Deriving the Half-Life Formula

What Half-Life Measures

A decaying quantity may decrease continuously rather than by the same fixed percentage at separate time intervals. In the continuous exponential model, its amount at time $t$ is $P(t)=P_0e^{kt}$, where $P_0$ is the amount at time $0$, $e$ is the base of the natural logarithm, and $k$ is the constant relative rate. For decay, $P_0>0$ and $k<0$.

The half-life is the elapsed time it takes this model to reach half of its initial amount. We denote it by $T$, so its defining condition is $P(T)=P_0/2$. The goal is to solve that condition for $T$, taking care that a time representing decay must be positive.

Definition: For the model $P(t)=P_0e^{kt}$ with $P_0>0$ and $k<0$, the half-life $T$ is the positive time satisfying $P(T)=P_0/2$.

The rate constant $k$ has units of inverse time. Thus, if time is measured in days, $k$ is measured in $\text{day}^{-1}$; the resulting half-life is measured in days. We will see that the initial amount does not affect $T$.

Deriving the Formula

Start with the model and impose the defining condition for half-life. Since $P_0>0$, dividing by $P_0$ is valid. The natural logarithm can then be applied because the exponential expression is positive.

Set the amount to half its initial value.
Substitute $t=T$ into the decay model and use the definition of half-life.

Cancel the initial amount.
Because $P_0>0$, divide both sides by $P_0$ to isolate the exponential factor.

Take natural logarithms.
Use that $\ln(e^x)=x$, then solve the resulting linear equation for $T$.

\begin{aligned} P(T)&=\frac{P_0}{2},\\ P_0e^{kT}&=\frac{P_0}{2},\\ e^{kT}&=\frac12,\\ kT&=\ln\left(\frac12\right)=-\ln 2,\\ T&=-\frac{\ln 2}{k}. \end{aligned}

The final expression is already positive when $k<0$: its numerator is negative and its denominator is negative. Equivalently, because $k<0$, its absolute value satisfies $|k|=-k$. Replacing $-k$ by $|k|$ gives the standard form.

Theorem: For $P(t)=P_0e^{kt}$, with $P_0>0$ and $k<0$, the half-life is $$ T=-\frac{\ln 2}{k}=\frac{\ln 2}{|k|}. $$ Since $\ln 2>0$ and $|k|>0$, this half-life is positive.

This derivation also explains why $P_0$ does not appear in the answer: it cancels when the amount is compared with half its initial value. The half-life depends on the decay rate, not on the size of the starting amount.

Why the Sign Produces a Positive Time

The logarithm of the remaining fraction $1/2$ is negative: $\ln(1/2)=-\ln 2<0$. In a decay model, $k<0$, so the equation $kT=\ln(1/2)$ divides one negative quantity by another. The resulting elapsed time is positive.

The absolute-value form makes that sign especially clear. The quantity $|k|$ is the positive magnitude of the decay rate. Dividing the positive number $\ln 2$ by this positive magnitude gives a positive time. It is not a change to the model or to the sign of $k$; it is an equivalent way to write the solution under the decay condition $k<0$.

The assumptions matter. If $k=0$, the model is constant and never falls to half of a positive initial amount. If $k>0$, the model grows; the equation $P(T)=P_0/2$ has a negative solution, not a nonnegative elapsed time after $t=0$. Thus, the half-life formula as a positive decay time applies when $P_0>0$ and $k<0$.

Worked Examples

Worked Example: A Rate of $-0.2$ per Hour

Suppose a quantity follows $P(t)=P_0e^{-0.2t}$, with $t$ measured in hours. Here $k=-0.2\ \text{hour}^{-1}$. Use the half-life formula:

\begin{aligned} T&=\frac{\ln 2}{|k|}\\ &=\frac{\ln 2}{|-0.2|}\ \text{hours}\\ &=\frac{\ln 2}{0.2}\ \text{hours}\\ &=5\ln 2\ \text{hours}\\ &\approx 3.466\ \text{hours}. \end{aligned}

The exact value is $5\ln 2$ hours; the decimal is rounded to three digits after the decimal point. To check it, substitute the exact time into the exponential factor:

e^{-0.2(5\ln 2)} =e^{-\ln 2} =\frac12.

Therefore $P(T)=P_0/2$, as required.

Worked Example: Starting with 240 Units

Suppose $P(t)=240e^{-0.08t}$, where $t$ is in days. The initial amount is $240$ units, so half of it is $120$ units. We can first solve the defining equation directly:

\begin{aligned} 240e^{-0.08T}&=120,\\ e^{-0.08T}&=\frac{120}{240}=\frac12,\\ -0.08T&=\ln\left(\frac12\right)=-\ln 2,\\ T&=\frac{\ln 2}{0.08} =\frac{25}{2}\ln 2\ \text{days}\\ &\approx 8.664\ \text{days}. \end{aligned}

The decimal is rounded to three digits after the decimal point. Check by substituting the exact value into the model:

\begin{aligned} P\left(\frac{25}{2}\ln 2\right) &=240e^{-0.08(\frac{25}{2}\ln 2)}\\ &=240e^{-\ln 2}\\ &=240\left(\frac12\right)=120. \end{aligned}

This confirms both the target amount and the cancellation of the initial amount in the general derivation.

