It's still 1958, and I am tasked to find a given number, less than 1, to the negative 1/3 power, using logarithms and the table, all other computations by hand. No pocket calculator, they haven't been invented yet. How should I proceed? I try to follow the example in the book.

First, state your negative exponent as a reciprocal fraction, that is, one over the number whose exponent is now positive, but still fractional. Then find the logarithm of the number. State the number in scientific notation, which will give me the "characteristic" that is the integer part of the logarithm. In this case it's negative 2. I find the mantissa (four-digit decimal part of the logarithm) in the table by interpolating between two entries on the chart. The logarithm is negative, which is a no-no, you can't work with that in this system. You have to re-state your logarithm as the difference between two positive numbers, for instance a negative 2 must be stated as "8 minus 10."

Now you want to divide your logarithm by three, since it's a fractional exponent, one third. You can either do this by upping your two positive numbers so that they can be divided by three and retain their original subtractive relationship coming out to negative 2. Or you can do more calculations and find the negative logarithm that is the reciprocal of the one you have, by subtracting the smaller from the larger and giving this remainder a negative sign. Then you divide that by three. Either way, it's supposed to get you a fractional exponent. Or a root, if that's what you're after. They're the same thing, sort of.

When you have the logarithm divided by three, then you must reverse it to its reciprocal again and find its antilogarithm, which is supposed to be the final number you were looking for in the first place. For any other number in this computation, perform the same process of logarithmizing, reciprocation, and antilogarithmizing, and then do the other computations necessary to get your totally pointless result.

This business takes me about 15 to 20 minutes for each problem. If I should peek back into the twenty-first century and use my little pocket calculator, it takes about a second.

I have been working on these problems for days now. Hours go by as I find logs in the table, reverse them, multiply or divide them, re-reverse them, negative or positive, reciprocal or original, and then find the anti-log in the table. The slide rule isn't good enough, it only gives about three significant digits, and the book asks for four. I have done one problem after another. The third root of (0.8210)² minus the fifth root of 2.927. Or (2.138)³ multiplied by (43.10)^-2. The hours go by and things get heavier and more involved and more soaked with the fog of logs.

And no matter how many of these things I do, I haven't gotten a single one of them right. There's always something I've missed. I miss a negative/positive sign. I miss a reversed or reciprocal logarithm. I miss a digit in hand calculation. I miss altogether, having inattentively switched or added some four-digit number with something else it wasn't supposed to be with. My eyes blur from looking so long at the small print in the tables.

I keep hoping that sooner or later I will master the twists and turns and reverses and finally get one right. For all I know, this type of calculation is necessary practice for my working with far more complex and twisty-turny negative-positive-negative problems later on. Or it could be a waste of time. Perhaps there is some virtue to mastering a technique which has not been practiced since the advent of far more efficient mechanical calculators. At this point, I just don't know. Maybe one of my Friendly Mathematicians will tell me what this is supposed to do.

Some physics formulas have logarithms in them, so I know I will need this somewhere. But how much logging do I have to do with the wooden tables? I know some high-minded types who insist that just because a computer does the work now does not excuse you from learning the original way it was done. But at this point it feels like it will be years before I get one of these problems right using the tables and the reciprocals. And I have such a long, long way ahead of me.

Posted at 2:38 am | link