2

Here is what I'm talking about:

Equation 1: $P(t) = P(0)e^{rt}$
Equation 2: $P(t) = P(0)(1+r)^t$

What is the difference? I've been seeing both used in modeling the size of bacteria populations and I am so confused.

Theo Bendit
  • 50,900
Benjamin Little
  • 37
  • 2
    "Exponential growth" describes a function which grows in a manner similar to $\mathrm{e}^t$, where "similar to" can be made mathematically rigorous in terms of, for example, asymptotic analysis or a particular type of differential equation. Note that both of your functions can be written as $C \mathrm{e}^{kt}$ for some constants $C$ and $k$ (in the second case, $$(1+r)^t = \mathrm{e}^{t \log(1+r)}, $$ from which my claim follows). – Xander Henderson Apr 12 '22 at 00:12
  • 1
    If $r$ is an interest rate in, say, percent per year, then the second formula tells you the principal after $t$ years, assuming interest is credited at the end of the year. The first formula tells you the principal if the interest is compounded continuously. @XanderHenderson 's comment tells you the effective continuous rate that corresponds to a nominal annual rate of $r$. – Ethan Bolker Apr 12 '22 at 00:18
  • @Xander Henderson: I think the confusion might be that the value of $r$ is not the same in these two formulas, since $r \neq \ln(1+r).$ (I'm using $\ln$ because at this level $\log$ probably means base-$10$ logarithm for the OP.) Since $r$ is often called the growth rate, or something similar, it appears that the growth rate varies according to which formula is used. When I taught this, I used the formulas $P(0)e^{kt}$ (continuous compounding) and $P(0)(1+r)^t,$ (continued) – Dave L. Renfro Apr 12 '22 at 06:59
  • and I tried to avoid giving descriptive names to $k$ and $r$ (aside from "$r$ as a percent" gives the percent increase per unit time). Note that things like doubling time and for what value of $t$ will the value of $P(t)$ equal $147$ are not affected by which formula is used. – Dave L. Renfro Apr 12 '22 at 07:00

0 Answers0