Math · College algebra · Concept

Exponential growth and decay

A quantity grows or decays exponentially when it changes by the same factor over equal time intervals. Continuous growth follows A = A₀e^(kt), interest compounded n times a year follows A = P(1 + r/n)^(nt), and a decaying sample halves every half-life T: A = A₀(1/2)^(t/T).

Equal times, equal factors

Linear change adds the same amount each period; exponential change multiplies by the same factor. At 10% growth a year, 100 becomes 110, then 121, then 133.1: each year adds 10% of a larger amount.

Adding 10 a year against growing 10% a year
Year	Linear: +10	Exponential: ×1.10
0	100	100
1	110	110
2	120	121
3	130	133.1
4	140	146.41

Continuous growth and decay

In A = A₀e^(kt), A₀ is the starting amount and k the continuous rate: k > 0 means growth and k < 0 means decay. The rate is a fraction, so 3% a year is k = 0.03.

A (t) = A_{0} e^{k t}

Compound interest

Interest at annual rate r compounded n times a year multiplies the balance by 1 + r/n each period, nt times in t years. As n grows without bound the formula becomes continuous compounding, Pe^(rt).

A = P {(1 + \frac{r}{n})}^{n t} → P e^{r t}

\begin{array}{c} A = P {(1 + \frac{r}{n})}^{n t} \\ → P e^{r t} \end{array}

Half-life and doubling time

A decaying amount halves every half-life T, whatever the starting amount. A growing amount doubles every doubling time, ln 2/k, which is about 70 divided by the percentage rate.

A (t) = A_{0} {(\frac{1}{2})}^{\frac{t}{T}}, t_{double} = \frac{ln 2}{k}

\begin{array}{c} A (t) = A_{0} {(\frac{1}{2})}^{\frac{t}{T}} \\ t_{double} = \frac{ln 2}{k} \end{array}

Solving for time

To find when an amount reaches a target, isolate the power and rewrite it in logarithmic form. The time appears in the exponent, so logarithms are the tool that brings it down.

Common mistakes

Using a percentage as a whole number: 5% is r = 0.05.
Adding the growth each period, which is linear, instead of multiplying by the growth factor.
Mixing time units: if the half-life is in hours, t must be in hours.
Thinking a whole sample is gone after two half-lives: a quarter still remains.

Key terms

Exponential growth: Growth by the same factor in each equal time interval, modeled by A = A₀e^(kt) with k > 0 or A = A₀bᵗ with b > 1. The amount added each period grows with the amount present.
Exponential decay: Decrease by the same factor in each equal time interval, modeled by A = A₀e^(kt) with k < 0 or A = A₀bᵗ with 0 < b < 1. Radioactive decay and drug elimination often follow it.
Compound interest: Interest added to the balance so that it earns interest itself. At annual rate r compounded n times a year, A = P(1 + r/n)^(nt); compounding continuously gives A = Pe^(rt).
Doubling time: The time an exponentially growing quantity takes to double: ln 2 / k for a continuous rate k. It does not depend on the starting amount.
Half-life: The time it takes for a quantity to drop to half its starting value. For a first-order reaction it is ln 2/k and does not depend on the starting concentration.
Exponential function: A function with the variable in the exponent, such as bˣ with a fixed base b > 0, b ≠ 1. The natural exponential eˣ is its own derivative.

Work through an example

A 400 mg dose of a drug has a half-life of 6.0 hours in the body. How much remains after 24 hours, and when does 30 mg remain?

Solve a half-life problem →

Compare compound and continuous interest →

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Graph the decay in Graph Check the amount after 24 hours in Math Solve for the time in Math Open worked example on a board Continuous growth in Math Reference

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