# Decimal Approximations of Logarithms (Part 3)

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.

To approximate $\log_{10} x$, look for integer powers of $x$ that are close to powers of 10. In the previous post in this series, we essentially used trial and error to find such powers of 3. We found $3^{153} \approx 9.989689095 \times 10^{72} \approx 10^{73}$,

from which we can conclude $\log_{10} 3^{153} \approx \log_{10} 10^{73}$ $153 \log_{10} 3 \approx 73$ $\log_{10} 3 \approx \displaystyle \frac{73}{153} \approx 0.477124$.

This approximation is accurate to five decimal places.

By now, I’d imagine that our student would be convinced that logarithms aren’t just a random jumble of digits… there’s a process (albeit a complicated process) for obtaining these decimal expansions. Of course, this process isn’t the best process, but it works and it only uses techniques at the level of an Algebra II student who’s learning about logarithms for the first time.

If nothing else, hopefully this lesson will give students a little more appreciation for their ancestors who had to perform these kinds of calculations without the benefit of modern computing.

We also saw in the previous post that larger powers can result in better and better approximation. Finding suitable powers gets harder and harder as the exponent gets larger. However, when a better approximation is found, the improvement can be dramatic. Indeed, the decimal expansion of a logarithm can be obtained up to the accuracy of a hand-held calculator with a little patience. For example, let’s compute $3^{323,641}$

Predictably, the complaint will arise: “How did you know to try $323,641$?” The flippant and awe-inspiring answer is, “I just kept multiplying by 3.”

I’ll give the real answer that question later in this series.

Postponing the answer to that question for now, there are a couple ways for students to compute this using readily available technology. Perhaps the most user-friendly is the free resource Wolfram Alpha: $3^{323,641} \approx 9.999970671 \times 10^{154,415} \approx 10^{154,416}$.

That said, students can also perform this computation by creatively using their handheld calculators. Most calculators will return an overflow error if a direct computation of $3^{323,641}$ is attempted; the number is simply too big. A way around this is by using the above approximation $3^{153} \approx 10^{73}$, so that $3^{153}/10^{73} \approx 1$. Therefore, we can take large powers of $3^{153}/10^{73}$ without worrying about an overflow error.

In particular, let’s divide $323,641$ by $153$. A little work shows that $\displaystyle \frac{323,641}{153} = \displaystyle 2115 \frac{46}{153}$,

or $323,641 = 153 \times 2115 + 46$.

This suggests that we try to compute $\displaystyle \left( \frac{3^{153}}{10^{73}} \right)^{2115} \times 3^{46}$,

and a hand-held calculator can be used to show that this expression is approximately equal to $10^{21}$. Some of the last few digits will be incorrect because of unavoidable round-off errors, but the approximation of $10^{21}$ — all that’s needed for the present exercise — will still be evident.

By the Laws of Exponents, we see that $\displaystyle \left( \frac{3^{153}}{10^{73}} \right)^{2115} \times 3^{46} \approx 10^{21}$ $\displaystyle \frac{3^{153 \times 2115 + 46}}{10^{73 \times 2115}} \approx 10^{21}$ $\displaystyle \frac{3^{323,641}}{10^{154,395}} \approx 10^{21}$ $3^{323,641} \approx 10^{154,395} \times 10^{21}$ $3^{323,641} \approx 10^{154,395+21}$ $3^{323,641} \approx 10^{154,416}$. Whichever technique is used, we can now use the Laws of Logarithms to approximate $\log_{10} 3$: $\log_{10} 3^{323,641} \approx \log_{10} 10^{154,416}$ $323,641 \log_{10} 3 \approx 154,416$ $\log_{10} 3 \approx \displaystyle \frac{154,416}{323,641} \approx 0.477121254723598\dots$.

This approximation matches the decimal expansion of $\log_{10} 3$  to an astounding ten decimal places: $\log_{10} 3 \approx 0.47712125471966\dots$

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. Summarizing, Algebra II students can find the decimal expansion of $\log_{10} x$ can be found up to the accuracy of a hand-held scientific calculator. The only tools that are needed are

• The Laws of Logarithms
• A hand-held scientific calculator
• A lot of patience multiply $x$ by itself repeatedly in a quest to find integer powers of $x$ that are close to powers of $10$.