Exponential Growth and Decay

Exponential growth occurs when the growth rate of the value of a mathematical function is proportional to the function's current value. Exponential decay occurs in the same way, providing the growth rate is negative.

In the long run, exponential growth of any kind will overtake linear growth of any kind as well as any polynomial growth.

Exponential Growth

This graph illustrates how exponential growth (green) surpasses both linear (red) and cubic (blue) growth.

The formula for exponential growth of a variable $x$ at the (positive or negative) growth rate $r$, as time $t$ goes on in discrete intervals (that is, at integer times 0, 1, 2, 3, ...), is: 

$x_{t} = x_{0}(1 + r^{t})$ 

where $x_0$ is the value of $x$ at time $0$. For example, with a growth rate of $r = 5 \% = 0.05$, going from any integer value of time to the next integer causes $x$ at the second time to be $1.05$ times (i.e., $5\%$ larger than) what it was at the previous time.

A quantity $x$ depends exponentially on time $b$ if:

 $\displaystyle{x_{t} = ab^{\left(\frac{t}{\tau }\right)}}$

where the constant $a$ is the initial value of $x$, $x(0) = a$, the constant $b$ is a positive growth factor, and $\tau$ is the time constant—the time required for $x$ to increase by one factor of $b$:

$x(\tau + t)= ab^{\left(\frac{\tau + t}{\tau }\right)} = ab^{\left(\frac{t}{\tau }\right)}b^{\left(\frac{\tau }{\tau}\right)} = x (t)b$

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

Let's assume that 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 $\tau = 10\text{ min}$.

$x (t) = ab^{\dfrac{t}{\tau}} = 1 \cdot 2^{\left(\frac{60\text{ min}}{10\text{ min}}\right)}$

$x (1\text{ hour}) = 1 \cdot 2^{(6)} = 64$

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