Worked Example: A Rate Given Using a Logarithm

Suppose $P(t)=P_0e^{-(\ln 3)t/6}$, with time $t$ measured in months. The rate constant is $k=-(\ln 3)/6\ \text{month}^{-1}$, which is negative. Applying the formula gives:

\begin{aligned} T&=\frac{\ln 2}{|k|}\\ &=\frac{\ln 2}{\ln 3/6}\ \text{months}\\ &=\frac{6\ln 2}{\ln 3}\ \text{months}\\ &\approx 3.786\ \text{months}. \end{aligned}

The decimal is rounded to three digits after the decimal point. For a check, substitute the exact time into the exponent:

\begin{aligned} e^{-(\ln 3/6)(6\ln 2/\ln 3)} &=e^{-\ln 2}\\ &=\frac12. \end{aligned}

Thus the amount at this time is $P_0/2$. The logarithms in the rate and the formula cancel as shown, leaving the required half-amount factor.

A Common Sign Error

A frequent mistake is to use $T=\ln 2/k$ without the minus sign or absolute value. For example, if $k=-0.1\ \text{minute}^{-1}$, that incorrect expression gives:

\frac{\ln 2}{k} =\frac{\ln 2}{-0.1} \approx -6.931\ \text{minutes}.

A negative answer cannot represent the elapsed time after the starting time. The error comes from dropping the negative sign in $\ln(1/2)=-\ln 2$. Keeping that sign gives the correct calculation:

T=-\frac{\ln 2}{k} =-\frac{\ln 2}{-0.1} =\frac{\ln 2}{0.1} \approx 6.931\ \text{minutes}.

Check the corrected value directly:

e^{(-0.1)(\ln 2/0.1)} =e^{-\ln 2} =\frac12.

Writing $T=\ln 2/|k|$ also prevents this particular sign error, provided the model is in fact a decay model with $k<0$.

Interpreting the Result

The formula $T=\ln 2/|k|$ has useful consequences that follow directly from its structure:

The half-life is positive. Both $\ln 2$ and $|k|$ are positive for decay.
The initial amount is irrelevant to the half-life. It cancels from the equation defining the time to reach half the start.
A larger decay-rate magnitude gives a shorter half-life. With $\ln 2$ fixed, increasing the positive denominator decreases the quotient.
The units are time. Since $|k|$ has units of inverse time, its reciprocal has units of time, while $\ln 2$ is dimensionless.

For example, a rate constant expressed per day must produce a half-life in days. If a computed result has units of inverse time, a unit conversion or algebraic step has likely gone wrong. Also, the formula uses the rate constant in the time unit of the model: changing the time unit requires expressing $k$ consistently in that unit.

Half-life is a statement about the model reaching half its initial amount at a particular elapsed time. It does not mean the amount becomes zero at that time; the model gives $P(T)=P_0/2$, still a positive amount. The derivation is useful because it turns a precise target into an explicit time while making the assumptions and sign of that time visible.

Check Your Understanding

Answer each question using the decay model and the conditions stated in the tutorial.

Starting from $P(T)=P_0/2$, what equation involving $e^{kT}$ results after dividing by $P_0$?
Why does $-\ln 2/k$ give a positive time when $k<0$?
If $k=-0.05\ \text{year}^{-1}$, write the exact half-life in years using the formula. What substitution would verify it?
Why does $P_0$ not appear in the half-life formula?
What does the equation $T=\ln 2/k$ predict for a negative $k$, and what sign error has occurred?

What You'll Learn

What Half-Life Measures

Deriving the Formula

Why the Sign Produces a Positive Time

Worked Examples

Worked Example: A Rate of \(-0.2\) per Hour

Worked Example: Starting with 240 Units

Worked Example: A Rate Given Using a Logarithm

A Common Sign Error

Interpreting the Result

Check Your Understanding

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Deriving the Half-Life Formula

What You'll Learn

What Half-Life Measures

Deriving the Formula

Why the Sign Produces a Positive Time

Worked Examples

Worked Example: A Rate of \(-0.2\) per Hour

Worked Example: Starting with 240 Units

Worked Example: A Rate Given Using a Logarithm

A Common Sign Error

Interpreting the Result

Check Your Understanding

Test yourself: Calculus I final exam practice

Want all 1,114 practice problems?

Get the Calc Pass App

Vol. 1: 1,114 Questions

Vol. 2: Solutions Manual

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