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Given a logarithm is true, if and only if, $y = \log_b{x}$ and $b^y = x$ (and $x$ and $b$ are positive, and $b$ is not equal to $1$)[1], are true, why aren't logarithms defined for negative numbers?

Why can't $b$ be negative? Take $(-2)^3 = -8$ for example. Turning that into a logarithm, we get $3 = \log_{(-2)}{(-8)}$ which is an invalid equation as $b$ and $x$ are not positive! Why is this?

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For the real, continuous exponentiation operator -- the used in the definition of the real, continuous logarithm -- $(-2)^3$ is undefined, because it has a negative base.

The motivation stems from continuity: If $(-2)^3$ is defined, then $(-2)^{\pi}$ and $(-2)^{22/7}$ should both be defined as well, and be "close" in value to $(-2)^3$, because $\pi$ and $22/7$ are "close" to 3. But the same line of reasoning says that $(-2)^{22/7}$ should be $8.8327...$, which isn't very close to $-8$ at all, and what could $(-2)^{\pi}$ possibly mean?!?!

You can define a "discrete logarithm" operator that corresponds to the discrete exponentiation operator (i.e. the one that defines $(-2)^3$ to be repeated multiplication, and thus $8$), but this has less general utility -- my expectation is that one would be much better off learning just the continuous logarithm and dealing with the signs "manually" in those odd cases where you want to solve questions involving discrete exponentiation with negative bases. (and eventually learn the complex logarithm)

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  • $\begingroup$ Great explanation! $\endgroup$
    – babakks
    Oct 1, 2016 at 9:43
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In complex analysis, $x$ can be negative. For example $e^{i\pi} = -1$, so $\ln{(-1)} = i\pi$.

I hadn't seen a log with a negative base, but I thought one could define it with the normal change of base formula: $\log_{b}{x} = \frac{\ln{x}}{\ln{b}}$. However, this turns out to be inconsistent Might be inconsistent, at the very least, it doesn't give 3:

$$\log_{(-2)}{(-8)} = \frac{\ln{(-8)}}{\ln{(-2)}} = \frac{\ln{8} + i\pi}{\ln{2} + i\pi} = \frac{3 \ln{2} + 3i\pi - 2i\pi}{\ln{2} + i\pi} = 3 - \frac{2i\pi}{\ln{2} + i\pi} \ne 3$$

This is because the complex log has a branch cut in it. For example: $e^{3i\pi} = -1$, but $\ln{(-1)} = i\pi$, not $3i\pi$. The cut is made so that $\mathrm{Im}(\ln{z}) \in \left(-\pi,\pi\right]$.

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Try taking $\log_{-2} (8)$ using your logarithm rule or for $\log_2 (-8)$. While the relation that you defined is valid, it is only valid in a particular set of cases. It's essentially useless as a mathematical tool.

And you actual can take logarithms of negative numbers, and through the change of base formula you can take logarithms of negative bases. You would have to use the Complex Logarithm to do it, which is not single valued (it actually has infinitely many values) and cannot be continuously defined on $\mathbb C$.

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You have to accept the possibility of a complex number and use a principal branch.

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In $\log_{−2}(8) = 1/2$, the way to solve it normally is to write: $(-2)^{1/2} = 8$. But in this case, $(-2)^{1/2}$ is undefined. Hence, negative bases for log cannot be used.

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  • $\begingroup$ This is an old and well answered question. Your answer contributes nothing to add value. $\endgroup$
    – Shailesh
    Oct 1, 2016 at 9:47

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