Suppose that you were stranded on a desert island without a computer or a calculator or a slide rule or a book with math tables. Could you put together a table of values for the common log of the numbers 1 through 10 (i.e., the log to the base 10 of 1 through 10)? Why you would want to do so, I'm not sure. But don't you remember the professor on Gilligan's Island? He seemed to need a lot of math and science that he had to conjure up from nothing but coconuts and cleverness.

The easy answers are the logs of 1 and 10, because they are 0 and 1. But the others are more of a challenge. My solution builds on a single observation about powers of 2:

2¹⁰ ≈ 10³

Taking the log of both sides and dividing both sides by 10, we get:

10 ⋅ log(2) ≈ 3
log(2) ≈ 0.3

This automatically gives us answers for 4 and 8 because they are powers of 2. So we'd say that they are 0.6 and 0.9 respectively. And it gives us an answer for 5 because:

log(10) = log(5 ⋅ 2) = log(5) + log(2)

The log of 10 is 1 and we know the log of 2 is approximately 0.3, so given the equation above, we know that the log of 5 has to be around 0.7.

For the other numbers we'd need some other trick. We know something about powers of 2 and 10, so we have to somehow relate other powers to that. Take 7 as an example. We could say that 7 is close to 8, but it's not all that close. But what about powers of 7? If we square 7 we get 49. What do we know about 49? It's very close to 50. And what is 50 equal to? It's one half of 100. So we get:

7² ≈ 50
7² ≈ 100 / 2
7² ≈ 10² / 2

If we take the log of both sides and do some algebra, we get:

log(7²) ≈ log(10² / 2)
2 ⋅ log(7) ≈ log(10²) - log(2)
2 ⋅ log(7) ≈ 2 - log(2)
2 ⋅ log(7) ≈ 2 - 0.3
2 ⋅ log(7) ≈ 1.7
log(7) ≈ 0.85

Get the idea? We basically have just one big piece missing. We need something similar for 3. The square of 3 is 9 and that's somewhat close to 10. But we can do better. The cube of 3 is 27, which is one-fourth of 100. That would give a very good approximation, but we can do slightly better. The fourth power of 3 is 81, which is extremely close to 8 times 10:

3⁴ ≈ 80
3⁴ ≈ 2³ ⋅ 10

So taking the log of both sides and doing some algebra:

log(3⁴) ≈ log(2³ ⋅ 10)
log(3⁴) ≈ log(2³) + log(10)
4 ⋅ log(3) ≈ 3 ⋅ log(2) + log(10)
4 ⋅ log(3) ≈ 3 ⋅ 0.3 + 1
4 ⋅ log(3) ≈ 1.9
log(3) ≈ 0.475

And once we have the log of 3, we know the log of 9 is twice that because it's the square of 3, so it's approximately 0.95. And because 6 is the product of 2 and 3, it's log is the sum of the logs of 2 and 3, so we'd say it's approximately equal to 0.775.

Here's a summary of our estimates and the actual values and how close we got:

Number	log approximation	actual log	percent error
1	0	0.000	0.00%
2	0.3	0.301	0.34%
3	0.475	0.477	0.44%
4	0.6	0.602	0.34%
5	0.7	0.699	0.15%
6	0.775	0.778	0.40%
7	0.85	0.845	0.58%
8	0.9	0.903	0.34%
9	0.95	0.954	0.44%
10	1	1.000	0.00%

So our biggest error was less than 0.6%. That's pretty good for coconuts and cleverness.

This page was developed as part of a talk on ballparking that was given at the CS4HS workshop at the University of Washington.

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Stuart Reges