# Relationship between a^x and e^(bx)

Is it true that any exponential $$a^x$$ can be represented using $$e^{bx}$$, assuming a suitable choice for b? For example, if we were to consider $$y = 2^x$$, what would be the equivalent $$y = e^{bx}$$ that gives the same curve as $$2^x$$? In my empirical studies, by plotting both functions, I can seem to always find a value for b that will give a curve $$e^{bx}$$ that looks like any $$a^x$$. This would make sense since growth functions are described using $$e^{bx}$$, hence $$e^{bx}$$ should be flexible enough to represent any exponential. Is this correct, and if so, what is the relationship between $$a^x$$ and $$e^{bx}$$?

Another way of asking is if I had an exponential such as $$3^x$$, that would be the value of $$b$$ in the equivalent $$e^{bx}$$?

• I think I just figured it out, b = ln (a), all I did was set a^x= e^{bx), take log on both sides, then solve for b in which case I get ln (a). Correct? Sep 11, 2022 at 21:20
• Probably this question is essentially a duplicate, but if $a^x = e^{bx} = (e^b)^x$, then $a = e^b$, so by definition $b = \log a$, where $\log$ is the natural logarithm function. en.wikipedia.org/wiki/Natural_logarithm Sep 11, 2022 at 21:21
• That's right. Sep 11, 2022 at 21:21
• There's a minor problem when $a\le0$ Sep 11, 2022 at 21:28
• Yes, I can see now. Sep 11, 2022 at 21:31