Exponential Growth Calculator

This exponential growth calculator solves the model \( \frac{dP}{dt} = kP \). Enter a starting amount \( P_0 \), a rate \( k \) and a time \( t \), and it returns the formula \( P(t) = P_0e^{kt} \), the value at your time, and the doubling time. Enter a negative \( k \) and it becomes an exponential decay calculator that reports the half-life instead.

Use it for population growth, bacteria cultures, continuously compounded interest, radioactive decay and drug elimination problems. Also, it’s free and needs no sign-up.

How to use the exponential growth calculator

1. Enter the Initial P₀, the amount at \( t = 0 \).
2. Enter the Rate k per unit of time. Use a positive \( k \) for growth and a negative \( k \) for decay. For example, you can type expressions: -ln(2)/6 is accepted, as is pi or e.
3. Enter the Time t in the same units as \( k \) (if \( k \) is per hour, \( t \) is in hours).
4. Press Model. As a result, you’ll see \( P(t) \) as a formula, the value \( P(t) \) at your time, and either the doubling time \( \frac{\ln 2}{k} \) or the half-life \( \frac{\ln 2}{|k|} \).

Formula

Exponential change means the rate of change is proportional to the current amount:

$$\frac{dP}{dt} = kP, \qquad P(0) = P_0$$

Separate the variables and integrate. The left side becomes \( \int \frac{dP}{P} = \ln|P| \) (see the integral of 1/x), which leads to

$$P(t) = P_0\,e^{kt}$$

You can check it by differentiating: the chain rule gives \( P'(t) = kP_0e^{kt} = kP \), the same pattern as the derivative of e^(2x). So, setting \( P = 2P_0 \) or \( P = \tfrac12 P_0 \) and solving gives

$$t_{\text{double}} = \frac{\ln 2}{k}, \qquad t_{\text{half}} = \frac{\ln 2}{|k|}$$

Worked example: bacterial growth

A culture starts with 500 bacteria and grows at a continuous rate of \( k = 0.3 \) per hour. How many are there after 8 hours, and how long does it take to double?

$$P(8) = 500\,e^{0.3 \cdot 8} = 500\,e^{2.4} \approx 5511.6$$

$$t_{\text{double}} = \frac{\ln 2}{0.3} \approx 2.31\ \text{hours}$$

These are the values in the calculator above. As a result, a useful sanity check: 8 hours is about 3.5 doubling times, and \( 500 \cdot 2^{3.5} \) is in the same ballpark.

Worked example: half-life and decay

A 200 mg dose of a drug has a half-life of 6 hours. How much remains after 24 hours?

Find \( k \). Rearranging the half-life formula gives \( k = -\frac{\ln 2}{6} \approx -0.1155 \) per hour.

Evaluate. \( P(24) = 200\,e^{-(\ln 2/6)\cdot 24} = 12.5 \) mg. That matches the shortcut: 24 hours is four half-lives, and \( 200 \cdot \left(\tfrac12\right)^4 = 12.5 \).

First, in the calculator, enter \( P_0 = 200 \), \( k = \) -ln(2)/6 and \( t = 24 \). The half-life row shows 6, confirming the setup.

Finding k from two data points

Many problems give two measurements instead of a rate. If a town grows from 1,200 to 1,800 people in 5 years, then \( 1800 = 1200\,e^{5k} \), so

$$k = \frac{\ln(1800/1200)}{5} = \frac{\ln 1.5}{5} \approx 0.0811$$

per year. Next, plug that \( k \) into the calculator to predict any future year; the doubling time is about 8.55 years. The derivative of ln x explains why taking logs is the natural way to undo \( e^{kt} \).

Tips and common mistakes

• Continuous vs periodic rates. A 5% annual rate compounded continuously is \( k = 0.05 \). Also, \( \$1000 \) grows to about \( \$1648.72 \) in 10 years. Compounded once a year it’s \( 1000(1.05)^{10} \), which is a slightly different model.
• Rule of 70. Doubling time is roughly \( 70 \) divided by the percentage rate. For example, for 5%, \( 70/5 = 14 \) years, close to the exact \( \ln 2/0.05 \approx 13.86 \).
• Mismatched units. A rate per day with a time in hours gives a wrong answer without any error message.
• Limited growth. Real populations level off. For logistic models, use the Euler’s method calculator instead; the Euler’s method guide explains how it works.

Continuous and yearly growth

Growth of \( r \) per year compounded yearly, \( P(1 + r)^t \), is the same as continuous growth \( Pe^{kt} \) with \( k = \ln(1 + r) \). A 5% yearly rise is a continuous rate of \( \ln 1.05 \approx 4.88\% \). Mixing the two is a common source of small but real errors.

Doubling time and half-life

For growth at continuous rate \( k \), the doubling time is

$$T_2 = \frac{\ln 2}{k}$$

At \( k = 0.05 \), \( T_2 \approx 13.86 \) years. For decay, the same formula gives the half-life. Carbon-14 has a half-life of about 5,730 years, so its decay constant is \( \ln 2 / 5730 \approx 0.000121 \) per year. See the half-life calculator.

When exponential growth stops

However, real populations cannot grow exponentially forever: food, space and competition slow them down. The logistic model

$$P(t) = \frac{K}{1 + Ae^{-kt}}$$

starts out almost exponential and levels off at the carrying capacity \( K \). However, growth is fastest at \( P = K/2 \), the curve’s inflection point. See the inflection point calculator.

Reading k

A positive \( k \) means growth, a negative \( k \) decay. The rate is per unit of time, so \( k = 0.05 \) per year is not the same as 0.05 per month.

Practice problems

1. A town of 5,000 grows continuously at 3% a year. So, what is its population after 10 years?
2. A sample decays from 80 g to 20 g in 12 hours. What is its half-life?

Answers: 1. \( 5000e^{0.3} \approx 6{,}749 \). 2. Two half-lives in 12 hours, so 6 hours.

Related guides

• Integral of 1/x: Why It’s ln|x| + C
• Derivative of e^(2x) Step by Step
• Derivative of ln(x)
• Euler’s Method: Formula and Example

Further reading

• 3.9 Derivatives of Exponential and Logarithmic Functions (OpenStax Calculus Volume 1) — why \( \frac{d}{dt}e^{kt} = ke^{kt} \), the fact the growth model is built on.
• Applications of Exponential and Logarithmic Functions (Paul’s Online Math Notes) — compound interest, growth and decay, and half-life examples.

For graphs, compound interest and the half-life equation, see exponential functions; for growth that levels off (logistic growth), see differential equations. Also, to solve for time, use the logarithm calculator.

FAQ

How do I calculate exponential decay?

Use the same calculator with a negative \( k \). The formula stays \( P(t) = P_0e^{kt} \), and the output shows the half-life instead of the doubling time.

How do you find the half-life from k?

First, divide \( \ln 2 \) by \( |k| \). For \( k = -0.1155 \) per hour, the half-life is about 6 hours.

What is the exponential growth formula?

\( P(t) = P_0e^{kt} \), where \( P_0 \) is the starting amount, \( k \) is the continuous rate, and \( t \) is time. Also, it’s the solution of \( \frac{dP}{dt} = kP \).

Can I solve for time instead of P(t)?

The calculator doesn’t solve for \( t \) directly. Rearrange by hand: \( t = \frac{1}{k}\ln\frac{P}{P_0} \), or try times in the calculator until \( P(t) \) reaches your target.