information:extraction
Basic formula

A quantity x depends exponentially on time t if x (t) = a ⋅ b t / τ {\displaystyle x(t)=a\cdot b^{t/\tau }} where the constant a is the initial value of x, x (0) = a, {\displaystyle x(0)=a\,,} the constant b is a positive growth factor, and τ is the time constant —the time required for x to increase by one factor of b: x (t + τ) = a ⋅ b t + τ τ = a ⋅ b t τ ⋅ b τ τ = x (t) ⋅ b. {\displaystyle x(t+\tau)=a\cdot b^{\frac {t+\tau }{\tau }}=a\cdot b^{\frac {t}{\tau }}\cdot b^{\frac {\tau }{\tau }}=x(t)\cdot b\,.}

If τ > 0 and b > 1, then x has exponential growth. If τ < 0 and b > 1, or τ > 0 and 0 < b < 1, then x has exponential decay.

Example: If a species of bacteria doubles every ten minutes, starting out with only one bacterium, how many bacteria would be present after one hour? The question implies a = 1, b = 2 and τ = 10 min.

x (t) = a ⋅ b t / τ = 1 ⋅ 2 t / (10 min) {\displaystyle x(t)=a\cdot b^{t/\tau }=1\cdot 2^{t/(10{\text{ min}})}} x (1 hr) = 1 ⋅ 2 (60 min) / (10 min) = 1 ⋅ 2 6 = 64. {\displaystyle x(1{\text{ hr}})=1\cdot 2^{(60{\text{ min}})/(10{\text{ min}})}=1\cdot 2^{6}=64.}

After one hour, or six ten-minute intervals, there would be sixty-four bacteria.

Many pairs (b, τ) of a dimensionless non-negative number b and an amount of time τ (a physical quantity which can be expressed as the product of a number of units and a unit of time) represent the same growth rate, with τ proportional to log b. For any fixed b not equal to 1 (e.g. e or 2), the growth rate is given by the non-zero time τ. For any non-zero time τ the growth rate is given by the dimensionless positive number b.

Thus the law of exponential growth can be written in different but mathematically equivalent forms, by using a different base. The most common forms are the following: x (t) = x 0 ⋅ e k t = x 0 ⋅ e t / τ = x 0 ⋅ 2 t / T = x 0 ⋅ (1 + r 100) t / p, {\displaystyle x(t)=x_{0}\cdot e^{kt}=x_{0}\cdot e^{t/\tau }=x_{0}\cdot 2^{t/T}=x_{0}\cdot \left(1+{\frac {r}{100}}\right)^{t/p},} where x expresses the initial quantity x (0).

Parameters (negative in the case of exponential decay):

The quantities k, τ, and T, and for a given p also r, have a one-to-one connection given by the following equation (which can be derived by taking the natural logarithm of the above): k = 1 τ = ln 2 T = ln (1 + r 100) p {\displaystyle k={\frac {1}{\tau }}={\frac {\ln 2}{T}}={\frac {\ln \left(1+{\frac {r}{100}}\right)}{p}}} where k = 0 corresponds to r = 0 and to τ and T being infinite.

If p is the unit of time the quotient t / p is simply the number of units of time. Using the notation t for the (dimensionless) number of units of time rather than the time itself, t / p can be replaced by t, but for uniformity this has been avoided here. In this case the division by p in the last formula is not a numerical division either, but converts a dimensionless number to the correct quantity including unit.

A popular approximated method for calculating the doubling time from the growth rate is the rule of 70,that is, T ≃ 70 / r {\displaystyle T\simeq 70/r}.