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 $e^{bx}$?

  • 2
    $\begingroup$ 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? $\endgroup$
    – rhody
    Sep 11, 2022 at 21:20
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    $\begingroup$ 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 $\endgroup$ Sep 11, 2022 at 21:21
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    $\begingroup$ That's right.$ $ $\endgroup$ Sep 11, 2022 at 21:21
  • $\begingroup$ There's a minor problem when $a\le0$ $\endgroup$
    – PM 2Ring
    Sep 11, 2022 at 21:28
  • $\begingroup$ Yes, I can see now. $\endgroup$
    – rhody
    Sep 11, 2022 at 21:31


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