# Decimal Approximations of Logarithms (Part 4)

While some common (i.e., base-10) logarithms work out evenly, like $\log_{10} 10,000$, most do not. Here is the typical output when a scientific calculator computes a logarithm: To a student first learning logarithms, the answer is just an apparently random jumble of digits; indeed, it can proven that the answer is irrational. With a little prompting, a teacher can get his/her students wondering about how people 50 years ago could have figured this out without a calculator. This leads to a natural pedagogical question:

Can good Algebra II students, using only the tools at their disposal, understand how decimal expansions of base-10 logarithms could have been found before computers were invented?

Here’s a trial-and-error technique — an exploration activity — that is within the grasp of Algebra II students. It’s simple to understand; it’s just a lot of work. The only tools that are needed are

• The Laws of Logarithms
• A hand-held scientific calculator
• A lot of patience multiplying $x$ by itself repeatedly in a quest to find integer powers of $x$ that are close to powers of $10$. In the previous post in this series, we found that $3^{153} \approx 10^{73}$

and $3^{323,641} \approx 10^{154,416}$.

Using the Laws of Logarithms on the latter provides an approximation of $\log_{10} 3$ that is accurate to an astounding ten decimal places: $\log_{10} 3^{323,641} \approx \log_{10} 10^{154,416}$ $323,641 \log_{10} 3 \approx 154,416$ $\log_{10} 3 \approx \displaystyle \frac{154,416}{323,641} \approx 0.477121254723598\dots$.

Compare with: $\log_{10} 3 \approx 0.47712125471966\dots$

Since hand-held calculators will generate identical outputs for these two expressions (up to the display capabilities of the calculator), this may lead to the misconception that the irrational number $\log_{10} 3$ is actually equal to the rational number $\displaystyle \frac{154,416}{323,641}$, so I’ll emphasize again that these two numbers are not equal but are instead really, really close to each other. We now turn to a question that was deferred in the previous post.

Student: How did you know to raise 3 to the 323,641st power?

Teacher: I just multiplied 3 by itself a few hundred thousand times.

Student: C’mon, really. How did you know?

While I don’t doubt that some of our ancestors used this technique to find logarithms — at least before the discovery of calculus — today’s students are not going to be that patient. Instead, to find suitable powers quickly, we will use ideas from the mathematical theory of continued fractions: see Wikipedia, Mathworld, or this excellent self-contained book for more details.

To approximate $\log_{10} x$, the technique outlined in this series suggests finding integers $m$ and $n$ so that $x^n \approx 10^m$,

or, equivalently, $\log_{10} x^n \approx \log_{10} 10^m$ $n \log_{10} x \approx m$ $\log_{10} x \approx \displaystyle \frac{m}{n}$.

In other words, we’re looking for rational numbers that are reasonable close to $\log_{10} x$. Terrific candidates for such rational numbers are the convergents to the continued fraction expansion of $\log_{10} x$. I’ll defer to the references above for how these convergents can be computed, so let me cut to the chase. One way these can be quickly obtained is the free website Wolfram Alpha. For example, the first few convergents of $\log_{10} 3$ are $\displaystyle \frac{1}{2}, \frac{10}{21}, \frac{21}{44}, \frac{52}{109}, and \frac{73}{153}$.

A larger convergent is $\frac{154,416}{323,641}$, our familiar friend from the previous post in this series.

As more terms are taken, these convergents get closer and closer to $\log_{10} 3$. In fact:

• Each convergent is the best possible rational approximation to $\log_{10} 3$ using a denominator that’s less than the denominator of the next convergent. For example, the second convergent $\displaystyle \frac{10}{21}$ is the closest rational number to $\log_{10} 3$ that has a denominator less than $44$, the denominator of the third convergent.
• The convergents alternate between slightly greater than $\log_{10} 3$ and slightly less than $\log_{10} 3$.
• Each convergent $\displaystyle \frac{m}{n}$ is guaranteed to be within $\displaystyle \frac{1}{n^2}$ of $\log_{10} 3$. (In fact, if $\displaystyle \frac{m}{n}$ and $\displaystyle \frac{p}{q}$ are consecutive convergents, then $\displaystyle \frac{m}{n}$ is guaranteed to be within $\displaystyle \frac{1}{nq}$ of $\log_{10} 3$.)
• As a practical upshot of the previous point: if the denominator of the convergent $\displaystyle \frac{m}{n}$ is at least six digits long (that is, greater than $10^5$), then $\displaystyle \frac{m}{n}$ must be within $\displaystyle \frac{1}{(10^5)^2} = 10^{-10}$ of $\log_{10} 3$… and it’ll probably be significantly closer than that.

So convergents provide a way for teachers to maintain the illusion that they found a power like $3^{323,641}$ by laborious calculation, when in fact they were quickly found through modern computing.

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