Dear Uncle Colin,

If $(-2)^3 = -8$, why doesn’t $\log_{-2}(-8) = 3$?

Exponential Rules: Rubbish Or Reasonable?

Hi, ERROR, and thanks for your message! As is so often the case, there are two (or possibly three) answers to this.

### A problem of definition

The way logarithms are usually ((by usually, I mean ‘in my head’ - other places of definition are available)) defined as $\log_{a}(x) = \frac{\ln(x)}{\ln(a)}$.

For this to work, it requires $\ln(a)$ to be defined (so $a$ has to be positive) - and non-zero (so $a$ can’t be 1).

That’s probably why your calculator complains.

### In the real numbers

The trouble with negative-based logarithms is that something like $f(x) = (-2)^x$ isn’t a very well-behaved function for real values of $x$: it only takes on real values when $x$ is an integer.

It’s possible to construct an inverse function for $f(x)$ all the same, but it’s only defined for exact powers of -2, and isn’t exactly all that useful - it has much the same behaviour from $\log_2(x)$ with signs alternating.

### In the complex numbers

If you want things to get really messy, you can think about logarithms in the complex numbers.

Even there, the answer isn’t quite 3. This is because, in the complex numbers, logarithm functions aren’t single-valued - as a result of Euler’s identity, $e^{i\pi} = -1$. This means that $\ln(-8)$, for example, would give all of the complex numbers $z$ such that $e^z = -8$ - which would be $z=3\ln(2) \pm (2n+1)\pi$, where $n$ is an integer.

Similarly, $\ln(-2)$ is $\ln(2) \pm (2m+1) \pi$, where $m$ is an integer.

That means that $\log_{-2}(-8)$, in the complex numbers, is $\frac{3 \ln(2) \pm (2n+1)\pi}{\ln(2) \pm (2m+1) \pi}$. In the very specific case where $n=1$ and $m=0$, this does work out to three! But there are infinitely many other answers, too.

Hope that helps!

- Uncle Colin

* Edited 2020-01-22 to fix a typo. Thanks, Rob!