I'm struggling to understand why a $\log_x y$ is only true for $y > 0$.

I know that if $y$ was $0$ , $x$ could only be $0$, but if it was $0$, then $y$ could be anything. So we'll leave that undefined.

I also know that, for example, $\log_2 -x$ doesn't make sense, because $2^a = -x$ has no real solution.

However, I just cannot understand why $\log_{-2} -8$ doesnt' make sense. The argument is negative, but it works, it's 3.

So why do we say that the argument of a logarithm cannot be negative? And also, do you confirm that the base of a logarithm can be anything but $0$? I have seen people state that the base also cannot be negative, which to me seems absurd.

  • $\begingroup$ In principle it makes sense on integer powers of a negative base, but all the nice properties for real (or even rational) powers become a real mess! What is log_{-2} (-5) ? Any solution will be complex values and messy. In conclusion, better consider a positive base. $\endgroup$ – H. H. Rugh Oct 10 '16 at 16:39
  • $\begingroup$ I know, but messy and hard doesn't mean impossible, I'm just trying to understand the underlying reason why log(-2)(-8) is not defined. When I look at the function graph, I struggle to understand why it's only defined in X > 0. The log of (-2)(-5) is 2.321928094887362 EDIT: It would be, but it's not, because log(-2)(-5) is not defined [WHY?!]. $\endgroup$ – Athamas Oct 10 '16 at 16:43
  • $\begingroup$ $(-b)^{odd} = neg$ and $(-b)^{even} = pos$ and $(-b)^{n/2}= \sqrt {(-b)^n}$ is undefined. $\log_{-2}(-5)$ is undefined because $(-2)^x = -5$ has no solution. $\endgroup$ – fleablood Oct 10 '16 at 16:52
  • $\begingroup$ Sir, -2^2.321928094887362 is -5, so it is defined. What am I doing wrong? (−b)^(n/2) = sqrt((-b)^n) is undefined only if (-b)^n < 0. Not to be confued with -(b^n). $\endgroup$ – Athamas Oct 10 '16 at 16:57

This difficulty in the definition of the logarithm arises due to the way exponentiation works.

If we allow solutions only in real numbers, not complex numbers, it is possible to define some non-integer powers of negative numbers. Specifically, the rule is,

For any real number $a$, for any integers $m$ and $n$ such that $n>1$ and $m$ and $n$ have no common factor, if $\sqrt[n]a$ is a real number then $$a^{m/n} = \left(\sqrt[n]a\right)^m.$$

You'll find this rule, or rules that imply it, in various places, including here, here, the Wikipedia page on Exponentiation, or any number of high-school algebra textbooks.

This rule depends on the assumption that we know when $\sqrt[n]a$ is defined (for an integer $n$) and what it is when it is defined. When $n$ is odd, $\sqrt[n]a$ is the unique real number $x$ such that $x^n = a$. But when $n$ is even, $\sqrt[n]a$ is defined only when $a \geq 0$, and it is defined then as the unique non-negative real number $x$ such that $x^n = a$.

We cannot define $(-4)^{1/2}$ in real numbers, because there is no real number $x$ such that $x^2 = -4$. There is not even a real number $x$ whose square is close to $-4$; all the squares of real numbers are zero or positive.

Now consider irrational exponents. We can define irrational powers of real numbers by assuming $a^x$ is a continuous function of $x$ when $a$ is positive. We can do this because the rational powers of positive real numbers "fit the curve" of a continuous function; if $p_1, p_2, p_3, \ldots$ is any sequence of rational numbers converging to a certain rational number $p$, then $2^{p_1}, 2^{p_2}, 2^{p_3}, \ldots$ is a sequence of real numbers converging to $2^p$. To extend this to irrational exponents, we define $2^\pi$ (for example) as the limit of $2^{p_1}, 2^{p_2}, 2^{p_3}, \ldots$ where $p_1, p_2, p_3, \ldots$ converges to $\pi$; and if for every sequence of rational numbers $q_1, q_2, q_3, \ldots$ that converges to a certain real number $r$, the powers $2^{q_1}, 2^{q_2}, 2^{q_3}, \ldots$ converge to $5$, then we say that $2^r = 5$.

This works fine for defining $\log_2 5$, because there is a unique real number $r$ such that for every rational sequence $q_1, q_2, q_3, \ldots$ that converges to $r$, the sequence $2^{q_1}, 2^{q_2}, 2^{q_3}, \ldots$ converge to $5$. It does not work for $\log_{-2} (-5)$, however.

The problem with $\log_{-2} (-5)$ (in particular, the reason it is not equal to $\log_2 5$) is that in any sequence of rational numbers $q_1, q_2, q_3, \ldots$ that converges to $\log_2 5$, it is possible that when reduced to lowest terms ($m/n$ where $m$ and $n$ have no common factor), the numerator of each $q_i$ might be positive or it might be negative. We can easily make a sequence where all the numerators are odd, in which case $(-2)^{q_1}, (-2)^{q_2}, (-2)^{q_3}, \ldots$ converges to $-5$, or a sequence where all the numerators are even, in which case $(-2)^{q_1}, (-2)^{q_2}, (-2)^{q_3}, \ldots$ converges to $5$, or a sequence in which the numerators alternate between odd and even, so $(-2)^{q_1}, (-2)^{q_2}, (-2)^{q_3}, \ldots$ alternates between $5$ and $-5$ and does not converge to anything.

