# Believe it or not… a log question which was briefly stumping us

Hi all,

A teacher approached me with the following question.

The function $\ln(x^{-2})$ has a graph that looks like: It makes sense that the function exists for all negative x values, because when you raise a negative number to the -2 power, you’re going to get a positive number. And you can take the natural log of a positive number.

Then the teacher said to consider the following function: $-2 \ln(x)$, and the graph looks like: Notice that you can’t input negative x values, because the domain of natural log doesn’t allow for it.

Here’s the question.

According to the log rules/properties, we know that: $\ln(a^b)=b\ln(a)$ (obviously).

So $\ln(x^{-2})=-2\ln(x)$. But the graphs are different.

We went a little crazy trying to figure out what’s going on… For about 3 minutes, we were having a great conversation. But we quickly converged on the little text that accompanies the log rules in any textbook… and this text says that these rules work but are only valid for $a>0$.

I kinda love this as an in-class exercise (I’ll probably forget this when I get to logarithms, but maybe posting it here will prevent me from forgetting it). Because it will force kids to (a) be confuzzled, (b) talk through ideas, (c) go back to the definition and qualifications for the log rules, and (d) see that these rules are indeed valid (we didn’t break math), but they are a bit more restrictive that we might have thought.

What I love is that $\ln(x^{-2})=-2\ln(x)$ isn’t actually an identity. But we are so used to using the rules blindly, robotically, that we never think about it. But for it to be a good mathematical statement, you need to qualify it! You need to say this is only an equivalence for $x>0$. This was a good reminder for us.

1. Coquejj says:

Hi Sam and congratulations for your blog. I’ve followed you for quite a long time (in google reader, so your traffic didn’t increase)
Actually, this wonderful fact was a controversial topic in the 18th century among mathematicians. Take a look at this book:
http://goo.gl/d2a7i

2. pamlpatterson says:

I believe it! This stumped us last year so, too. I made a point to focus on it this year because the part about a being a positive real is lost in the listing of the actual properties.

3. John Berray says:

Awesome to show students! It gives them a glimpse into why all the math legalese is important, without just saying “it’s important.” Graphs don’t lie. Mostly.
It makes me think of this…every blue moon a student wonders why we can’t multiply two negative radicands, get a positive product, and then circumvent complex numbers altogether.

1. samjshah says:

HAAAAAAAA! I just talked about that in a class I subbed yesterday!

4. Kendall Root says:

Curiously, the advanced algebra text and precalc texts that our school uses says that *a* can be any real number. Furthermore, the way they define an identity would allow -2 ln x = ln (x^-2) to be an identity. However, they would still be able to explain this dilemma because their definition says that the equation only needs to be true for x values in both the domain of -2 ln x and ln (x^-2). Negative x values are not in the domain of -2 ln x. This is also how they justify saying cos x = 1/sec x is an identity, even though cosine is defined at x = pi/2, but secant isn’t.

5. peeterjoot says:

I’d look at it this way: \begin{aligned}\ln(-{\left\lvert{x}\right\rvert}^{-2})&=- 2 \ln( -{\left\lvert{x}\right\rvert} ) \\ &=- 2 ( i \pi + \ln( {\left\lvert{x}\right\rvert} ) ) \\ &=- 2 i \pi - 2 \ln( {\left\lvert{x}\right\rvert} ) \\ \end{aligned}

So, you get your expected result from “plugging” into the formula, with the addition of a $2 i \pi$ term which doesn’t change the result if re-exponentiated (ie: $e^{z + 2 \pi i} = e^z$.)

1. Dario says:

I agree with you! The way I would explain this to students (no need of the complex analysis here, since our argument is positive), is using the fact that . Something like this:

2. Dario says:

I agree with you! The way I would explain this to students (no need of the complex analysis here, since our argument is positive), is using the fact that $\sqrt{x^2}=|x|$. Something like this:
$\ln(x^{-2})=-2\ln(\sqrt{x^{-2}})=-2\ln(|x|^{-1}).$

6. Sue VanHattum says:

Hmm, wonder if I can use this…

7. Bowen Kerins says:

Yup. You can get there without the negatives too: ln(x^2) versus 2 ln(x).

One of my usual log test questions was sketching accurate graphs of e^(ln x) and ln(e^x), and explaining how / why they were different / same. This after an earlier test question about graphing sqrt(x^2) and (sqrt(x))^2.

1. Jim Doherty says:

Great example. I remember similar conversations with the early TI calculators when we would talk about how the machine treated sqrt(x^3) differently than x^(3/2) Good conversations…

8. LSquared says:

It reminds me a bit of all of the rules about infinite series I learned for an analysis class long ago. In my memory, then all boiled down to: if everything converges, you can split up the infinite series the way you want to, but if any of the parts diverges, then you can’t use that algebraic manipulation.

1. gasstationwithoutpumps says:

If I remember right (from 35 years ago) convergence is not enough. It has to converge absolutely in order to be free with rearrangements.

9. Pablo says:

That’s a really interesting little thing to show students why proofs matter. The goal of learning how to prove the log properties shouldn’t be just to learn them by rote, but to understand better how logarithms work. (At least in this case).

Congratulations for the blog,
Pablo