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I've read through all answers to the similar questions suggested on here, but still I don't understand why the base can't be negative.

To use the most common example, what's wrong with log-₂(4) = 2 ?

2^-2 = 4 , so surely the equation is consistent and correct. I'd really appreciate if someone could explain to me why I'm wrong. Thanks in advance.

Beezo
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    https://en.wikipedia.org/wiki/Complex_logarithm TL;DR: It's much more hairy since there is no continuous logarithm defined on all non-zero reals (let alone complex). Thus, there is no one "log" function any more once you leave the positive reals. – Rushabh Mehta Sep 27 '21 at 14:45
  • $2^{-2} \ne 4$... – Xander Henderson Sep 27 '21 at 15:26

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It's not that the base can't be negative: it's that you lose the ability to guarantee your answers are consistent and real once you do.

To give a different example, think of the input function to the function $f(x) = \log_{-2}(x)$ when $x = 8$. There is no real number that works as a result of this function.

However, there IS a complex number: $\frac{3\ln(2)}{ln(2)+\pi i}$. So just define it this way and we're done, right?

Wrong. See, the issue is that, yes, $\frac{3\ln(2)}{ln(2)+\pi i}$ works. But so does $\frac{3\ln(2)+2\pi i}{ln(2)+\pi i}$. And $\frac{3\ln(2)+4\pi i}{ln(2)+\pi i}$. And $\frac{3\ln(2)+6\pi i}{ln(2)+\pi i}$. And even $\frac{3\ln(2)-2\pi i}{ln(2)+\pi i}$. Any $f(8) = \frac{3\ln(2)+2k\pi i}{ln(2)+\pi i}$ for some integer $k$ works.

The long story short is that "logarithm" isn't a function on the complex plane (because one input can map to multiple outputs), and there's rarely a real answer when dealing with negative logarithm bases. So in most applications, we refuse to use negative bases.

HOWEVER, just because the concept of "logarithm" isn't a function on the complex plane, does not mean that we can't USE complex logarithms: it's just that we have to do something called a "branch cut", which basically picks one output for all values in a consistent way.

Breaking Bioinformatics
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  • Thank you, this makes sense. To clarify, you're saying that although when using logarithms with negative bases you lose the ability to guarantee your answers are consistent and real, logarithms with negative bases do exist. For example, log-₂(4) does truly represent a real number? – Beezo Sep 27 '21 at 15:16
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    It CAN. But the problem becomes that while $\log_{-2}(4) = 2$, it also can be equal any number $\frac{2(\ln(2) + k\pi i)}{\ln(2) + \pi i} $ for any integer $k$ – Breaking Bioinformatics Sep 27 '21 at 15:29
  • Got it, it's not a function but rather a one-to-many mapping – Beezo Sep 27 '21 at 15:36