# Engaging students: Computing the composition of two functions

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission again comes from my former student Alexandria Johnson. Her topic, from Algebra II/Precalculus: computing the composition of two functions.

The following link is to a worksheet over composition of functions. The worksheet allows students to explore composition of functions without outright telling them what composition of functions is. Instead, the students are working on real world problems about shopping in a store that is having a 20% sale with mystery coupons. In the worksheet, students explore whether or not it matters which discount is applied first and the equations that go along with each scenario. This worksheet is interesting because it approaches composition of functions in an explorative way and it is using a real-world situation students in high school may find relatable, which can help hook students that are math-phobic.

https://betterlesson.com/community/document/1326462/going-shopping-student-materials-docx

Computing the composition of two functions requires prior knowledge of basic operations and combining like terms. This topic will expand upon their knowledge of basic operations by applying them to functions. Students will be able to add, subtract, multiply, and divide functions. Students should be able to use the distribution property; this is important when students are writing (fog)(x) and (gof)(x). During this topic, students should be able to expand upon their knowledge of creating functions from real world problems, which can be seen in the worksheet from the link above.

Musical composition is a way this topic can appear in high culture. Musical composition is the process of combining notes, chords, and melodies in a particular way. Arranging the notes, chords, or melodies in different ways can change the composition. Function composition is the combining of different functions f(x) and g(x) in different ways like addition, subtraction, multiplication, and division. Order usually matters in function composition just like in musical composition. If you have several band students, or musically inclined students, this would be a good hook to grab students interest.

# Engaging students: Using a recursively defined sequence

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission comes from my former student Austin DeLoach. His topic, from Precalculus: using a recursively defined sequence.

How could you as a teacher create an activity or project that involves your topic?
One activity that would be interesting to introduce recursion would be Fibonacci’s rabbit problem. In his book, Liber Abaci, Fibonacci introduced a problem where you start with one young pair of rabbits and try to find out how many rabbits you would have after a year. Every month, a grown pair of rabbits can give birth to a new pair, and it only takes one month for a young pair to grow up and be able to reproduce on their own, and the rabbits also never die. This is one of the most popular recursive sequences (the Fibonacci sequence), and, by itself, can be solved without a prior knowledge of recursion, but is a very good way to introduce the idea once the students begin to analyze the pattern of how many pairs of rabbits there are after each month. This problem is laid out in this video, https://youtu.be/sjQlW6cH3Ko but it is not necessary to show the video to introduce the problem.

How can this topic be used in your students’ future courses in mathematics or science?

One major place that a solid grasp of recursion can be used is in computer programming courses. Although not everyone takes these, they are becoming increasingly popular and the field is not likely to shrink any time soon. In programming, there are certain things that can either only be written recursively (as opposed to explicitly) or at least ones that are simpler to write and understand with recursion than with an explicit algorithm. There are also times, depending on the language and content, that a recursive function can be more efficient. Because of this, an understanding of recursion is becoming increasingly important for more people, and the ability to write and understand how it works is practically becoming necessary. So, even though not every student will go on to take computer science, many will, and the basic idea is still important to understand.

How can technology be used to effectively engage students with this topic?

There is a series of Khan Academy videos on recursively defined sequences online. The first one is https://youtu.be/lBtb30SjU2Q and it shows how to read and understand what the basic frame for recursion is. Although Khan Academy videos are not always the most engaging for all students, they do work for many because of their consistent structure. This video in particular is about recursive formulas for arithmetic sequences. Without mentioning the vocabulary yet, the video does introduce the idea of a base case and the method for finding subsequent values. The video both shows how to look at a list of values and determine the recursive definition, as well as how to understand the recursive definition if that is what you are given. For a three minute video, it does a very good job of introducing important topics for recursive series and explaining the basic ideas so that students have a framework to build on later when more complex recursively defined sequences are introduced.

# Slightly Incorrect Ugly Mathematical Christmas T-Shirts: Part 2

This was another T-shirt that I found in my search for the perfect ugly mathematical Christmas sweater: https://www.amazon.com/Pascals-Triangle-Math-Christmas-shirt/dp/B07KJS5SM2/I love the artistry of this shirt; the “ornaments” at the corners of the hexagons and the presents under the tree are nice touches.

There’s only one small problem:

$\displaystyle {8 \choose 3} = \displaystyle {8 \choose 5} = \displaystyle \frac{8!}{3! \times 5!} = 56$.

Oops.

# Engaging students: Using Pascal’s triangle

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission comes from my former student Rachel Delflache. Her topic, from Precalculus: using Pascal’s triangle.

How does this topic expand what your students would have learned in previous courses?

