# Square roots and logarithms without a calculator (Part 3)

I’m in the middle of a series of posts concerning the elementary operation of computing a square root. This is such an elementary operation because nearly every calculator has a $\sqrt{~~}$ button, and so students today are accustomed to quickly getting an answer without giving much thought to (1) what the answer means or (2) what magic the calculator uses to find square roots. I like to show my future secondary teachers a brief history on this topic… partially to deepen their knowledge about what they likely think is a simple concept, but also to give them a little appreciation for their elders.

Today’s topic is the use of log tables. I’m guessing that many readers have either forgotten how to use a log table or else were never even taught how to use them. After showing how log tables were used in the past, I’ll conclude with some thoughts about its effectiveness for teaching students logarithms for the first time.

This will be a fairly long post about log tables. In the next post, I’ll discuss how log tables can be used to compute square roots.

To begin, let’s again go back to a time before the advent of pocket calculators… say, 1912.

Before the advent of pocket calculators, most professional scientists and engineers had mathematical tables for keeping the values of logarithms, trigonometric functions, and the like. The following images come from one of my prized possessions: College Mathematics, by Kaj L. Nielsen (Barnes & Noble, New York, 1958). Some saint gave this book to me as a child in the late 1970s; trust me, it was well-worn by the time I actually got to college.

With the advent of cheap pocket calculators, mathematical tables are a relic of the past. The only place that any kind of mathematical table common appears in modern use are in statistics textbooks for providing areas and critical values of the normal distribution, the Student $t$ distribution, and the like.

That said, mathematical tables are not a relic of the remote past. When I was learning logarithms and trigonometric functions at school in the early 1980s — one generation ago — I distinctly remember that my school textbook had these tables in the back of the book.

And it’s my firm opinion that, as an exercise in history, log tables can still be used today to deepen students’ facility with logarithms. In this post and Part 4 of this series, I discuss how the log table can be used to compute logarithms and (using the language of past generations) antilogarithms without a calculator. In Part 5, I’ll discuss my opinions about the pedagogical usefulness of log tables, even if logarithms can be computed more easily nowadays with scientific calculators. In Part 6, I’ll return to square roots — specifically, how log tables can be used to find square roots.

How to use the table, Part 1. How do you read this table? The left-most column shows the ones digit and the tenths digit, while the top row shows the hundredths digit. So, for example, the bottom row shows ten different base-10 logarithms: $\log_{10} 9.90 \approx 0.9956, \log_{10} 9.91 \approx 0.9961, \log_{10} 9.92 \approx 0.9965,$ $\log_{10} 9.93 \approx 0.9969, \log_{10} 9.94 \approx 0.9974, \log_{10} 9.95 \approx 0.9978,$ $\log_{10} 9.96 \approx 0.9983, \log_{10} 9.97 \approx 0.9987, \log_{10} 9.98 \approx 0.9991,$ $\log_{10} 9.99 \approx 0.9996$

So, rather than punching numbers into a calculator, the table was used to find these logarithms. You’ll notice that these values match, to four decimal places, the values found on a modern calculator. How to use the table, Part 2. What if we’re trying to take the logarithm of a number between $1$ and $10$ which has more than two digits after the decimal point, like $\log_{10} 5.1264$? From the table, we know that the value has to lie between $\log_{10} 5.12 \approx 0.7093$ and $\log_{10} 5.13 \approx 0.7101$

So, to estimate $\log_{10} 5.1264$, we will employ linear interpolation. That’s a fancy way of saying “Find the line connecting $(5.12,0.7093)$ and $(5.13,0.7101)$, and find the point on the line whose $x-$coordinate is $5.1264$. The graph of $y = \log_{10} x$ is not a straight line, of course, but hopefully this linear interpolation will be reasonably close to the correct answer.

Finding this line is a straightforward exercise in the point-slope form of a line: $m = \displaystyle \frac{0.7101-0.7093}{5.13-5.12} = 0.08$ $y - 0.7093= 0.08 (x - 5.12)$ $y = 0.7093+ 0.08 (5.1264-5.12)$ $y = 0.7093+ 0.08(0.0064) = 0.7093 + 0.000512 = 0.709812$

Remembering that this log table is only good to four significant digits, we estimate $\log_{10} 5.1264 \approx 0.7098$.

