James Bloody Grime and his blasted logarithms

It’s typical of James Grime to ask a really interesting question just as I’m going to bed. I was going to sleep like a log, but suddenly I was awake liking logarithms.

Now, the question isn’t quite perfect as specified: in fact, there are an infinite number of bases that work perfectly well, and an infinite number of those aren’t particularly interesting. There’s only one interesting answer: $a = e^{1/e}$, which is about 1.445. It’s also an unexpected answer, and I’ll show you how to work it out.

Before I tell you more, I suggest (a) you try to solve it on paper using MATHS, and/or (b) go to my handy-dandy Desmos graph and play around with it.

Clue #1: inverses make this much easier

The first thing to notice is that the two functions $f(x) = \log_a (x)$ and $g(x) = a^x$ are inverses of each other: $fg(x) = gf(x) = x$, for all $x$ in the appropriate domains. If you know your C3, you’ll know that that means the curves are reflections in the line $y=x$ – which means in turn that when the curves cross, they also have to cross the line $y=x$.

So, instead of solving the fairly awful $\log_a(x) = a^x$, we can just say $a^x = x$. Taking logs, $x \ln(a) = \ln(x)$, or $\ln(a) = \frac{1}{x}\ln(x)$.

Clue #2: draw a graph

How does that help? Well, we can think about where the graph of $y = \frac{1}{x}\ln(x)$ has only one solution. You can plot it with Desmos, of course (which is precisely what I did), or you can figure out its behaviour, which is what I’ll pretend I did.

First up, the domain is clearly $x>0$: you can’t put a non-positive number into logarithms (really). There’s an obvious solution to $y=0$ when $x=1$, where $\ln(x) = 0$, but no others. How about the behaviour at each end of the domain? As $x \rightarrow +\infty$, $\frac{1}{x}$ dominates – but both parts are positive, so the curve stays above the $x$-axis: $y \rightarrow 0_+$. As $x \rightarrow 0_+$, $\frac{1}{x}$ becomes a huge positive number and $\ln(x)$ a huge negative number, so $y \rightarrow -\infty$.

That means we have a graph that comes up from (very loosely) $(0,-\infty)$, crosses the $x$-axis at $(1,0)$, and then (again, loosely) ends up at $(+\infty, 0)$.

How many solutions?

To find out how many solutions there are for any value, you can just draw a horizontal line ($y=k$, for whatever value of $k$ you’re interested in) – across the page. However many times that line crosses the curve is the number of solutions.

If you pick any $k \le 0$, there’s only one solution to $y=k$, and it’s for $0 \lt x \le 1$. All of those give valid answers for $a$ except $x=1$, which would give us $a=1$ – not a valid log base. Those are the infinitely many uninteresting solutions. Forget about them.

You gotta Rolle with it

Back to the curve! A wise man – some claim it was Newton [1] – once said “What goes up, must come down.” That means, if you’ve got a continuous function (this one doesn’t have gaps in) and two places where it has the same value (for instance, $f(1) = 0$ and $\lim_{x \rightarrow +\infty} f(x) = 0$), it must have at least one turning point in between. Let’s find it!

You can differentiate $y = \frac{1}{x} \ln(x)$ using the quotient rule to get $\frac{dy}{dx} = \frac{1 – \ln(x)}{x^2}$, which has a turning point when the top of the fraction is 0 – when $1 – \ln(x) = 0$, which gives $x=e$. Neat!

Why am I bothering with the turning point? It’s because that’s the other place where $y=k$ has only one solution – the line just grazes the curve there. For all other $y$ values, the line either cuts the curve twice, or not at all.

When $x = e$, $y = \frac{1}{e} \ln(e)$, which is just $\frac{1}{e}$. So, the only interesting solution is when $\ln(a) = \frac{1}{e}$, giving $a = e^{\frac{1}{e}} \simeq 1.4447$.

[1] Rolle got the credit, though.

Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

Share

4 comments on “James Bloody Grime and his blasted logarithms

  • Jan Van lent

    What about the decaying exponential?
    $a = e^{-e} \approx 0.065988$

    [CB: edited to correct LaTeX]

    • Colin

      Thanks for your comment, Jan – any $a < 1$ is a valid solution to this, corresponding to $k<0$. $e^{\frac{1}{e}}$ is the only solution for $a>1$.

      • Jan Van lent

        I do not think that is right. Your simplification of the equation misses out some solutions. The value $a=e^{-e}$ I singled out is the other value where $a^x$ and $\log_a(x)$ touch. For $a < e^{-e}$ there are three intersections. So I think the answer is $a \in [e^{-e}, 1) \cup \{e^{e^{-1}}\}$.

        • Colin

          Oo, I see what you mean! That is interesting 🙂

Leave a Reply

Your email address will not be published. Required fields are marked *

Sign up for the Sum Comfort newsletter and get a free e-book of mathematical quotations.

No spam ever, obviously.

Where do you teach?

I teach in my home in Abbotsbury Road, Weymouth.

It's a 15-minute walk from Weymouth station, and it's on bus routes 3, 8 and X53. On-road parking is available nearby.

On twitter