# Can a logarithm have a negative base?

Nov 21, 2015

Sort of yes, but it's not very useful...

#### Explanation:

This is a really interesting question.

The answer is basically yes, but it's not generally very useful.

First let's look at logarithms with positive bases.

If $a > 0$ and $a \ne 1$, then:

${a}^{x} : \left(0 , \infty\right) \to \mathbb{R}$

is a one-one function with inverse:

${\log}_{a} : \mathbb{R} \to \left(0 , \infty\right)$

Does this idea extend to negative bases?

If $a < 0$ and $n \in \mathbb{N}$, then we can quite happily define:

${a}^{n} = \stackrel{\text{n times}}{\overbrace{\left(a\right) \left(a\right) \ldots \left(a\right)}}$

giving us a well defined one to one function:

${a}^{n} : \mathbb{N} \to \left\{{a}^{n} : n \in \mathbb{N}\right\} \subset \mathbb{R}$

which has a well defined inverse

${\log}_{a} : \left\{{a}^{n} : n \in \mathbb{N}\right\} \to \mathbb{N}$

So in this sense we can say things like ${\log}_{- 2} - 8 = 3$

Things get more complicated and Complex once we start dealing with fractional exponents.

Suppose $y = {a}^{z}$ for some $y \in \mathbb{R}$

What values of $z$ work?

Taking natural logs of both sides:

$\ln y = \ln {a}^{z} = z \ln a = z \left(\ln \left(- a\right) + i \pi \left(2 k + 1\right)\right)$

So $z = \ln \frac{y}{\ln \left(- a\right) + i \pi \left(2 k + 1\right)}$ for some $k \in \mathbb{Z}$

That is, we can define:

${\log}_{a} \left(y\right) = \ln \frac{y}{\ln \left(- a\right) + i \pi \left(2 k + 1\right)}$ for some $k \in \mathbb{Z}$

Regardless of what value of $k$ we choose, this will have a non-Real Complex value most of the time, but at least we can restrict the domain of ${a}^{z}$ to get a well defined inverse.