1

So as far as I understand $e^x$ describes the growth after some time $x$. Where $ln(x)$ is the time needed to grow to $x$. Where both assume 100% continuous compounding growth. What decides the units of time $x$? Is my understanding correct that it would be the time it takes to achieve 100% continuous compounding growth in whatever phenomena we're modelling?

PassageTime
  • 11
  • 2
    $e^x$ is a function independent of any units. The units in use are supplied by the context that $e^x$ is being used in. – abiessu Nov 19 '23 at 22:45
  • 1
    Also, both $e^x$ and $\ln x$ have no upper bound, so there is no "$100$%" for either function. – abiessu Nov 19 '23 at 22:47
  • 2
    @abiessu I think OP comes from a finance background and by "100% continuous compounding growth" they meant "continuous compounding" rather than some fixed-term compounding like monthly compounding of interest. – Benjamin Wang Nov 19 '23 at 22:51
  • 2
    When we are using exponential growth to model a real-world phenomenon such as continuously compounded interest in a bank account or the growth of a colony of bacteria, we don't write $e^x$. We write something like $e^{rx}$ or $e^{rt}$ where $r$ is a constant (as in the linked question). Sometimes we use a letter or expression other than $r$. But if we know $r$, and choose $t$ such that $rt=1$, then $t$ is the amount of time that it would take the quantity to double if it grew at a constant rate relative to the original amount rather than having compound growth. – David K Nov 20 '23 at 00:50

2 Answers2

1

The following classic formula: $$ e^x = \lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^n $$ is the one that relates $x \mapsto e^x$ to compound interest.

If you invest $\$1$ at an interest rate of $x$ over some given time interval, say 1 year, and the interest is compounded $n$ times, then at the end of the year you will have accrued this many dollars: $$ \left(1 + \frac{x}{n}\right)^n $$

$e^x$ gives you the precise upper bound on the advantage that compound interest can achieve for you.

I have used units of dollars and years above for concreteness but the actual values of those units don't affect the exponential function. It assumes an initial investment of $1$ monetary unit and an interest rate of $x$ over some time unit. You fill in the units in a specific example and it gives you the right answer.

See the Wikipedia article on $e$ for more information and some of the history.

Rob Arthan
  • 48,577
0

The $x$ and time $t$ are same. So, the second and third sentences can perhaps be deleted from the question.

You know the compound interest formula. If compounding takes place every second or even at shorter periods compared to the annual, half year and quarterly instalments of payment, the amount still is bounded:

$$ \text{Amount } = Pe^{n r} = Pe^{t r}$$

where P is principal, $n$ or $t$ is time or number of years and $r$ the rate of interest.

If $ A, P,r$ are given the time can found using logarithms. So, $t~r= \log~ 2$ for money to double with time and interest by above relation.

Narasimham
  • 40,495