the natural history of abstract objects
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Can you find all the possible solutions to this equation?

\[\Huge a^{1+\log_{10}(a)} = 100\]

When I ran this as a test question in Fall 2019, perhaps a third to half of the 50ish kids managed to figure out one of the solutions. Of those who did, nearly all of them found it by some variant on guess-and-check. Only one (two?) people found both solutions. These are my notes on the problem!

So. If we stare at this equation for long enough, and test random numbers, or semi-random numbers, we might realize that \(a=10\) works as a solution. We can verify it works by plugging it in: \[\begin{align*} &a^{1+\log_{10}(a)} \\ &=10^{1+\log_{10}(10)} \\ &=10^{1+1} \\ &=10^{2} \\ &=100 \end{align*}\] So, indeed, \(a=10\) is one solution to this equation. But there are two problems here. First of all, the “stare at it” method is not a great one. I mean, sometimes it works, and apparently it worked for a lot of people in this case. If you don’t have any better ideas, then by all means, stare at it and test random numbers. Guess and check all you want! Appeal to the mathematical muses—to their benevolence, and to their charity! Sometimes the gods of mathematics are generous, and will bequeath us with an answer. And for many of the kids, that worked in this case.

HOWEVER, we DON’T want to have to rely on the benevolence and charity of the gods. They are fickle. Sometimes they give, and sometimes they take away. We want to be Nietzscheans here—we want to make an end-run around the gods! We’d like to have some sort of procedure and system and algorithm for solving problems. We’d like to have a set of rules that always works.

It’s sort of like how when we factor quadratic equations, sometimes it’s fine to just, like, look at it and come up with a factorization: \[\begin{align*} x^2 + 9x + 14 &= (x \,+\, ???)(x \,+\,???) \\ &=\text{ hmmm hmmm hmmm}\\ &=\text{ A BLINDING INSIGHT FROM THE GODS!!!!}\\ &= (x+2)(x+7) \end{align*}\] But if the gods are ever parsimonious, we know that we have a MACHINE that will answer our prayers. We have a FORMULA that ALWAYS WORKS—the quadratic formula! \[\begin{align*} x^2 + 9x + 14 &= \left(x - \frac{-9 + \sqrt{9^2-4\cdot1\cdot14}}{2\cdot1}\right)\left(x- \frac{-9 - \sqrt{9^2-4\cdot1\cdot14}}{2\cdot1}\right) \\ &= \left(x - \frac{-9 + \sqrt{81-56}}{2}\right)\left(x- \frac{-9 - \sqrt{81-56}}{2}\right) \\ &= \left(x - \frac{-9 + \sqrt{25}}{2}\right)\left(x- \frac{-9 - \sqrt{25}}{2}\right) \\ &= \left(x - \frac{-9 + 5}{2}\right)\left(x- \frac{-9 -5}{2}\right) \\ &= \left(x - \frac{-4}{2}\right)\left(x- \frac{-14}{2}\right) \\ &= (x+2)(x+7) \end{align*}\] Thanks to the mightiness of the quadratic formula, we no longer have to rely on the benevolence and charity of the mathematical gods when trying to factor quadratics!!! We can factor them—ANY of them—with CERTAINTY—with CONFIDENCE—just using this machine. No need to hope that the gods are generous. The gods are dead. We killed them.

Back to the problem at hand: \[a^{1+\log_{10}(a)} = 100\] We want to solve this problem in a systematic way. We’d prefer not to solve it via guess-and-check, or via the “stare at it really hard” method. We don’t want to have to rely on the whims of the gods. We don’t want to hope that the muse sings to us.

I have heard the mermaids singing, each to each.

I do not think they will sing to me.

Relatedly, there’s a second problem here. Namely: this equation actually has two solutions!!! The mathematical gods may have been kind enough to give us (or to give some of you) ONE of them—but only two of you found both solutions.

So, is there a way we can solve this equation, that’s systematic and procedural and algebraic, that doesn’t require us to appeal to the mathematical gods, and which will beget both solutions? Can we use algebra and our knowledge of logarithms to solve this?

A lot of you did try to do this in an algebraic/systematic way, and that was great. Most of you who took such an approach started by taking the \(\log_a\) of both sides. That was a great idea! It’s systematic and algebraic. So, let’s see what happens when we do that. We have: \[a^{1+\log_{10}(a)} = 100\] Taking the \(\log_a\) of both sides, we get: \[\log_a\left(a^{1+\log_{10}(a)}\right) = \log_a(100)\] On the left side, the \(\log_a\) and \(a^\text{stuff}\) cancel out, so this becomes: \[1+\log_{10}(a) = \log_a(100)\] Hooray! The \(a^\text{stuff}\) is gone. That’s an accomplishment. It’s why I assume most of you chose to take the \(\log_a\). But we’re not done—we want to solve this for \(a\). As our next step, I guess we could move the \(1\) over: \[\log_{10}(a) = \log_a(100) - 1\] And then raise both sides by \(10\) to get rid of the \(\log_{10}\): \[10^{\log_{10}(a) }= 10^{\log_a(100) - 1}\] \[a= 10^{\log_a(100) - 1}\] So we have an equation for \(a\)… but \(a\) is also on the right side. We haven’t really solved for \(a\). Aaaagh. How do we get rid of that \(\log_a\) on the right side?!?? No clue.

