What if k is negative?

If k < 0, the quantity is decaying, so it will never double. Use the Half-Life Formula instead to find when it shrinks to half its original value. The Half-Life Formula has the same structure but applies to decay.

Why is it specifically \ln 2 in the formula?

Doubling means the final amount is 2A_0. Starting from the exponential model A_0 e^{kt_2} = 2A_0, divide by A_0 to get e^{kt_2} = 2. Take the natural logarithm of both sides: kt_2 = \ln 2. Solve for t_2 to get the formula.

What if I know the doubling time but not the growth rate?

Rearrange the formula to solve for k: k = \frac{\ln 2}{t_2}. If a population doubles every 5 years, then k = \frac{\ln 2}{5} \approx 0.1386 per year.

Do different starting amounts change the doubling time?

No. Doubling time depends only on the growth rate k, not on the initial amount. Whether you start with 100 bacteria or 1 million, at rate k = 0.1 per hour, both double in about 6.9 hours.

Doubling Time

Q: What if k is negative?

If k < 0, the quantity is decaying, so it will never double. Use the Half-Life Formula instead to find when it shrinks to half its original value. The Half-Life Formula has the same structure but applies to decay.

Q: Why is it specifically \ln 2 in the formula?

Doubling means the final amount is 2A_0. Starting from the exponential model A_0 e^{kt_2} = 2A_0, divide by A_0 to get e^{kt_2} = 2. Take the natural logarithm of both sides: kt_2 = \ln 2. Solve for t_2 to get the formula.

Q: What if I know the doubling time but not the growth rate?

Rearrange the formula to solve for k: k = \frac{\ln 2}{t_2}. If a population doubles every 5 years, then k = \frac{\ln 2}{5} \approx 0.1386 per year.

Q: Do different starting amounts change the doubling time?

No. Doubling time depends only on the growth rate k, not on the initial amount. Whether you start with 100 bacteria or 1 million, at rate k = 0.1 per hour, both double in about 6.9 hours.

Doubling Time calculates how long it takes for a quantity to double in exponential growth when you know the continuous growth rate.

t_2 = \frac{\ln 2}{k}

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What each symbol means

What Doubling Time takes

Symbol	Meaning
$t_2$	The time it takes for the quantity to double, measured in the same time units as the growth rate $k$ ; always positive since doubling takes finite time in growth processes.
$k$	The continuous growth rate expressed as a decimal (not a percentage), which must be positive for doubling to occur; if $k$ is negative, the quantity shrinks and never doubles.

When to use it

When you have an exponential growth model with a known growth rate and need to find when a quantity reaches twice its starting value.

Level

Usually taught in: Algebra II

Worked examples

1. Find doubling time with a simple decimal growth rate

Problem

A bacterial culture grows at a continuous rate of 0.12 per hour. Find the doubling time.

$t_2 = \frac{\ln 2}{k}$
Write the doubling time formula where $t_2$ is time to double and $k$ is the growth rate.
$t_2 = \frac{\ln 2}{0.12}$
Substitute $k = 0.12$ per hour into the formula.
$\ln 2 \approx 0.693$
Evaluate $\ln 2$ using a calculator.
$t_2 = \frac{0.693}{0.12} = 5.775$
Divide to find the doubling time is approximately 5.8 hours.

Answer: $t_2 \approx 5.8 \text{ hours}$

This straightforward example shows the basic process: substitute the growth rate into the formula and divide. Since doubling happens relatively quickly at this growth rate, the bacteria population will double about six times in a full day.

2. Handle a logarithmic expression for the growth rate

Problem

A protein concentration grows at a rate

k = \frac{\ln 4}{3}

per minute. Calculate the doubling time.

$t_2 = \frac{\ln 2}{k}$
Start with the doubling time formula.
$t_2 = \frac{\ln 2}{\frac{\ln 4}{3}}$
Substitute $k = \frac{\ln 4}{3}$ into the formula.
$t_2 = \frac{\ln 2}{\frac{\ln 4}{3}} = \frac{3\ln 2}{\ln 4}$
To divide by a fraction, multiply by its reciprocal.
$\ln 4 = \ln 2^2 = 2\ln 2$
Simplify using the power rule for logarithms: $\ln 4 = 2\ln 2$ .
$t_2 = \frac{3\ln 2}{2\ln 2} = \frac{3}{2} = 1.5$
Cancel $\ln 2$ from numerator and denominator to get the doubling time.

Answer: $t_2 = 1.5 \text{ minutes}$

When $k$ is expressed using logarithms, use logarithm rules to simplify before dividing. Here, recognizing that $\ln 4 = 2\ln 2$ lets us cancel terms and find an exact answer without decimals.

3. Find doubling time from an exponential model in a real context

Problem

A coffee shop's daily revenue follows

R(t) = 5000 e^{0.06t}

dollars, where

t

is days. At this growth rate, how many days until the daily revenue doubles?