In previous courses students have learned how to expand binomials, however after $(x+y)^3$ the process of expanding the binomial by hand can become tedious. Pascal’s triangle allows for a simpler way to expand binomials. When counting the rows, the top row is row 0, and is equal to one. This correlates to $(x+y)^0 =1$. Similarly, row 2 is 1 2 1, correlating to $(x+y)^2 = 1x^2 + 2xy + 1y^2$. The pattern can be used to find any binomial expansion, as long as the correct row is found. The powers in each term also follow a pattern, for example look at $(x+y)^4$:

$1x^4y^0 + 4x^3y^1 + 6x^2y^2 + 4x^1y^3 + 1x^0y^4$

In this expansion it can be seen that in the first term of the expansion the first monomial is raised to the original power, and in each term the power of the first monomial decreases by one. Conversely, the second monomial is raised to the power of 0 in the first term of the expansion, and increases by a power of 1 for each subsequent term in the expansion until it is equal to the original power of the binomial.

Sierpinski’s Triangle is triangle that was characterized by Wacław Sieriński in 1915. Sierpinski’s triangle is a fractal of an equilateral triangle which is subdivided recursively. A fractal is a design that is geometrically constructed so that it is similar to itself at different angles. In this particular construction, the original shape is an equilateral triangle which is subdivided into four smaller triangles. Then the middle triangle is whited out. Each black triangle is then subdivided again, and the patter continues as illustrated below.

Sierpinski’s triangle can be created using Pascal’s triangle by shading in the odd numbers and leaving the even numbers white. The following video shows this creation in practice.

What are the contributions of various cultures to this topic?

The pattern of Pascal’s triangle can be seen as far back as the 11th century. In the 11th century Pascal’s triangle was studied in both Persia and China by Oman Khayyam and Jia Xian, respectively. While Xian did not study Pascal’s triangle exactly, he did study a triangular representation of coefficients. Xian’s triangle was further studied in 13th century China by Yang Hui, who made it more widely known, which is why Pascal’s triangle is commonly called the Yanghui triangle in China. Pascal’s triangle was later studies in the 17th century by Blaise Pascal, for whom it was named for. While Pascal did not discover the number patter, he did discover many new uses for the pattern which were published in his book Traité du Triangle Arithméthique. It is due to the discovery of these uses that the triangle was named for Pascal.

# Happy Fibonacci Day!

Today is 11/23, and 1, 1, 2, 3 are the first four terms of the Fibonacci sequence.

# 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.

# 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.

# 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 $10$.

While I don’t have a specific reference, I’d be stunned if none of our ancestors tried something along these lines in the years between the discovery of logarithms (1614) and calculus (1666 or 1684).

# Decimal Approximations of Logarithms (Part 2)

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. While I don’t have a specific reference, I’d be stunned if none of our ancestors tried something along these lines in the years between the discovery of logarithms (1614) and calculus (1666 or 1684).

To approximate $\log_{10} x$, look for integer powers of $x$ that are close to powers of 10.

I’ll illustrate this idea with $\log_{10} 3$.

$3^1 = 3$

$3^2 = 9$

Not bad… already, we’ve come across a power of 3 that’s decently close to a power of 10. We see that

$3^2 = 9 < 10^1$

and therefore

$\log_{10} 3^2 < 1$

$2 \log_{10} 3< 1$

$\log_{10} 3< \displaystyle \frac{1}{2} = 0.5$

Let’s keep going. We just keep multiplying by 3 until we find something close to a power of 10. In principle, these calculations could be done by hand, but Algebra II students can speed things up a bit by using their scientific calculators.

$3^3 = 27$

$3^4 = 81$

$3^5 = 243$

$3^6 = 729$

$3^7 = 2,187$

$3^8 = 6,561$

$3^9 = 19,683$

$3^{10} = 59,049$

$3^{11} = 177,147$

$3^{12} = 531,441$

$3^{13} = 1,594,323$

$3^{14} = 4,782,969$

$3^{15} = 14,348,907$

$3^{16} = 43,046,721$

$3^{17} = 129,140,163$

$3^{18} = 387,420,489$

$3^{19} = 1,162,261,467$

$3^{20} = 3,486,784,401$

$3^{21} = 10,460,353,203$

This looks pretty good too. (Students using a standard ten-digit scientific calculator, of course, won’t be able to see all 11 digits.) We see that

$3^{21} > 10^{10}$

and therefore

$\log_{10} 3^{21} > \log_{10} 10^{10}$

$21 \log_{10} 3 > 10$

$\log_{10} 3 > \displaystyle \frac{10}{21} = 0.476190\dots$

Summarizing our work so far, we have

$0.476190\dots < \log_{10} 3 < 0.5$.