We wanted to make an end run around the gods. Instead we made an Enron around the gods.1

This method was a good attempt, but it doesn’t lead us to the solution (to either solution). However, a lot of the thinking was right—we want to somehow solve this algebraically and systematically. Let’s try it again. What if, instead of taking the \(\log_a\), we take a DIFFERENT log? This is just stabbing in the dark here, but notice that this equation also involves a \(\log_{10}\), so what if we use that? We start with: \[a^{1+\log_{10}(a)} = 100\] And then taking the \(\log_{10}\) of both sides, we get: \[\log_{10}\left(a^{1+\log_{10}(a)}\right) = \log_{10}(100)\] On the right side, \(\log_{10}(100)\) just becomes \(2\): \[\log_{10}\left(a^{1+\log_{10}(a)}\right) =2\] On the left side, \(\log_{10}\) and \(a^\text{stuff}\) won’t cancel out—the bases are different (\(10\) and \(a\)), so they’re not inverses. BUT!!! We have \(a\) raised to a bunch of stuff. We can use that law about how an exponent inside ANY logarithm can move to the outside, as a coefficient!!! \[\text{this law: }\log_a\left(b^c\right) = c\log_a(b)\] Applying that, and pulling out the exponent on the \(a\), we get:

\[\big(1+\log_{10}(a)\big) \cdot \log_{10}\left(a\right) =2\] This doesn’t seem THAT promising, but at least the \(\log_{10}\)’s are gone! Let’s multiply this out:

\[ \log_{10}\left(a\right) + \big(\log_{10}(a)\big)^2 =2\] Re-arranging: \[ \big(\log_{10}(a)\big)^2 +\log_{10}\left(a\right) - 2 =0\] What do we do with this? It looks complicated. So many details!

Hmm. I have bad vision. I squint a lot. If I’m not wearing glasses, that line is all blurry, and just looks like: \[(\text{SOMETHING})^2 + (\text{SOMETHING}) - 2 = 0\] But… that’s a quadratic! Beneath all the specific details, this is really just a quadratic! It’s just: \[x^2 + x - 2 = 0\] The “\(x\)” in a quadratic can be anything. It can be a single variable/number, or it can be something more complicated! In this case, “\(x\)” is \(\log_{10}(a)\).

Here’s part of the moral of the story here: sometimes blurriness brings clarity. Sometimes we need to step back and squint and see fewer details. Sometimes we miss the forest for the trees. If we’re too obsessed with the details, we miss the big picture.

Anyway, can we solve this using the quadratic equation?!?!?!? Or, even easier, maybe we can just factor it by hand??!? It looks a lot like: \[x^2 + x - 2 = 0\] and we can factor that as: \[(x-1)(x+2) = 0\] But in this case, our “variable” isn’t \(x\); it’s \(\log_{10}(a)\). So really, we have: \[\big( \log_{10}(a) - 1\big)\big( \log_{10}(a) +2\big) = 0\] So we get two equations out of this: \[\underbrace{\big( \log_{10}(a) -1\big)}_{ \log_{10}(a) -1 \,= \,0} \underbrace{\big( \log_{10}(a) +2\big)}_{\log_{10}(a) +2 \,=\, 0} = 0\] Let’s solve them! Let’s do the left one first. We have: \[\underbrace{\big( \log_{10}(a) -1\big)}_{ \substack{ \log_{10}(a) -1 \,= \,0 \\ \log_{10}(a) \,= \, 1 \\\\ 10^{\log_{10}(a)} \,= \, 10^1 \\\\ a \,=\, 10^1 \\\\ a\,=\,10 }} \big( \log_{10}(a) +2\big) = 0\] So we get that \(a\) can be \(10\)! That’s great—that’s what we already found. It’s some nice confirmation that this procedure works. We found that \(a\) can be \(10\), without having to rely on the benevolence of the gods! What about the other factor? Let’s find out what that tells us. \[\big( \log_{10}(a) - 1\big) \underbrace{\big( \log_{10}(a) + 2\big)}_{\substack{\log_{10}(a) + 2 \,=\, 0 \\\\ \log_{10}(a) \,=\, -2 \\\\ 10^{\log_{10}(a)} \,=\, 10^{-2} \\\\ a \,=\, 10^{-2}\\\\ a \,=\, \frac{1}{100} }} = 0\] So \(a\) can be \(1/100\)!!! That’s the other, missing solution!!!

Let’s check just to make sure it works. We’ll plug \(a=1/100=10^{-2}\) back into the original equation. We’ll plug it back into the left side of the equation, and after some simplification, we should get the right side (i.e., \(100\)): \[\begin{align*} &a^{1+\log_{10}(a)} \\ &=\left(10^{-2}\right)^{1+\log_{10}\left(10^{-2}\right)} \\ &=\left(10^{-2}\right)^{1+(-2)} \\ &=\left(10^{-2}\right)^{-1} \\ &=10^{(-2)(-1)} \\ &=10^{2} \\ &=100 \end{align*}\] We get \(100\), which is the right side of the equation, which is what we wanted. So \(a=\frac{1}{100}\) does indeed work as another solution!

\[ a^{1+\log_{10}(a)} = 100\] \[\text{begets}\] \[a = \Bigg\{10, \, \frac{1}{100}\Bigg\}\]

There are two big ideas here: