Ambiguous conditions for $x$ in logarithm?

Question

I have just realised that there is a little ambiguity in defining conditions for $x$ in logarithm. Let me illustrate it on a simple example:

$\log{x}$ is valid for $x>0$ ,

$2\log{x}$ is also valid only for $x>0$ ,

but $\log{x^2}$ is valid for both $x>0$ and $x<0$ .

How is that possible when $2\log{x} = \log{x^2}$ ?

That is right if you look at it from right to left but not from left to right. And also this is unfortunately not what they teach students at school as I realized today while tutoring... the basic formula is $c\log{a}=\log{a^c}$ without any comment on $c$. — Václav Pavlík, Jun 03 '16 at 10:27
Every decent source cites the appropriate condition for that formula to hold. Even Wikipedia https://en.wikipedia.org/wiki/List_of_logarithmic_identities ['..where $x$, $b$ and $y$ are positive real numbers'].... — Pierpaolo Vivo, Jun 03 '16 at 10:30

Noam · Answer 1 · 2016-06-03T10:45:18.837

1

Note that logarithm identities are true only when both sides of the equation are defined, so $2\log x = \log x^2$ for $x > 0$, and not when $x < 0$.

This thinking could lead to things like $-1 = \sqrt{-1} \cdot \sqrt{-1} = \sqrt{-1 \cdot -1} = \sqrt{1} = 1$, which is obviously false. The problem here is that $\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$ only when $a,b>0$.

edited Jun 03 '16 at 10:45

answered Jun 03 '16 at 10:40

Noam

439

score 1 · Accepted Answer · answered Jun 03 '16 at 11:09

There is no ambiguity:

it is true that $\log(x^2)$ is defined for all $x\ne0$;
it is false that $\log(x^2)=2\log x$ for all $x\ne0$;
it is true that $\log(x^2)=2\log x$ for all $x>0$;
it is true that $\log(x^2)=2\log|x|$ for all $x\ne0$.

You can't compare expressions when one of them is undefined.

The identity $\log(a^c)=c\log a$ is valid for $a>0$. If students are not aware of this, it is the tutor's/teacher's duty to make them being conscious that math formulas are not magic formulas and equalities apply only when both sides are defined.

score -1 · Answer 3 · edited Jun 03 '16 at 18:54

As well known language do not follow the rules of logics. "If you behave well you will have an ice cream" according to logic does not imply that if you behave badly you will not have the ice cream, but any child knows that in language this is precisely what it means. One could argue that language reflect the way we reason, logics the way we should reason. No students, even the ones in math would not understand what you mean when you say $\log(x)$ is defined for positive $x$ and then $\log x^2 =2 \log x$.

Ambiguous conditions for $x$ in logarithm?

3 Answers3