In short, it's really not justifiable to say that $(-2)^{\log_2 5} = -5$. There isn't a really good reason to say $(-2)^{\log_2 5}$ is a real number at all.

Nor is there any better candidate to be the real number $x$ that solves the equation $(-2)^x = -5$. That leaves us without a good way to define $\log_{-2}(-5)$, nor the log base $-2$ of most other numbers. We end up not being able to use log base $-2$ for just about any of the things we find logarithms really useful for, so we don't even try to define it for the cases where it might possibly make sense.

  • $\begingroup$ So would you say that $\log_a b$ is defined for $a > 0$ and $b > 0$ OR for $b > 0$ and $a != 0$ ? Because text books often only say "The condition of existence is that the argument be positive". $\endgroup$ – Athamas Oct 10 '16 at 21:15
  • $\begingroup$ No, that's not what I would say at all. I said there is not any good way to define the log base $-2$ of most numbers (which includes most positive numbers, if you follow the reasoning up to that point), and the same thing can be said about log base $a$ whenever $a < 0$. $\endgroup$ – David K Oct 11 '16 at 4:22
  • $\begingroup$ Ok, but in the end one must come to a strict, precise conclusion, because the log might be part of a bigger equation, where one needs to define a precise field of existence. Am I to assume that $\log_a(b)$ has a > 0 and b > 0 when I outline the solution to an equation that contains a log? For example an equation might be verified for a = -3 and b = 6. If it contains a $\log_a(b)$ somewhere, the only actual solution is 6? $\endgroup$ – Athamas Oct 11 '16 at 15:36
  • $\begingroup$ The conclusion in real analysis is that $\log_a$ is defined for $a > 0$ and only for $a>0$. I don't know what you mean by an equation that might be verified for $a=-3$ and $b=6$ that contains $\log_a b$ somewhere. An equation that uses $\log_a b$ cannot be satisfied by $a=-3$ and $b=6$. Moreover, when we rule out $a=-3,b=6$ as a solution, we rule out the entire solution--we don't just ignore the problem with the $a=-3$ part and say $b=6$ is a solution. $\endgroup$ – David K Oct 11 '16 at 16:20
  • $\begingroup$ If you have a specific mathematical problem in mind that you think needs to be solved by logarithms with negative bases, I suggest you post it as a new question. The answer to this question is that logarithms with positive bases are extremely useful in real analysis while logarithms with negative bases are practically useless for real analysis, so it saves a lot of trouble if, when doing real analysis, we just define logarithms on positive bases and leave negative bases undefined. $\endgroup$ – David K Oct 11 '16 at 16:24

The usual definition of $\log_ab=x$ in $\mathbb{R}$ is that $x$ is the number such that $a^x=b$.

This definition works well if $a$ and $b$ are positive numbers because:

1) For $a<0$ we can have situations as $\log_{-4}2=x$ that is $(-4)^x=2$ that is impossible in $\mathbb{R}$

2) for $b<0$ we can have situations as $\log_{4}(-2)=x$ that is $(4)^x=-2$ that is impossible in $\mathbb{R}$

In some case, and for suitable values of $a$ and $b$ we can find a value for $x$ also in these cases, but a complete and coherent definition of the logarithm function also in these case can be done only on the field $\mathbb{C}$ where we have no limitations (but some trouble with multivalued functions).

  • $\begingroup$ If given one specific case when these conditions do not apply, what am I to do? Write the answer or say that since th argument is <0 the log is not defined? $\endgroup$ – Athamas Oct 10 '16 at 16:51
  • $\begingroup$ The graph of $y=\log x$ has negative $y$ for $0<x<1$ because $a^y=b \Rightarrow a^{-y}=\frac{1}{b} $, but note that we have noting for $x<0$. For the other question: in these situation it is better to find the complex solutions and, eventually, take the real solution, if we are interested only on real values. $\endgroup$ – Emilio Novati Oct 10 '16 at 17:00
  • $\begingroup$ You solve it without evoking logarithm If you have $(-8)^x = 4$ you do not evoke $\log_{-8} 4 = ???$. You do something else instead. $(-8)^x = (-1)^x 8^x = 4$ so if $-1^x = 1$ so can declare $x = \log_8 4 = 2/3$. As $(-1)^{2/3} = 1$ we can conclude $x = 2/3$ but in general we may not have be able to do so. Example $(-8)^x = -4$ will have not be solvable. $\endgroup$ – fleablood Oct 10 '16 at 17:01
  • $\begingroup$ (-1)^x = 1? why? if x = 3, -1^3 = -1 $\endgroup$ – Athamas Oct 10 '16 at 17:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.