Less Fun with Logarithms

Today I did more exercises and reading in my calculus book. I also proved the converse of the proposition from last week’s post Fun with Logarithms. You can see the proof below. As I expected, all that was required was a little algebra. Finally, I listened to some of the BBC radio series A Brief History of Mathematics.

It was struggling to listen to that series at the beginning of this year that inspired me to try doing math with the aid of CBD gummies (as described in Take Drugs, Do Math). The series is intended for a broad audience and is not intellectually demanding. Yet even a short time listening to it left me completely exhausted. That served as strong evidence that the overwhelming difficulties I was still having when I tried to do math were at least partly emotional, as opposed to cognitive. Before, it had never been so clear. The new certainty, combined with an overall improvement in my mental health around that time, led me to seek new ways to overcome the emotional block.

Today I was able to enjoy the series with no ill effects, even without CBD gummies, which I am now using infrequently.

Proposition: Given two natural numbers $a$ and $b$ that are both greater than $1$, if both $a$ and $b$ are powers of a natural number $c$, then $\log_a b$ is rational.

Suppose that $a$ and $b$ are natural numbers greater than $1$ that are both powers of a natural number $c$.

Then, by definition, $a=c^m$ and $b=c^n$ for some natural numbers $m$ and $n$.

Since $a\neq 1$, it follows that $m\neq 0$.

And since $m\neq0$, $a^{\frac{n}{m}}=(c^m)^{\frac{n}{m}}=c^n=b$.

Thus $\log_a b = \frac{n}{m}$, which is rational, and the proposition is proven.