Can a logarithm have a negative base?

1 Answer
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 != 1, then:

a^x: (0, oo)->RR

is a one-one function with inverse:

log_a: RR->(0, oo)

Does this idea extend to negative bases?

If a < 0 and n in NN, then we can quite happily define:

a^n = stackrel "n times" overbrace ((a)(a)...(a))

giving us a well defined one to one function:

a^n:NN->{a^n: n in NN} sub RR

which has a well defined inverse

log_a:{a^n: n in NN}->NN

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 RR

What values of z work?

Taking natural logs of both sides:

ln y = ln a^z = z ln a = z (ln (-a) + i pi (2k+1))

So z = ln y / (ln (-a) + i pi (2k+1)) for some k in ZZ

That is, we can define:

log_a(y) = ln y / (ln (-a) + i pi (2k+1)) for some k in ZZ

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.