Revisiting, and something light

Well, today was the fourth day of my official employment with Wolfram. This job is absolutely amazing; I couldn’t be more stoked. It’s saddening, of course, that I’m not spending my days engulfed in the books I’d been looking at earlier in the summer; it’s also a bit saddening that I have less time to spend with you beautiful people. Regardless, things are pretty amazing and overall, I couldn’t be happier.

I wanted to take some time to swing by here and say something, though, and fortunately for me, my TA duties this semester have yielded me something of precisely the right balance of depth (or lack thereof) and length (or brevity) to be fitting for tonight’s (this morning’s) pit stop.

On Wednesday (July 3), I was sitting in a precalculus class, doing Wolfram stuff and vaguely listening to what the instructor was talking about at the time. The topic? Logarithms. As someone who’s solo-taught precalculus before, I know precisely how little students understand – or like – or care – about logarithms. I also know how much we try to convince them to believe without their understanding which – among others – has to be a primary cause for their confusion and disdain.

One thing we try to get them to believe? The change of base formula. The change of base formula says that given a base b>0, b\neq 1, the quantity \log_b(x) is equal to the quantity

\log_b(x)=\displaystyle\frac{\log_c(x)}{\log_c(b)} where c>0, c\neq 1.

This information is shared with students at that level largely so they can feel comfortable evaluating an expression like \log_{15}(31) in their calculators given only the capacity to utilize \log(x)=\log_{10}(x) and \ln(x)=\log_e(x) functionality. Surely, they never really need to know it.

And then I realized…

In all my years in mathematics, I’ve never actually seen this rule proven before. That, of course, sparked my interest, and so I went back to my office and jotted the (surprisingly simple) proof on my whiteboard just to appease my curiosity. Here’s the way that goes:

Proof of The Change of Base Formula.
Let y=\log_b(x) so that b^y=x. In particular, then, it follows that for c>0, c\neq 1,

\displaystyle\frac{\log_c(x)}{\log_c(b)} = \frac{\log_c\left(b^y\right)}{\log_c(b)}=\frac{y\cdot\log_c(b)}{\log_c(b)}=y=\log_b(x). \square

I think I may force my next round of precalculus students to know that. It keeps ’em fresh, on their toes, where they gotta baayayaeeee….

Did anyone just catch my reference to ‘Heat’? Or, rather, my reference to Aries Spears’ reference to ‘Heat’?

I hope everyone’s 4th was safe and that there were only minimal injuries due to inebriation, explosives, and general tomfoolery.

Until next time….