$R(t) = 5000 e^{0.06t}$
The revenue model is in the form $A(t) = A_0 e^{kt}$ , where $k = 0.06$ per day is the continuous growth rate.
$t_2 = \frac{\ln 2}{k}$
Apply the doubling time formula to find when revenue doubles.
$t_2 = \frac{\ln 2}{0.06}$
Substitute the growth rate $k = 0.06$ extracted from the exponential model.
$\ln 2 \approx 0.693$
Evaluate the natural logarithm of 2 using a calculator.
$t_2 = \frac{0.693}{0.06} \approx 11.55$
Divide to find that revenue doubles in approximately 11.55 days.

Answer: $t_2 \approx 11.55 \text{ days}$

Word problems give you the exponential model in context, from which you extract the growth rate $k$ . Then apply the doubling time formula directly. Notice that the starting revenue amount ($5000) does not affect the answer—doubling time depends only on the growth rate, not on the initial quantity.

Common mistakes

Where Doubling Time usually goes wrong

The mistake	Why it is wrong	The fix
Using a growth rate from the wrong time period without converting it.	The growth rate $k$ must have the same time units as your answer for $t_2$ . If $k$ is per year but you want $t_2$ in months, the calculation will give you the wrong time interval.	If $k = 0.06$ per year and you want doubling time in months, first convert to a monthly rate: $k_{\text{monthly}} = \frac{0.06}{12} \approx 0.005$ per month. Then calculate $t_2 = \frac{\ln 2}{0.005} \approx 138.6$ months. Alternatively, calculate $t_2$ in years first, then convert to months.
Confusing percentage growth with the rate $k$ and writing $k = 50$ for a 50% growth rate.	The continuous growth rate $k$ in the formula $e^{kt}$ is not a percentage—it is a decimal exponent. A 50% growth per period means the multiplier is $1.5$ , so $e^k = 1.5$ and thus $k = \ln(1.5) \approx 0.405$ , not $0.5$ or $50$ .	Convert percentage growth to a rate using $k = \ln(1 + \frac{r}{100})$ where $r$ is the percent. For 50% growth, $k = \ln(1.5) \approx 0.405$ . Use this decimal value in the doubling time formula.
Calculating only $\ln 2 \approx 0.693$ and stopping, forgetting to divide by $k$ .	The formula is $t_2 = \frac{\ln 2}{k}$ , not just $\ln 2$ . Stopping at $0.693$ gives the wrong value with wrong units, and the denominator $k$ is essential—it scales the doubling time to match the actual growth rate.	Always complete the formula: divide $\ln 2$ by $k$ . For example, if $k = 0.06$ , then $t_2 = \frac{0.693}{0.06} = 11.55$ , not just $0.693$ .

Tips and when to use something else

Double-check that your growth rate $k$ has units matching the time units you want in your answer: if $k$ is per hour, then $t_2$ will be in hours.
The answer $t_2$ is always positive. If you calculate a negative or zero answer, check that $k > 0$ and review your arithmetic.
For decay or shrinking quantities (negative $k$ ), use the Half-Life Formula instead, which finds when a quantity shrinks to half: $t_{1/2} = \frac{\ln 2}{|k|}$ .
Verify your work: start with any initial amount $A_0$ , wait time $t_2$ , and calculate $A_0 e^{k \cdot t_2}$ . This should equal $2A_0$ if your doubling time is correct.

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.Continuous Compound InterestCalculate how much money grows continuously over time, or solve how long growth takes at a given rate of compound interest.Change of Base FormulaConvert logarithms between different bases so you can evaluate them on a calculator or combine logarithms with different bases.

Frequently asked questions

What if $k$ is negative?: If $k < 0$ , the quantity is decaying, so it will never double. Use the Half-Life Formula instead to find when it shrinks to half its original value. The Half-Life Formula has the same structure but applies to decay.
Why is it specifically $\ln 2$ in the formula?: Doubling means the final amount is $2A_0$ . Starting from the exponential model $A_0 e^{kt_2} = 2A_0$ , divide by $A_0$ to get $e^{kt_2} = 2$ . Take the natural logarithm of both sides: $kt_2 = \ln 2$ . Solve for $t_2$ to get the formula.
What if I know the doubling time but not the growth rate?: Rearrange the formula to solve for $k$ : $k = \frac{\ln 2}{t_2}$ . If a population doubles every 5 years, then $k = \frac{\ln 2}{5} \approx 0.1386$ per year.
Do different starting amounts change the doubling time?: No. Doubling time depends only on the growth rate $k$ , not on the initial amount. Whether you start with 100 bacteria or 1 million, at rate $k = 0.1$ per hour, both double in about $6.9$ hours.

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Reviewed 2026-09-18