Exponential growth and decay,and the constant k

Question

In the equation $$A=Be^{kt}$$,where $B$ is the initial amount and $t$ is the time taken what is $k$,I know it's a constant of proportionality ,but is it the same as the number of time a certain amount of money gets compounded every year?

For example, if an amount of $\$~500$ is getting compounded four times at the rate of $5~\%$ per year ,here if they ask what is the amount if the money is compounded every instant ,the equation will the somewhat similar to exponential growth equation,is the number 4 here same as k?

Answer 1

As Ben Grossmann also has explained in the comments, in your example, $k$ refers to the annual rate (provided $t$ is measured in years). Then your formula can be used to calculate the annual yield - the ratio of the total amount at the end of the year to the total amount at the beginning of the year.

Specifically, the rate is

$$r=e^k-1$$

See Effective interest rate for details.

Answer 2

$k$ is often described as the continuously compounding rate or the logarithmic return.

If we took your example of $\$ 500$ compounded four times at the rate of $5\%$ per year and wrote it as $A=500 \times (1.05)^4 \approx 607.75$ then we could write this as $A=Be^{kt}$ where

• $B=500$, the initial amount
• $t=4$, the time in years
• $k=\log_e(1.05)\approx 0.04879$

and get the same result. The same approach would work for integer $t$, and we could see it as a reasonable approach for non-integer time $t$.

To me in this example $e^k-1 = 0.05$ is the annual rate of growth, but some people use that phrase to mean $k$ itself and so it is important that two people communicating are clear about what they are talking about.