# Decimal Approximations of Logarithms (Part 5)

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:

In today’s post, I’ll summarize the past few posts to describe how talented Algebra II students, who have just been introduced to logarithms, can develop proficiency with the Laws of Logarithms while also understanding that the above answer is not just a meaningless jumble of digits. The only tools students will need are

To estimate $\log_{10} 5.1264$, Algebra II students can try to find a power of 5.1264 that is close to a power of 10. In principle, this can be done by just multiplying by $5.1264$ until an answer decently close to $5.1264$ arises. For the teacher who’s guiding students through this exploration, it might be helpful to know the answer ahead of time.

One way to do this is to use Wolfram Alpha to find the convergents of $\log_{10} 5.1264$. If you click this link, you’ll see that I entered

Convergents[Log[10,5.1264],15]

A little explanation is in order:

• Convergents, predictably, is the Mathematica command for finding the convergents of a given number.
• Log[10,5.1264] is the base-10 logarithm of 5.1264. By contrast, Log[5.1264] is the natural logarithm of 5.1264. Mathematica employs the convention of that $\log$ should be used for natural logarithms instead of $\ln$, as base-10 logarithms are next to useless for mathematical researchers. That said, I freely concede that this convention is confusing to students who grew up thinking that $\log$ should be used for base-10 logarithms and $\ln$ for natural logarithms. (See also my standard joke about using natural logarithms.) Naturally, the $5.1264$ can be changed for other logarithms.
• The 15 means that I want Wolfram Alpha to give me the first 15 convergents of $\log_{10} 5.1264$. In most cases, that’s enough terms to provide a convergent whose denominator is at least six digits long. In the rare instance when this doesn’t happen, a number larger than 15 can be entered.

From Wolfram Alpha, I see that $\displaystyle \frac{22}{31}$ is the last convergent with a numerator less than 100. For the purposes of this exploration, I interpret these fractions as follows:

• The best suitable power of $5.1264$ for an easy approximation on a scientific calculation will be $(5.1264)^{31}$. In this context, “best” means something that’s close to a power of 10 but less than $10^{100}$. Students entering $(5.1264)^{31}$ into a calculator will find

$(5.1264)^{31} \approx 1.009687994 \times 10^{22}$

$(5.1264)^{31} \approx 10^{22}$

In other words, the denominator of the convergent $\displaystyle \frac{22}{31}$ gives the exponent for $5.1264$, while the numerator gives the exponent for the approximated power of 10. Continuing with the Laws of Logarithms,

$\log_{10} (5.1264)^{31} \approx \log_{10} 10^{22}$

$31 \log_{10} 5.1264 \approx 22$

$\log_{10} 5.1264 \approx \displaystyle \frac{22}{31}$

$\log_{10} 5.1264 \approx 0.709677\dots$

A quick check with a calculator shows that this approximation is accurate to three decimal places. This alone should convince many students that the above apparently random jumble of digits is not so random after all.

While the above discussion should be enough for many students, some students may want to know how to find the rest of the decimal places with this technique. To answer this question, we again turn to the convergents of $\log_{10} 5.1264$ from Wolfram Alpha. From this list, we see that $\displaystyle \frac{89,337}{125,860}$ is the first convergent with a denominator at least six digits long. The student therefore has two options:

Option #1. Ask the student to use Wolfram Alpha to raise $5.1264$ to the denominator of this convergent. Surprisingly to the student, but not surprisingly to the teacher who knows about this convergent, the answer is very close to a power of 10: $10^{89,337}$. The student can then use the Laws of Logarithms as before:

$\log_{10} (5.1264)^{125,860} \approx \log_{10} 10^{89,337}$

$125,860 \log_{10} 5.1264 \approx 89,337$

$\log_{10} 5.1264 \approx \displaystyle \frac{89,337}{125,860}$

$\log_{10} 5.1264 \approx 0.70981249006\dots$,

which matches the output of the calculator.

Option #2. Ask the student to “trick” a hand-held calculator into finding $(5.1264)^{125,860}$. This option requires the use of the convergent with the largest numerator less than 100, which was $\displaystyle \frac{22}{31}$.

• Option #2A: Use the Microsoft Excel spreadsheet that I’ve written to perform the calculations that follow.
• Option #2B: The student divides the smaller denominators into the larger denominator and finds the quotient and remainder. It turns out that $125,860 = 31 \times 4060 + 0$. (This is a rare case where there happens to be no remainder.) Next, the student uses a hand-held calculate to compute

$\displaystyle \left( \frac{(5.1264)^{31}}{10^{22}} \right)^{4060} \times (5.1264)^0$

In this example, the $\times (5.1264)^0$ is of course superfluous, but I include it here to show where the remainder should be placed. Entering this in a calculator yields a result that is close to $10^{17}$. (The teacher should be aware that some of the last few digits may differ from the more precise result given by Wolfram Alpha due to round-off error, but this discrepancy won’t matter for the purposes of the student’s explorations.) In other words,

$\displaystyle \left( \frac{(5.1264)^{31}}{10^{22}} \right)^{4060} \times (5.1264)^0 \approx 10^{17}$,

which may be rearranged as

$(5.1264)^{125,860} \approx 10^{89,337}$

after using the Laws of Exponents. From this point, the derivation follows the steps in Option #1.

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