In your head, can you find the average of 2741, 2742 and 2746?

Yes

Yes. Just averaged 1, 2, & 6.

In my head, I can do anything 0.o

Why make things so complicated? Add the last digit of each number and divide by 3.

Yes, I could easily figure it out in my head, but when you got to the “log” portion of your details, I stopped reading.

I did the same as @filmfann. So yes.

But how is step 1 (both steps 1) either necessary or useful? Here it obviously makes the computation easier because you can see at a glance that the numbers are close together. But what if you just follow the normal rules for averaging? You get

2741 + 2742 + 2746 = 8229
8229 / 3 = 2743

This is a corner case, right? If the starting numbers were, say, 2741, 87, and 953, you wouldn’t perform such a transformation to begin with.

I see what you’re getting at with your analogy, even though I had logs removed from my operating system decades ago, but I’m sorry to say that I don’t think this one works.

Incidentally, you introduced the term isomorphism but didn’t tell us what it means. So it does not add anything in the way of clarity here.

Also, if the numbers were 2746, 2750, and 2751, which is the same range of values, the answer becomes less obvious. (It’s 2749.) Would we all do that so easily in our heads? I know I wouldn’t.

I avoided defining isomorphism because the definition is a bit abstract and I wanted to just give the general idea. But I will give it a shot. Consider the case of the log function. We have log(x*y)=log(x)+log(y). In place of * and +, we can substitute other operators and denote them as & and %. In place of log we can use some other transformation T. So we have an isomorphism when T(x&y)=T(x)%T(y)

As for working on 2746,2750 and 2751, we could subtract 2740 from each, giving 6, 10 and 11, which add to 27. 27/3=9, so the answer is 2749. Another approach is to list the numbers relative to 2750, giving -4, 0 and 1. -4 +0+1=-3 and -3/3=-1, so the average is2750–1=2749

Ok. But your details imply that this first step, subtraction, is necessary to determining the average. It isn’t. Would anyone really look at the problem of averaging 2746, 2750, and 2751 and say, “Well, now, let’s start by subtracting 2740 from each”? Would you? It’s completely arbitrary. Why not subtract 2746 from each? or, for that matter, why not 1963?

How would that work with the other example, 2741, 87, and 953? I don’t see how this approach is going to help explain your concept.

My assumption is that you’re trying out on us various exercises that you want to use with students. No problem with that. In the present instance, though, it’s my opinion that this is a forced analogy that will cast more shadow than light.

easy peasey lemon squeezy

@Jeruba, In my example, you can replace subtraction and addition with removing the leading digits, finding the average of the last digits, and then restoring the leading digits. The key idea is that the first and last operations are the opposite of each other, if not explicitly then at least implicitly.

Subtraction works when the numbers are close to one other, which would include the case where the initial digits are all the same.

@LostInParadise, I’m not saying it doesn’t work. Of course it works. But why do this? Why not just use the numbers as they are? Because it isn’t necessary to getting a correct answer, I don’t see that it serves as an apt analogy for your explanation about logs. I’m focusing on the problem as stated:

The purpose of this exercise is to see if it can be used to explain the very useful mathematical concept of isomorphism. Let me know if this makes any sense.

That’s what I’m trying to do. And I’d say that it doesn’t. That is, it makes sense strictly literally and logically, but it’s not realistic. Won’t students hang up on that and miss that it’s being used as a (contrived) analogy? Just my opinion here, and not trying to prove anything. My last math class was a very long time ago, and it was never my best subject.

I’m wondering if you might do better to look outside of math for a suitable analogy. Some processes do end with a reversal of the first step.

What if I replace the example that I used with one of the ones that you suggested? It seems to me that the problem using averages is a good intuitive demonstration of ending by reversing the last step. Another simple example is to multiply two Roman numerals by converting them to regular numbers, which are then multiplied, and finally convert the answer to a Roman numeral.

The idea of ending by reversing the last step shows up in a lot of places, for example creating a mathematical or physical model, the manipulation of which can be used to test and refine whatever it is that is being modeled.

Oh, I think the Roman numerals example is much better! It’s clear and relevant, and the transformation to Arabic is necessary, not an arbitrary contrivance. You do have to decode and then encode the values (or vice versa) to get an answer, much like what you’re trying to illustrate. Your students will thank you.

And I would have thanked my 11th-grade math teacher if he’d explained logarithms this way.

Sure, even I could do that. :)

The Roman numeral example also helps to illustrate the essential idea of isomorphism. Isomorphism means “same form”. Multiplying Roman numerals is essentially the same thing as multiplying Arabic numerals – different names for the same operation. For example, does V * VI equal VI * V? V*VI gets translated to 5*6 and VI*V gets translated to 6*5. 5*6=6*5, so the same must be true of V*VI and VI*V.

The multiplication formula using logarithms translates multiplication of positive numbers into addition of all numbers. In some sense, these two operations must be structurally identical. Does 5*6 = 6*5? The first gets translated to log(5) + log(6) and the other to log(6)+log(5). If we assume that order does not matter for addition then the same must hold for multiplication.