With a little practice, one can do the above calculations with relative ease. Also, many log tables of the past had a column called “proportional parts” that essentially replaced the step of linear interpolation, thus speeding the use of the table considerably.

Again, this matches the result of a modern calculator to four decimal places: How to use the table, Part 3. So far, we’ve discussed taking the logarithms of numbers between $1$ and $10$ and the antilogarithms of numbers between $0$ and $1$. Let’s now consider what happens if we pick a number outside of these intervals.

To find $\log_{10} 12,345$, we observe that $\log_{10}12345 = \log_{10} (10,000 \times 1.2345)$ $\log_{10} 12345 = \log_{10} 10,000 + \log_{10} 1.2345$ $\log_{10} 12345 = 4 + \log_{10} 1.2345$

More intuitively, we know that the answer must lie between $\log_{10} 10,000 = 4$ and $100,000 = 5$, so the answer must be $4.\hbox{something}$. The value of $\log_{10} 1.2345$ is the necessary $\hbox{something}$.

We then find $\log_{10} 1.2345$ by linear interpolation. From the table, we see that $\log_{10} 1.23 \approx 0.0899$ and $\log_{10} 1.24 \approx 0.0934$

Employing linear interpolation, we find $m = \displaystyle \frac{0.0934-0.0899}{1.24-1.23} = 0.35$ $y - 0.0899= 0.35 (x - 1.23)$ $y = 0.0899+ 0.35 (1.2345-1.23)$ $y = 0.0899+ 0.35(0.0045) = 0.0899 + 0.001575 = 0.091475$

Remembering that this log table is only good to four significant digits, we estimate $\log_{10} 1.2345 \approx 0.0915$, so that $\log_{10} 12,345 \approx 4.0915$.

Again, this matches the result of a modern calculator to four decimal places (in this case, five significant digits): How to use the table, Part 4.  Let’s now consider what happens if we pick a positive number less than $1$. To find $\log_{10} 0.00012345$, we observe that $\log_{10}0.00012345 = \log_{10} \left( 1.2345 \times 10^{-4} \right)$ $\log_{10}0.00012345 = \log_{10} 1.2345 + \log_{10} 10^{-4}$ $\log_{10} 0.00012345 = -4 + \log_{10} 1.2345$

We have already found $\log_{10} 1.2345 \approx 0.0915$ by linear interpolation. We therefore conclude that $\log_{10} 12,345 \approx 0.0915 - 4 = -3.9085$. Again, this matches the result of a modern calculator to four decimal places (in this case, five significant digits):  So that’s how to compute logarithms without a calculator: we rely on somebody else’s hard work to compute these logarithms (which were found in the back of every precalculus textbook a generation ago), and we make clever use of the laws of logarithms and linear interpolation.

Log tables are of course subject to roundoff errors. (For that matter, so are pocket calculators, but the roundoff happens so deep in the decimal expansion — the 12th or 13th digit — that students hardly ever notice the roundoff error and thus can develop the unfortunate habit of thinking that the result of a calculator is always exactly correct.)

For a two-page table found in a student’s textbook, the results were typically accurate to four significant digits. Professional engineers and scientists, however, needed more accuracy than that, and so they had entire books of tables. A table showing 5 places of accuracy would require about 20 printed pages, while a table showing 6 places of accuracy requires about 200 printed pages. Indeed, if you go to the old and dusty books of any decent university library, you should be able to find these old books of mathematical tables.

In other words, that’s how the Brooklyn Bridge got built in an era before pocket calculators.

At this point you may be asking, “OK, I don’t need to use a calculator to use a log table. But let’s back up a step. How were the values in the log table computed without a calculator?” That’s a perfectly reasonable question, but this post is getting long enough as it is. Perhaps I’ll address this issue in a future post.

1. #### zma

/  March 10, 2016

How to approach:
resolve sqrt(3) without caclulator?
log(sqrt(3))=1/2*log3?
br