The fireworks are out and it's back to math. I'm working my way through lots of logarithm problems. As you may remember from last time, I spoke in a rather agricultural way about "raising" and "rooting" numbers. How would I describe a logarithm, then? It's a number seed which when planted, both raises and roots at the same time.

As I understand it, a logarithm is the number which designates how one number, called the "base," can be transformed into another number, (can't find the term for this one), not by plain old multiplication, but by either multiplying the base by itself (raising) or taking the root of it, whether square, cube, or whatever (rooting). The root is the number that has to be multiplied by itself in order to get your base number.

The logarithm is that fractional distillation of all the raising and rooting that has to be done to turn one number into another. I have no idea (yet) how the mathematicians of history came up with the logarithms that fill the tables at the back of the book, but logarithms, like sines and cosines, are something that you look up, or nowadays, punch into your calculator.

My problem set comes from Algebra 1958. Like all math problem sets, the examples start simply and then progress one by one to greater complexity. With more than a couple of mistakes here and there, which I rectified by the somewhat dishonest process of working backwards from the given answer, I found my way to problem number 63. "Find the logarithm, to the base 10, of 25." OK, to you mathematicians and scientists out there, this is child's play. But I've never seen this kind of thing before. I couldn't just go look it up, because the book specified that I could only use the logarithms (to base 10) of 2, 3, and 7, which were given in the problem set.

So far I had been solving 'em by factoring out whatever number they wanted, into combinations of 2's, 3's, and 7's. But what to do with 25? Sure, it's a perfect square of 5, but they didn't give me any logarithm for 5. I pondered over it for a while, trying to figure out how to get 25 to be a combination of 2, 3, or 7. Factoring didn't work, because that 5 stood in the way. What I needed was something that 10 could be raised to, since I was already working with a base of 10, and also something that 2 could be raised to, since I knew the logarithm of 2.

So finally I thought it out. How else could I express number 25 so it would be in the realm of 10's and 2's? Hey, isn't 25 also 100 divided by 4? What if I translated 25 into 100/4? Then it works out just fine, 'cause four is 2 squared and I have the logarithm of 2. Sure enough, 100/4 did the trick, and the rest of the calculation, as long as I followed the logarithm rules, was quick and easy. With trepidation I went to the back of the book to check the answer for problem number 63. I know, real mathematicians don't bother with the answers at the back of the book. So I'm not real (am I imaginary?). The book's answer was exactly mine, so I was RIGHT. Yeah! It's kind of the feeling that you get when you land a crunched-up piece of paper into a narrow wastebasket from 10 feet away. It's trivial, but it's satisfying.

Posted at 2:33 am | link