Despite this being mainly a blog on using technology to teach maths, I still have a deep appreciation for the old ways. So here we go!

Logarithms are a topic that can often confuse students. Even after a few lessons on logs, many students are shaky on the basics, don’t like the word logarithm and see it as just another thing to learn. The past two years I have taught logs using a different approach and found it worked quite well. I taught it with a historical angle and avoided using the word logarithm until I absolutely had to.

To begin, I asked my Year 12 class to compute 32^{2} in their heads. We share methods. If any don’t spot it, I share with them that they may notice that 32 = 2^{5}. So 32^{2} = 2^{5} x 2^{5} = 2^{10}. And you just may happen to know that 2^{10} = 1024.

But what if you didn’t know your powers of 2 up to 2^10 and beyond?! What if it was 31 x 32? What is 31 written as a power of 2? About 4.9 perhaps? We need a shift of focus from the ‘answer’ (the 1024 from above) to the power (whilst keeping the base constant). If you had a table of every number and its equivalence as a power to the base 2, that would be very useful. (Obviously, calculators are not permitted at this stage!) We could make this table with a base of 2, but as our number system is base 10, it is much more efficient to use 10 as our base. Using a base of 10 would mean that numbers that are a factor of ten apart would have the same mantissa (decimal part of the number). For example, log 4.6=0.6627578…, log 46 = 1.6627578…, and log 460=2.6627578… This repetition means that your log table is not infinitely big. It’s also a good illustration of the fact that logarithms are a “series of numbers in arithmetical progression which correspond to others in geometrical progression” (Encyclopaedia Britannica 1797).

I normally now set the task of the students making a log table for base 10. Once they realise how much work this is (even with a calculator!) they look at me despairingly and I stop them. Thankfully, someone already thought that making these tables was a good idea. His name was John Napier and it took him 20 years to construct his tables in the early 17^{th} century.

Say we want to do 23.64 x 57.3. Now we could do this with pen and paper and get a perfectly accurate answer within about a minute. So although this is perhaps not the best illustration of the power of log tables but it is a relatively simple one to get to grips with using the tables.

Note that I am using 4 figure log tables here (available online here from Page 209). Books containing tables of more figures than this are available, I have a good one from here.

So from the tables we start by looking up the mantissa for 23.64. We do this by finding the 23 on the left column then the 6 from the top row. Because we have a fourth figure, we need to add on the proportional part. So we have 3729 then the proportional part is 7 so our mantissa is 3729+7=3736. We know the integer part of the power will be 1 (because 23.64 is between 10^{1} and 10^{2}) so we can say that 23.64 is approximately 10^{1.3736}.

Looking up the other number is the calculation, 57.3, tells us that 57.3 ≈ 10^{1.7582}.

Therefore, 10^{1.3736} x 10^{1.7582} = 10^{3.1318}

Then we use the anti-log table to convert back the mantissa. We do this by picking out the integer part of the power 10^{3.1318} = 10^{3} x 10^{0.1318}

From the anti-log table, we find 0.1318 from reading across from .13 and down from 1 to get 1352. We must then add on the proportional part of 3 gained from the fourth figure (8). Making sure to get the magnitude correct, we get that 10^{0.1318} ≈ 1.355

So, our full calculation now looks like this:

23.64 x 57.3

≈ 10^{1.3736 }x 10^{1.7582}

≈ 10^{3.1318}

≈ 10^{3} x 10^{0.1318}

≈ 1000 x 1.355

≈ 1355

So this was possible to do without a calculator because we had the log tables and because generally adding is easier than multiplying. We then have the very rewarding task of seeing how accurate we were! Two seconds on any calculator tells us the actual answer is 1354.572. Therefore, our answer calculated from the log tables was only 0.03% off – pretty good!

At this stage I usually give the students a couple to try on their own and throw in a division too. They can normally figure out that division is accomplished by subtracting the powers instead of adding them. It takes the students a while to do the calculations but they generally can see that this would be useful if we lived in the days before calculators. Further demonstrations of the power of log tables extends to calculations such as 1.6^{9}, which would be very tedious to calculate by hand. However, this really needs a log law (which I haven’t introduced yet) so I would probably come back to such examples.

I then show students the slide rule in all its glory. With the aid of a visualiser I show them the how it works. I then also show them one of my all time favourite mathematical instruments and that’s my Otis King pocket calculator. I’ve had some great reactions to this as it isn’t at all obvious what it is or how it works! It’s a beautifully made item and is basically a slide rule wrapped around a cylinder. Someone asked if they could take it into a non-calculator exam – which I don’t know the answer to!?

At the risk of implying that logarithms are an antiquated idea, I do make sure to tell them that logs are still important today! I generally talk about log scales on graphs, solving equations such as 3^{x }= 20, differential equations and applications in chemistry. At the end of the lesson the students haven’t written much (if anything) down but they have a good sense of what a logarithm is, an appreciation of its historical use, and a deep sense of gratitude for their calculators! If you have any comments or ideas as how to develop any of the ideas above I would love to hear from you – thanks.