# 5.3: Exponential Solution


To think about this mathematically, first set s to zero, meaning no density dependence. The differential equation then reduces to dN /dt = r, and if you replace s in the equation above with 0 you get this:

$N(t)\,=\frac{1}{(\frac{0}{r}\,+\frac{1}{N_0})\,e^{-rt}\,-\frac{0}{r}}$

$\,=\frac{1}{\frac{1}{N_0}\,e^{-rt}\,}$

$\,=\,N_0e^{rt}$

A measure often used for exponential growth, and that we will apply later in this book, is “doubling time”—the time that must elapse for the population to double. For exponential growth, this is always the same, no matter how large or small the population. For exponential growth, the equation above is

$N(t)\,=\,N_0\,e^{rt}$

N0 is the starting population at time 0, N (t ) is the population at any time t, and r is the constant growth rate—the “intrinsic rate of natural increase.” How much time, τ, will elapse before the population doubles? At some time t, the population will be N (t ), and at the later time t+τ, the population will be N (t+τ). The question to be answered is this: for what τ will the ratio of those two populations be 2?

$\frac{N(t\,+\,τ)}{N(t)}\,=\,2$

Substituting the right-hand side of the exponential growth equation gives

$\frac{N_0\,e^{r\,(t\,+\,τ)}}{N_0\,e^{rt}}\,=\,2$

The factor N0 cancels out, and taking natural logarithms of both sides gives

$\ln \frac{e^{r(t\,+\,τ)}}{e^{rt}}\,=\,\ln \,2$

Since the log of a ratio is the difference of the logs, this yields

$\\ln \,e^{r(t\,+\,τ)}\,-\,\ln \,e^{rt}\,=\,\ln \,2$

Since logarithms and exponentials are inverse processes— each one undoes the other—the natural logarithm of ex is simply x. That gives

$r(t\,+\,τ)\,-\,rt\,=\,\ln \,2$

$r\,τ\,=\,\ln \,2$

and finally, the doubling time τ is

$τ\,=\frac{\ln \,2}{r}$

In other words, the doubling time for exponential growth, where r is positive and s is 0, is just the natural logarithm of 2 (0.69314718...) divided by the growth rate r.

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