Content-Type: text/shitpost

Subject: Estimating logarithms
Path: you​!your-host​!walldrug​!epicac​!qwerty​!fpuzhpx​!plovergw​!plovervax​!shitpost​!mjd
Date: 2018-07-15T19:42:16
Newsgroup: talk.mjd.logarithm-estimationg
Message-ID: <29be9a46a97f1e50@shitpost.plover.com>
Content-Type: text/shitpost

Estimating quanities like !!\log_2 3!! is easier than it looks, if you know a couple of tricks. In the previous article I said:

I estimated that as 1.6 off the top of my head just by half-assing it: !!2^{3/2} \approx 2.828!!, so it must be a little bigger than !!\frac32!!, which it is.

I really did do it in this ridiculous way, starting from !!1.5 \Rightarrow 2.828!! and then adding a fudge factor to get !!1.6 \stackrel?\Rightarrow 3!!.

(The !!2.828!! is because !!2^{3/2} = 2^{1/2}\cdot 2 \approx 1.414 \cdot 2 = 2.828!!, and memorizing !!\sqrt2\approx 1.414!! is something everyone does in high school.)

I could have arrived at the same answer by a slightly less fudge-flavored route: \begin{align} 256 = 2^8 & \approx 3^5 = 243\\ 8\log 2 & \approx 5 \log 3 \\ \frac85 & \approx \frac{\log3}{\log2} = \log_2 3. \end{align}

Using a slightly less cruddy initial approximation, !!2048 = 2^{11}\approx 3^7 = 2187!!, we get !!\log_2 3 \approx \frac{11}7 = 1.5714\ldots!!. The correct value is !!1.58496\ldots!!. Incidentally the computer informs me that !!524288 = 2^{19} \approx 3^{12} = 531441!!, yielding the approximation !!\frac{19}{12} = 1.5833\ldots!!, but I did not know either of those off the top of my head. I only know the powers of !!3!! up to !!3^7!!, or maybe !!3^8!! on a good day.

We can similarly notice that !!125 = 5^3 \approx 2^7 = 128!! and so estimate that !!\log_2 5 \approx \frac 73!!; the correct value is !!2.322\ldots!!.

File this under “lower mathematics”.