We also note that this latest approximation actually gives the first two digits in the decimal expansion of $\log_{10} 3$.

To get a better approximation of $\log_{10} 3$, we keep going. I wouldn’t blame Algebra II students a bit if they use their scientific calculators for these computations — but, ideally, they should realize that these calculations could be done by hand by someone very persistent.

$3^{22} = 31,381,059,609$

$3^{23} = 94,143,178,827$

$3^{24} = 282,429,536,481$

$3^{25} = 847,288,609,443$

$3^{26} = 2,541,865,828,329$

$3^{27} = 7,625,597,484,987$

$3^{28} = 22,876,792,454,961$

$3^{29} = 68,630,377,364,883$

$3^{30} = 205,891,132,094,649$

$3^{31} = 617,673,396,283,947$

$3^{32} = 1,853,020,188,851,841$

$3^{33} = 5,559,060,566,555,523$

$3^{34} = 16,677,181,699,666,569$

$3^{35} = 50,031,545,098,999,707$

$3^{36} = 150,094,635,296,999,121$

$3^{37} = 450,283,905,890,997,363$

$3^{38} = 1,350,851,717,672,992,089$

$3^{39} = 4,052,555,153,018,976,267$

$3^{40} = 12,157,665,459,056,928,801$

$3^{41} = 36,472,996,377,170,786,403$

$3^{42} = 109,418,989,131,512,359,209$

$3^{43} = 328,256,967,394,537,077,627$

$3^{44} = 984,770,902,183,611,232,881$

Using this last line, we obtain

$3^{44} < 10^{21}$

and therefore

$\log_{10} 3^{44} < \log_{10} 10^{21}$

$44 \log_{10} 3 < 21$

$\log_{10} 3 < \displaystyle \frac{21}{44} = 0.477273\dots$

Summarizing our work so far, we have

$0.476190\dots < \log_{10} 3 < 0.477273\dots$.

A quick check with a calculator shows that $\log_{10} 3 = 0.477121\dots$. In other words,

• This technique actually works!
• This last approximation of $0.477273\dots$ actually produced the first three decimal places of the correct answer!

With a little more work, the approximations

$3^{109} \approx 1.014417574 \times 10^{52} > 10^{52}$

$3^{153} \approx 9.989689095 \times 10^{72} < 10^{73}$

can be found, yielding the tighter inequalities

$\displaystyle \frac{52}{109} < \log_{10} 3 < \displaystyle \frac{73}{153}$,

or

$0.477064\dots < \log_{10} 3 < 0.477124$.

Now we’re really getting close… the last approximation is accurate to five decimal places.

# Decimal Approximations of Logarithms (Part 1)

My latest article on mathematics education, titled “Developing Intuition for Logarithms,” was published this month in the “My Favorite Lesson” section of the September 2018 issue of the journal Mathematics Teacher. This is a lesson that I taught for years to my Precalculus students, and I teach it currently to math majors who are aspiring high school teachers. Per copyright law, I can’t reproduce the article here, though the gist of the article appeared in an earlier blog post from five years ago.

Rather than repeat the article here, I thought I would write about some extra thoughts on developing intuition for logarithms that, due to space limitations, I was not able to include in the published article.

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?

Students who know calculus, of course, can do these computations since

$\log_{10} x = \displaystyle \frac{\ln x}{\ln 10}$,

$\ln (1+t) = t - \displaystyle \frac{t^2}{2} + \frac{t^3}{3} - \frac{t^4}{4} + \dots$,

a standard topic in second-semester calculus, can be used to calculate $\ln x$ for values of $x$ close to 1. However, a calculation using a power series is probably inaccessible to bright Algebra II students, no matter how precocious they are. (Besides, in real life, calculators don’t actually use Taylor series to perform these calculations; see the article CORDIC: How Hand Calculators Calculate, which appeared in College Mathematics Journal, for more details.)

In this series, I’ll discuss a technique that Algebra II students can use to find the decimal expansions of base-10 logarithms to surprisingly high precision using only tools that they’ve learned in Algebra II. This technique won’t be very efficient, but it should be completely accessible to students who are learning about base-10 logarithms for the first time. All that will be required are the Laws of Logarithms and a standard scientific calculator. A little bit of patience can yield the first few decimal places. And either a lot of patience, a teacher who knows how to use Wolfram Alpha appropriately, or a spreadsheet that I wrote can be used to obtain the decimal approximations of logarithms up to the digits displayed on a scientific calculator.

I’ll start this discussion in my next post.