What if the problem gives me a decay rate like '10% per year' instead of a half-life?

Use the general exponential decay formula A = A_0(b)^t instead, where b is the remaining fraction per time period. For 10% decay, b = 0.9. The Half-Life Formula is specifically for when you know the half-life—the time for 50% to decay.

Can I find the half-life if I only know the initial amount and a later amount?

Only if you also know how much time has passed. With both the time and the amounts, you can rearrange the formula and solve for t_{1/2} by dividing out A_0, leaving an equation with only t_{1/2} as an unknown (usually needing logarithms to finish).

What if the substance never fully disappears in reality?

The Half-Life Formula is a mathematical model that assumes the substance never truly reaches zero—it just keeps halving forever. In real life, physical limitations stop decay eventually, but the formula works perfectly for predicting amounts as long as the half-life remains constant.

Why can't I just multiply the half-life by the number of half-lives instead of dividing?

You're right that if you know the number of half-lives, you can jump straight to the exponent. But the formula is written to find that automatically by dividing t by t_{1/2}, so it works for any time value—not just whole multiples. This makes it more flexible and general.

Half-Life Formula

The Half-Life Formula determines how much of a radioactive or decaying substance remains after a given time period has passed.

A = A_0 \left(\tfrac{1}{2}\right)^{t / t_{1/2}}

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What each symbol means

What Half-Life Formula takes

Symbol	Meaning
$A$	The amount of the substance remaining after time $t$ has passed, measured in the same units as $A_0$ (grams, milligrams, micrograms, etc.).
$A_0$	The initial amount of the substance before any decay occurs; often called the starting amount, original quantity, or amount at time zero.
$t$	The elapsed time since decay started, measured in whatever units match $t_{1/2}$ (hours, days, years, minutes, etc.)—must use the same time unit as the half-life.
$t_{1/2}$	The half-life: the time required for exactly half of any starting amount to decay away (e.g., Carbon-14 has a half-life of 5,730 years).

When to use it

Use this formula when you know how long it takes for a substance to decay to half its amount (the half-life) and want to find what remains after time passes.

Level

Usually taught in: Algebra II

Worked examples

1. Finding remaining amount after whole number of half-lives

Problem

A radioactive sample contains 128 grams initially. After 2 half-lives pass, how much remains?

$A = A_0 \left(\tfrac{1}{2}\right)^{t / t_{1/2}}$
Write the Half-Life Formula.
$A = 128 \left(\tfrac{1}{2}\right)^{2}$
Substitute $A_0 = 128$ and $t / t_{1/2} = 2$ (since 2 half-lives have passed).
$A = 128 \cdot \tfrac{1}{4}$
Simplify the exponent: $(\tfrac{1}{2})^2 = \tfrac{1}{4}$ .
$A = 32 \text{ grams}$
Multiply: $128 \times \tfrac{1}{4} = 32$ .

Answer: $A = 32 \text{ grams}$

Each half-life cuts the amount in half, so after 2 half-lives, the amount becomes one-quarter of the original. The formula correctly models this by raising $\tfrac{1}{2}$ to the power of 2.

2. Finding half-life when given decay data

Problem

A 64 mg sample of a radioactive element decays to 1 mg after 18 hours. What is the half-life?

$1 = 64 \left(\tfrac{1}{2}\right)^{18 / t_{1/2}}$
Substitute known values into the formula: $A = 1$ , $A_0 = 64$ , $t = 18$ .
$\tfrac{1}{64} = \left(\tfrac{1}{2}\right)^{18 / t_{1/2}}$
Divide both sides by 64.
$\left(\tfrac{1}{2}\right)^{6} = \left(\tfrac{1}{2}\right)^{18 / t_{1/2}}$
Recognize that $\tfrac{1}{64} = \tfrac{1}{2^6} = (\tfrac{1}{2})^6$ since $2^6 = 64$ .
$6 = \tfrac{18}{t_{1/2}}$
Since the bases are equal, the exponents must be equal: $6 = \tfrac{18}{t_{1/2}}$ .
$t_{1/2} = \tfrac{18}{6} = 3 \text{ hours}$
Divide both sides by 6 to solve for $t_{1/2}$ .

Answer: $t_{1/2} = 3 \text{ hours}$

This example shows how to rearrange the formula when the half-life is unknown. By recognizing that both sides are powers of $\tfrac{1}{2}$ , we can match exponents and solve algebraically.

3. Real-world problem: medication clearance

Problem

A patient receives an injection of 200 mg of a medication. The medication has a half-life of 8 hours in the bloodstream. How much remains 24 hours later?

