A week ago, also, the maths internet lost its mind over this
log troll, I guess.
And the reason people got so excited about it is because even though this is technically correct,
the log of 6 does equal the log of 1 plus the log of 2 plus the log of 3,
this doesn't work in general. It looks like you can generalize it,
but if these were A B and C, instead of 1 2 3, it simply would not hold.
Although, it's very close to a real log law!
Instead of adding at the top there, that should be...
This works in general; The previous one does not.
So how did somebody go from this *legitimate* log equation to the fake one?
Well, it's just using the fact that 1 times 2 times 3 happens to equal 1 plus 2 plus 3.
Because of that particular fact, you're able to make the fake log law. And that's why people got so emotional.
What I found extra interesting is wherever this meme was discussed, someone mentioned perfect numbers!
Perfect numbers kept coming up, even though they don't explain why it works.
Although, they're close. So 6 is a perfect number, which means if you take all of its proper factors—
numbers that divide into it which are not the number itself—
if you add 1 + 2 + 3 it equals 6, if you multiply 1 times 2 times 3 it equals 6.
But for any other perfect number, let's say 28.
Yeah, sure, if you add all the proper factors of 28 they give you 28,
But if you multiply all the factors you don't get 28. You'd have to multiply just 2 times 14, or maybe 4 times 7.
All of them, though, doesn't work. So 6 only works in the log troll because if you multiply all of its factors you get itself.
So being a perfect number doesn't mean that the number will work in the log troll,
and some numbers will work in the log troll that aren't perfect numbers.
So this is a total distraction.
What we're actually looking for are sets of numbers such that if you add all of the numbers together—
So Sigma for summing across X 1 X 2 up to some X, and that's just some set of numbers I'm calling X's—
then you get the same answer as when you multiply them all together. That's a capital Pi,
so same as a capital Sigma means to sum all the numbers, capital PI means that you find the product. You multiply all the numbers.
So, if the sum of a set of numbers equals the product of that set of numbers, then you can get a log troll.
And solutions to this are easy to find, if you don't care about having whole numbers.
So my favorite that I found after a quick search was 101 and 1.01
So, because 101 times 1.01 gives you the same answer as 101 plus 1.01, this holds! What a... Yeah.
Whole number solutions are much harder to find, though.
In fact, I would argue there is only one valid whole number solution to the log troll problem and it's 4,
because 2 plus 2 equals 4 and 2 times 2 equals 4. And so this—
Even though this shouldn't work by all known log laws—
It does work because you can replace the two plus two or 2 times 2 and then the whole thing would work, in terms of logs.
I reckon I can pseudo prove via a bit of hand waving that this is the only non-trivial solution to the log troll problem
So, to start with, I have plotted, for all the numbers—
starting down at 1 over the end of the origin up to a thousand over here—
the maximum value you can get when you multiply together a... factor decomposition of the number.
So you find out which set of factors, which multiply to give the original number, happen to sum to the maximum possible value,
and the numbers, they're not very big. They're nowhere near the original numbers.
So if you've got something like, I don't know, like, I think it was like, 400, like over there?
There's 400, and you can see- Well, for an even number, the biggest factor it can have is half of it, and then the other factor is 2, so 400 goes up to 202.
In fact that top line you can see going down, that's the line of half the number plus 2.
For the numbers which aren't even, but do have a factor of 3, the next line down. That's a third of the number, plus 3
And then you've got, you know, a fifth of the number plus 5, and each of these lines is a different prime factor.
And for a number it falls on its biggest prime factors line. So this only works for composite numbers,
If it's a prime number, you're gonna end up one bigger than the original number.
But for all other composite numbers you will end up with something... smaller.
And it doesn't matter if you try and have more factors because then you're just splitting the two you start with into smaller ones,
and, if every time you split it into two factors you get a smaller result, you're just gonna get smaller and smaller and smaller.
So two factors is your best case, and these are the maximum values you're gonna get.
None of them are near the original number except for t̶w̶o̶ [FOUR!], which is on the top line,
and half of t̶w̶o̶ [FOUR!] is two! and so Down near the origin somewhere— you can't see it on this scale— t̶w̶o̶ [FOUR!] is on that line.
The rest of them will never give you a whole numbered solution to the log troll problem.
Thank you for following along. I think I got there, hand waving!
"Wait a minute," you say! If I'm arguing that this is the only non-trivial solution to the log troll problem,
then what's up with the *original* solution to the log troll problem?
This one I found particularly amazing when it came out and people were discussing it online. People were talking about perfect numbers,
a semi obscure part of recreational mathematics, and not something particularly relevant to the conversation!
And no one was saying, "Hang on! Why have you got the log of one? That equals zero!"
You're just- You're adding zero to that half. That's ju-!
I mean, adding zero changes nothing, when you're doing addition,
just like multiplying by one changes nothing, when you're doing multiplication
And, well, actually that kind of makes sense because in a crude, hand-wavy sense logs take multiplying and turn it into addition.
So it takes the identity from multiplying, one, which changes nothing, to the identity for addition, o̶n̶e̶ [ZERO!], which changes nothing when you're adding.
And so, actually, if you want to, you can add as many log ones as you want, and it won't change it whatsoever.
They're- they're pointless. They're trivial! They're just- They aren't doing anything.
And we can use this to create a log troll for any composite number. For example, 10.
The first step is to take your composite number of choosing and turn it into some factors.
I've picked 2 and 5. We can now make a perfectly valid log statement about these factors.
The log of the factors equals the sum of the individual logs of the factors and that's true, that's valid...
But it's not gonna troll anyone, now, is it?
Our issue is that 2 and 5, while they multiply to give ten, they don't *add* to give ten.
So what we do, make a bit more space over there, and because multiplying by 1 doesn't change anything,
we can chuck as many ones as we need in there. Three more ones!
And then down on the bottom, I'm gonna add three more zeroes, or "log ones," as they're called.
So, both sides here still equal each other; they're both still equal to the same thing.
But because the set of numbers in the top log-
Because if you multiply them all together you get the same answers if you add them all together,
we can take those cheeky multiplication signs, give them a little twist, and now we're adding them!
This is a log troll!
WHa - WWHAaaat (accompanied by air horn noises)
We've had some fun today. We've learned about logs, we've learned about maths memes.
But can we be serious, for a moment, and take what we've learned, and use it to spice up the original log troll?
I think we can! So this one uses sequential numbers starting at 1, 1 2 3. That's good. Catches the eye.
But it's not- It's not top-shelf memeing. We can do better. A good maths meme should use the golden ratio.
So I'm now going to take the logs of all of the digits of one point six one eight—
the golden ratio, as far as most people are concerned—
and then show that is equal to the log of
a triangle of three golden ratios put together!
Try and prove me wrong! You know what? we better slap "The Power of the Golden Ratio" there.
Use some rainbow text. I reckon we need a Golden Spiral.
Maybe a sunflower! Now! Now we're memeing!
You know what? I reckon if we crank up the JPEG-ing artifacts, maybe slap on a few "Emotional Indicators."
And we're done! That - is a log troll.
Finally, I wanted to mention that this "super serious" educational maths video is brought to you by the fine people at Jane Street.
Jane Street are actually the new principal sponsor of my entire channel,
And they are perfect, because they don't want you to do anything. They don't want you to buy anything, sign up to anything,
They just like maths! And they use maths to solve complex problems in the financial world. And so they wanted to support the channel.
So there you are. Jane Street! They are great, and you can totally ignore them.
Unless, I guess, you're a mathematician looking for a job.