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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 $e^{bx}$?

rhody
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    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? – rhody Sep 11 '22 at 21:20
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    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. https://en.wikipedia.org/wiki/Natural_logarithm – Travis Willse Sep 11 '22 at 21:21
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    That's right.$ $ – Travis Willse Sep 11 '22 at 21:21
  • There's a minor problem when $a\le0$ – PM 2Ring Sep 11 '22 at 21:28
  • Yes, I can see now. – rhody Sep 11 '22 at 21:31
  • @PM2Ring - When dealing with real numbers, for $a < 0, a^x$ is only defined when $x$ is an integer. If you allow complex numbers for logarithm and exponent, then for $a < 0, \ln a = \ln |a| + \pi i$ (at least that is the most common definition), and $a^x = e^x\ln a$ still holds for integer $x$. (And for some non-integer $x$, but there are limitations.) – Paul Sinclair Sep 13 '22 at 12:28

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