$A_0 = 200 \text{ mg}, \quad t_{1/2} = 8 \text{ hours}, \quad t = 24 \text{ hours}$
Identify and organize the given information: initial amount, half-life, and elapsed time.
$\tfrac{t}{t_{1/2}} = \tfrac{24}{8} = 3$
Calculate how many half-lives have passed: $\tfrac{24 \text{ hours}}{8 \text{ hours per half-life}} = 3$ half-lives.
$A = 200 \left(\tfrac{1}{2}\right)^{3}$
Substitute into the Half-Life Formula with $A_0 = 200$ and the exponent 3.
$A = 200 \cdot \tfrac{1}{8}$
Evaluate: $(\tfrac{1}{2})^3 = \tfrac{1}{2} \cdot \tfrac{1}{2} \cdot \tfrac{1}{2} = \tfrac{1}{8}$ .
$A = 25 \text{ mg}$
Multiply: $200 \div 8 = 25$ .

Answer: $A = 25 \text{ mg}$

After 24 hours, three complete half-lives have passed (each 8 hours long), leaving one-eighth of the original 200 mg. This is why medical dosing intervals often align with half-lives—after one half-life, the dose is low enough for a second injection.

Common mistakes

Where Half-Life Formula usually goes wrong

The mistake	Why it is wrong	The fix
Writing $A = A_0 (\tfrac{1}{2})^t$ without dividing $t$ by $t_{1/2}$	This treats time and half-life as if they use the same scale, leading to wrong exponents and nonsensical answers when $t$ and $t_{1/2}$ have different units.	Always compute $t / t_{1/2}$ first to find the dimensionless exponent. The exponent must be the ratio of time elapsed to half-life.
Using base 2 instead of base $\tfrac{1}{2}$ , such as $A = A_0 (2)^{-t/t_{1/2}}$ or $A = A_0 (2)^{t/t_{1/2}}$	Using 2 as the base makes the substance grow exponentially (if positive exponent) or requires a negative exponent (if written correctly), which students often forget or apply incorrectly.	The base must be exactly $\tfrac{1}{2}$ (or equivalently $2^{-1}$ ). This base naturally models halving: $(\tfrac{1}{2})^1 = \tfrac{1}{2}$ , $(\tfrac{1}{2})^2 = \tfrac{1}{4}$ , etc.
Swapping the numerator and denominator of the exponent: $A = A_0 (\tfrac{1}{2})^{t_{1/2} / t}$ instead of $A = A_0 (\tfrac{1}{2})^{t / t_{1/2}}$	This reverses the decay rate: substances with long half-lives would decay faster, and substances would start growing as time increases—the opposite of reality.	The exponent is always time elapsed divided by half-life: $t / t_{1/2}$ , not $t_{1/2} / t$ . A larger $t$ in the numerator means a larger exponent and faster decay.

Tips and when to use something else

The base $\tfrac{1}{2}$ means 'lose half each time'—visualize dividing by 2 repeatedly rather than computing decimals. After $n$ half-lives, the amount is $A_0 \cdot (\tfrac{1}{2})^n$ .
When $t$ is an exact multiple of $t_{1/2}$ (like 2, 3, or 4 half-lives), you can skip the fraction and count directly: after 3 half-lives, the amount is always $A_0 \cdot \tfrac{1}{8}$ regardless of the actual time units.
If the problem gives you a different decay rate instead of a half-life—such as 'loses 5% per year'—use the general exponential decay formula $A = A_0(b)^t$ instead, where $b = 0.95$ .
Half-life problems often require logarithms to solve. If you need to find $t$ or $t_{1/2}$ and the exponent is not a whole number, take $\log(\tfrac{1}{2})$ of both sides and use logarithm rules to isolate your variable.

Related formulas

Exponential DecayExponential Decay models quantities that shrink at a rate proportional to their current amount, used for radioactive decay, cooling, and depreciation.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.Doubling TimeDoubling Time calculates how long it takes for a quantity to double in exponential growth when you know the continuous growth rate.Power Rule for ExponentsSimplify nested exponents by multiplying the exponents together: this is the fundamental tool you need whenever one exponent is itself raised to a power.

Frequently asked questions

What if the problem gives me a decay rate like '10% per year' instead of a half-life?: Use the general exponential decay formula $A = A_0(b)^t$ instead, where $b$ is the remaining fraction per time period. For 10% decay, $b = 0.9$ . The Half-Life Formula is specifically for when you know the half-life—the time for 50% to decay.
Can I find the half-life if I only know the initial amount and a later amount?: Only if you also know how much time has passed. With both the time and the amounts, you can rearrange the formula and solve for $t_{1/2}$ by dividing out $A_0$ , leaving an equation with only $t_{1/2}$ as an unknown (usually needing logarithms to finish).
What if the substance never fully disappears in reality?: The Half-Life Formula is a mathematical model that assumes the substance never truly reaches zero—it just keeps halving forever. In real life, physical limitations stop decay eventually, but the formula works perfectly for predicting amounts as long as the half-life remains constant.
Why can't I just multiply the half-life by the number of half-lives instead of dividing?: You're right that if you know the number of half-lives, you can jump straight to the exponent. But the formula is written to find that automatically by dividing $t$ by $t_{1/2}$ , so it works for any time value—not just whole multiples. This makes it more flexible and general.

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Reviewed 2026-09-18