Why logarithm of a negative number to a negative base is not defined?

Question

I have studied that the numbers of the form N=aˣ can be written as $\log_aN=x \;\forall N>0, a>0, a\not=1.$

Let's take $N=-8$ and $a=-2$ then $-8=(-2)^3$ and $\log_{(-2)}-8 = 3$.

Why there are restrictions on the number and the base even though $\log_{(-2)}-8 = 3$ is defined? Is there something I'm missing as I'm only learning the propaedeutics of mathematics?

Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. — José Carlos Santos, Sep 18 '21 at 07:26
It does make sense to assert that $\log_{(-2)} [-8] = 3$, but what about $\log_{(-2)} [-4]$ or, for that matter $\log_{(-2)} [-8.01]$? The point of logarithms is to solve problems. What problem solving is facilitated by defining base $(-2)$ logarithms for isolated numbers, such as $-2, -8,$ or $-32$? — user2661923, Sep 18 '21 at 07:33
@user2661923 Logarithms exist regardless of whether they're of any use to us. — Robert Soupe, Sep 18 '21 at 07:44
Full understanding of problem you will get, if you read about complex logarithms. — zkutch, Sep 18 '21 at 07:48
Questions about logs with negative bases have been asked and answered here many times already. Please see the list under Related for some of them, doubtless there are many others you can find by doing a search. — Gerry Myerson, Sep 18 '21 at 07:48
If you're now unable to see my answer here, and you still want to read it, let me know by pinging me here. — user21820, Jan 28 '22 at 18:19

score 2 · Answer 1 · answered Sep 18 '21 at 07:34

2

If $a>0$, then $a^x$ is defined for every real number $x$ and, if $y>0$, what we denote by $\log_ay$ is the only $x\in\Bbb R$ such that $a^x=y$.

But if $a<0$ then $a^x$ is defined only for some numbers $x$; for instance, $a^{1/2}$ is undefined. That's why we don't talk about $\log_ay$; it's undefined for most numbers $y$.

answered Sep 18 '21 at 07:34

José Carlos Santos

427,504

@user21820 It's gone. – José Carlos Santos Jan 28 '22 at 18:30
Thanks a lot! =) – user21820 Jan 28 '22 at 18:30

Narasimham · Answer 2 · 2021-09-19T07:54:06.710

0

In the special case when the given number and base both are real negative, log is well defined in terms of complex numbers and there are no restrictions.

$$ \log _{-a}(-N)=\frac{\log (-N)}{\log (-a)}= \frac{\log (N) + i \pi}{\log (a)+ i \pi} $$

$$=\frac{\pi-i\log (N)}{\pi-i\log (a)}$$

which is a complex number. The new unlabeled base can be again any real number b.

EDIT1:

If new base $b$ is positive then there is no problem but if negative we again have

$$ b= - |b|,\quad \log _{\;b} N =\frac {\log N }{\log b}=\frac {\log N }{i \pi+ \log |b|} $$

and the nestings / branch cuts of log can propogate continuously depending on the sign of newly chosen/introduced bases.

edited Sep 19 '21 at 07:54

answered Sep 18 '21 at 07:50

Narasimham

40,495

I think you will find that the final expression you have written down is NOT independent of the choice of (positive, real) base. They also depend on the choice of branch of logarithm you chose when you wrote $\log(-a)=\log(a)+i\pi$. – Gerry Myerson Sep 19 '21 at 03:40
Thanks, edited to include choice of sign of new base, hope OK. – Narasimham Sep 19 '21 at 07:57
I guess I wasn't clear. Pick some positive numbers $a$ and $N$. Pick two different positive bases, $b$ and $b'$. Calculate $(\pi-i\log_bN)/(\pi-i\log_ba)$ and $(\pi-i\log_{b'}N)/(\pi-i\log_{b'}a)$. Are they the same? – Gerry Myerson Sep 19 '21 at 10:19
And there's still the matter of choosing a branch for $\log_b(-N)$. – Gerry Myerson Sep 19 '21 at 10:31
In short, this answer is wrong, Narasimham. Why is it still here? – Gerry Myerson Sep 21 '21 at 13:30

Why logarithm of a negative number to a negative base is not defined?

2 Answers2