Exponents in bases other than e

You can relate exponential functions that are in bases other than \( e \) using the natural lograithm.

\[ base^x = e^{x \times ln(base)} \]

Example

Consider the function \( 2^x \). From the graph below we can see that it passes through \( 2 \) at \( x = 1 \) then \( 4 \) at \( x = 2 \) and so on. But how does this relate the base 100% compounding rate of \( e \)?

If we consider \( e^x \) we know that at \( x = 1 \) the graph will pass through, well, \( e \) itself, which is \( 2.7182~ \).

This means that the rate at which the graph \(2^x \) is changing is less than that of the graph of \( e^x \).

We’re kind of asking a new question then: What \(rate \times time \) combination to we need to raise \( e \) to get the graph to pass through \( 2 \) at \( x = 1 \) and \( 4 \) at \( x = 2 \)?

What rate can I pick such that a compounding function with \( e \) will overlay exactly the graph of \( 2^ x \)?

The answer is the natural log of the base you want \( e \) to be in.

For \( 2 \) this would be \( 0.6931~ \).

Essentially then to have a process that compounds with a base of \( 2 \), so that, after \( 1 \) period the growth is \( 2 \) and after \( 2 \) periods \( 4 \) etc. would be

\[ e^{ln(2)} \]

Which is the same as

\[ e^{0.6931} \]

Which is the same as a saying, the rate for this process is just \( 0.6931 \).

Below is the same graph of \( 2^x \) but this time with the \( e^{x \times ln(2)} \) overlayed on top of it.