$ .	The negative sign in the exponent is crucial: taking $\ln$ of both sides of $0.5 = e^{-20k}$ gives $\ln(0.5) = -20k$ , not $20k$ . Dropping the negative sign flips the sign of the final answer for $k$ .	When taking the natural log of $e^{-20k}$ , the result is $-20k$ (the exponent itself). Then divide by -20: $k = \ln(0.5) / (-20) \approx 0.0347$ .
Using $k$ measured in per year with $t$ measured in days without unit conversion (e.g., $A = 100 e^{-0.05 \cdot 30}$ when $k = 0.05$ per year and $t = 30$ days).	The units of $k$ and $t$ must be compatible: if $k$ is per year, then $t$ must be in years; if $t$ is in days, $k$ must be per day. Mixing units makes the exponent meaningless and gives wildly wrong answers.	Convert to consistent units first; for example, convert 30 days to $30/365 \approx 0.082$ years, then $A = 100 e^{-0.05 \cdot 0.082}$ .

Tips and when to use something else

The negative sign in $A = A_0 e^{-kt}$ is essential for decay. If you see $e^{kt}$ with $k > 0$ , that is exponential growth—use a different approach.
To find $k$ from data, set up the equation with known values of $A$ , $A_0$ , and $t$ , then solve using natural logarithms: take $\ln$ of both sides to bring the exponent down as a coefficient.
If you know the half-life (time for the amount to halve), you can use the Half-Life Formula $A = A_0 (\tfrac{1}{2})^{t/t_{1/2}}$ , which often avoids calculators for simple cases.
Exponential decay applies when the rate of decrease is proportional to the current amount (radioactive decay, cooling, population decline). If the amount decreases by a fixed number per time step, use Linear Decay $A = A_0 - ct$ instead.

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.Exponential GrowthUse Exponential Growth when a quantity grows by multiplying by a constant factor each time period, like bacteria populations or money earning compound interest.Half-Life FormulaThe Half-Life Formula determines how much of a radioactive or decaying substance remains after a given time period has passed.Euler's NumberEuler's Number e is a fundamental constant approximately equal to 2.71828 that models continuous exponential growth and decay.Power Rule for LogarithmsThe Power Rule for Logarithms lets you pull an exponent out of a logarithm as a multiplier, making complex logs easier to solve and simplify.

Frequently asked questions

When should I use exponential decay vs. the half-life formula?: Exponential decay $A = A_0 e^{-kt}$ is the general formula and works when you know (or can calculate) the decay constant $k$ . The half-life formula $A = A_0 (\tfrac{1}{2})^{t/t_{1/2}}$ is a convenient shortcut when you know how long it takes for the amount to halve. Both describe the same process; choose whichever matches your given information.
Why is $k$ always positive in the formula $A = A_0 e^{-kt}$ ?: The negative sign in front of $k$ is what makes the exponent negative, causing the quantity to shrink. If $k$ itself were negative, the formula would become $A = A_0 e^{kt}$ , which is growth. By convention, $k$ is defined as a positive rate, and the decay formula incorporates the negative sign.
How do I find $k$ if I only know the half-life?: If the half-life is $h$ (the time for the amount to halve), then at time $t = h$ , you have $A = \tfrac{1}{2} A_0$ . Substituting into the formula: $\tfrac{1}{2} A_0 = A_0 e^{-kh}$ , which simplifies to $0.5 = e^{-kh}$ . Taking $\ln$ : $\ln(0.5) = -kh$ , so $k = -\ln(0.5)/h = \ln(2)/h \approx 0.693/h$ .
Can I use exponential decay for real populations?: Yes, but only for declining populations where the number of deaths is proportional to the current population (like radioactive atoms decaying). For real human or animal populations, births must also be accounted for, making the model more complex. Exponential decay works best for declining populations under environmental stress with no reproduction, such as bacteria under starvation.

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Reviewed 2026-09-18